Ch 07 fluid and thermal systems

8/25/2013 System Dynamics 7.01 Fluid and Thermal Systems Fluid and Thermal Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.03 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.02 Part Fluid Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.04 §1.Conservation of Mass For incompressible fluids conservation of mass ⟺ conservation of volume The mass flow rate 𝑞𝑚 = 𝜌𝑞𝑣 𝜌: fluid density, 𝑘𝑔/𝑚3 𝑞𝑚 : the mass flow rates, 𝑘𝑔/𝑠 𝑞𝑣 : the volume flow rate, 𝑚3 /𝑠 1.Density and Pressure - Density (mass density): mass per unit volume, 𝑘𝑔/𝑚3 - Pressure: force per unit area that is exerted by the fluid, 𝑁/𝑚2 - Hydrostatic pressure: the pressure that exists in a fluid at rest §1.Conservation of Mass - Example 7.1.1 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.05 Nguyen Tan Tien Fluid and Thermal Systems §1.Conservation of Mass Solution The forces are related to the pressures and the piston areas 𝑓1 = 𝑝1 𝐴1 , 𝑓2 = 𝑝2 𝐴2 Assuming the system is in static equilibrium after the brake pedal has been pushed 𝑝1 = 𝑝2 + 𝜌𝑔ℎ ≈ 𝑝2 𝑓1 𝑓2 𝐴2 ⟹ 𝑝1 = = 𝑝2 = ⟹ 𝑓2 = 𝑓1 𝐴1 𝐴2 𝐴1 The force 𝑓3 𝐿1 𝐴2 𝐿1 𝑓3 = 𝑓2 = 𝑓 𝐿2 𝐴1 𝐿2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Fluid and Thermal Systems Nguyen Tan Tien Fluid and Thermal Systems A Hydraulic Brake System The figure is a representation of a hydraulic brake system The piston in the master cylinder moves in response to the foot pedal The resulting motion of the piston in the slave cylinder causes the brake pad to be pressed against the brake drum with a force 𝑓3 The force 𝑓1 depends on the force 𝑓4 applied by the driver’s foot The precise relation between 𝑓1 and 𝑓4 depends on the geometry of the pedal arm Obtain the expression for the force 𝑓3 with the force 𝑓1 as the input System Dynamics 7.06 Nguyen Tan Tien Fluid and Thermal Systems §1.Conservation of Mass - Conservation of mass 𝑚 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑞𝑚𝑖 : the mass inflow rate, 𝑘𝑔/𝑠 𝑞𝑚𝑜 : the mass outflow rate , 𝑘𝑔/𝑠 𝑞𝑚𝑖 = 𝜌𝑞𝑣𝑖 𝑞𝑚𝑜 = 𝜌𝑞𝑣𝑜 𝑞𝑣𝑖 : total volume inflow rate, 𝑚3 /𝑠 𝑞𝑣𝑜 : total volume inflow rate, 𝑚3 /𝑠 - The fluid mass 𝑚 is related to the container volume 𝑉 𝑚 = 𝜌𝑉 ⟹ 𝑚 = 𝜌𝑉 then 𝜌𝑉 = 𝜌𝑞𝑣𝑖 − 𝜌𝑞𝑣𝑜 ⟹ 𝑉 = 𝑞𝑣𝑖 − 𝑞𝑣𝑜 This is a statement of conservation of volume for the fluid HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics §1.Conservation of Mass - Example 7.1.2 7.07 A Water Supply Tank Water is pumped at the mass flow rate 𝑞𝑚𝑜 (𝑡) from the tank Replacement water is pumped from a well at the mass flow rate 𝑞𝑚𝑖 (𝑡) Determine the water height ℎ(𝑡), assuming that the tank is cylindrical with a cross section 𝐴 Solution From conservation of mass 𝑑 𝜌𝐴ℎ = 𝑞𝑚𝑖 𝑡 − 𝑞𝑚𝑜 𝑡 𝑑𝑡 𝑑ℎ ⟹ 𝜌𝐴 = 𝑞𝑚𝑖 𝑡 − 𝑞𝑚𝑜 𝑡 𝑑𝑡 𝑡 ⟹ℎ 𝑡 =ℎ + [𝑞 𝑢 − 𝑞𝑚𝑜(𝑢)]𝑑𝑢 𝜌𝐴 𝑚𝑖 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Fluid and Thermal Systems 7.09 Nguyen Tan Tien Fluid and Thermal Systems §1.Conservation of Mass Solution a.Model of the motion for figure (a) Assuming that 𝑝1 > 𝑝2 , the net force acting on the piston and mass 𝑚 is (𝑝1 − 𝑝2 )𝐴, and thus from Newton’s law 𝑚𝑥 = (𝑝1 − 𝑝2 )𝐴 Integrate this equation once to obtain the velocity 𝐴 𝑡 𝑥 𝑡 =𝑥 + [𝑝 𝑢 − 𝑝2 (𝑢)]𝑑𝑢 𝑚 The rate at which fluid volume is swept out by the piston is 𝐴𝑥, and thus if 𝑥 > 0, the pump providing pressure 𝑝1 must supply fluid at the mass rate 𝜌𝐴𝑥, and the pump providing pressure 𝑝2 must absorb fluid at the same mass rate HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.11 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.08 Fluid and Thermal Systems §1.Conservation of Mass - Example 7.1.3 A Hydraulic Cylinder Fig.(a): a cylinder and piston connected to a load mass 𝑚 Fig.(b): the piston rod connected to a rack-and-pinion gear The pressures 𝑝1 and 𝑝2 are applied to each side of the piston by two pumps Neglect the piston rod diameter and assume that the piston and rod mass have been lumped into 𝑚 a.Develop a model of the motion of the displacement 𝑥 of the mass in fig.(a) Also, obtain the expression for the mass flow rate that must be delivered or absorbed by the two pumps b.Develop a model of the displacement 𝑥 in fig.(b) The inertia of the pinion and the load connected to the pinion is 𝐼 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.10 Nguyen Tan Tien Fluid and Thermal Systems §1.Conservation of Mass Solution b.Model of the motion for figure (b) Firstly, obtain an expression for the equivalent mass of the rack, pinion, and load The kinetic energy of the system is 1 𝐼 𝐾𝐸 = 𝑚𝑥 + 𝐼𝜃 = 𝑚 + 𝑥 2 2 𝑅 because 𝑅𝜃 = 𝑥 Thus the equivalent mass is 𝐼 𝑚𝑒 = 𝑚 + 𝑅 Then, the required model can now be obtained by replacing 𝑚 with 𝑚𝑒 in the model developed in part (a) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.12 Nguyen Tan Tien Fluid and Thermal Systems §1.Conservation of Mass - Example 7.1.4 A Mixing Process A mixing tank is shown in the figure Pure water flows into the tank of volume 𝑉 = 600𝑚3 at the constant volume rate of 5𝑚3 /𝑠 A solution with a salt concentration of 𝑠𝑖 𝑘𝑔/𝑚3 flows into the tank at a constant volume rate of 2𝑚3 /𝑠 Assume that the solution in the tank is well mixed so that the salt concentration in the tank is uniform Assume also that the salt dissolves completely so that the volume of the mixture remains the same The salt concentration 𝑠𝑜 𝑘𝑔/𝑚3 in the outflow is the same as the concentration in the tank The input is the concentration 𝑠𝑖 (𝑡) , whose value may change during the process, thus changing the value of 𝑠𝑜 Obtain a dynamic model of the concentration 𝑠𝑜 §1.Conservation of Mass Solution Two mass species are conserved here: water mass and salt mass The tank is always full, so the mass of water 𝑚𝑤 in the tank is constant, and thus conservation of water mass gives 𝑑𝑚𝑤 = 5𝜌𝑤 + 2𝜌𝑤 − 𝜌𝑤 𝑞𝑣𝑜 = ⟹ 𝑞𝑣𝑜 = 7𝑚3 /𝑠 𝑑𝑡 𝜌𝑤 : the mass density of fresh water 𝑞𝑣𝑜 : the volume outflow rate of the mixed solution The salt mass in the tank is 𝑠𝑜 𝑉, and conservation of salt mass gives 𝑑 𝑑𝑠𝑜 2𝑠𝑖 − 7𝑠𝑜 𝑠 𝑉 = + 2𝑠𝑖 − 𝑠𝑜 𝑞𝑣𝑜 = 2𝑠𝑖 − 7𝑠𝑜 ⟹ = 𝑑𝑡 𝑜 𝑑𝑡 600 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 8/25/2013 System Dynamics 7.13 Fluid and Thermal Systems §2.Fluid Capacitance - Sometimes it is very useful to think of fluid systems in terms of electrical circuits Fluid mass, Mass flow rate, 𝑚 𝑞𝑚 Charge, Current, Pressure, 𝑝 Fluid linear resistance,𝑅 = 𝑝/𝑞𝑚 Fluid capacitance, 𝐶 = 𝑚/𝑝 Fluid inertance, 𝑄 𝑖 Voltage, 𝑣 Electrical resistance, 𝑅 = 𝑣/𝑖 Electrical capacitance, 𝐶 = 𝑄/𝑣 𝐼 = 𝑝/(𝑑𝑞𝑚/𝑑𝑡) Electrical inductance, 𝐿 = 𝑣/(𝑑𝑖/𝑑𝑡) - Fluid resistance is the relation between pressure and mass flow rate Fluid resistance relates to energy dissipation - Fluid capacitance is the relation between pressure and stored mass Fluid capacitance relates to potential energy - Fluid inertance relates to fluid acceleration and kinetic energy HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 7.15 Fluid and Thermal Systems §2.Fluid Capacitance 1.Fluid Symbols and Source - Resistance Both linear and nonlinear fixed resistances, for example, pipe flow, orifice flow, or a restriction - Valve • manually adjusted valve: faucet • actuated valve: driven by an electric motor or a pneumatic device System Dynamics 7.14 Fluid and Thermal Systems §2.Fluid Capacitance - Fluid systems obey two laws that are analogous to Kirchhoff’s current and voltage laws • The continuity law (conservation of fluid mass): the total mass flow into a junction must equal the total flow out of the junction • The compatibility law (conservation of energy): the sum of signed pressure differences around a closed loop must be zero - Note: the flow is through flexible tubes that can expand and contract under pressure, then the outflow rate is not the sum of the inflow rates This is an example where fluid mass can accumulate within the system and is analogous to having a capacitor in an electrical circuit HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.16 Nguyen Tan Tien Fluid and Thermal Systems §2.Fluid Capacitance - Ideal pressure source Supplying the specified pressure at any flow rate - Ideal flow source Supplying the specified flow - Pump Ideal sources are approximations to real devices such as pumps HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 7.17 Fluid and Thermal Systems §2.Fluid Capacitance 2.Capacitance Relations - The figure illustrates the relation between stored fluid mass and the resulting pressure caused by the stored mass - Fluid capacitance 𝐶 : the ratio of the change in stored mass to the change in pressure 𝑑𝑚 𝐶≡ 𝑑𝑝 𝑝=𝑝 𝑟 𝑚: the stored fluid mass, 𝑘𝑔 𝑝: the resulting pressure, 𝑁/𝑚2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics §2.Fluid Capacitance - Example 7.2.1 7.18 Nguyen Tan Tien Fluid and Thermal Systems Capacitance of a Storage Tank Consider the tank shown in the figure Assume that the cross sectional area 𝐴 is constant Derive the expression for the tank’s capacitance Solution The liquid mass in the tank: The total pressure at the bottom of the tank: Pressure due only to the stored fluid mass: The pressure function of the mass 𝑚: The capacitance of the tank is given by 𝑑𝑚 𝐴 𝐶≡ = 𝑑𝑝 𝑔 HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑚 = 𝜌𝐴ℎ 𝜌𝑔ℎ + 𝑝𝑎 𝑝 = 𝜌𝑔ℎ 𝑝 = 𝑚𝑔/𝐴 Nguyen Tan Tien 8/25/2013 System Dynamics 7.19 Fluid and Thermal Systems §2.Fluid Capacitance - When the container does not have vertical sides, the crosssectional area 𝐴 is a function of the liquid height ℎ , and the relations between 𝑚 and ℎ and between 𝑝 and 𝑚 are nonlinear - The fluid mass stored in the container ℎ 𝑑𝑚 𝑚 = 𝜌𝑉 = 𝜌 𝐴 𝑥 𝑑𝑥 ⟹ = 𝜌𝐴 𝑑ℎ - For such a container, conservation of mass gives 𝑑𝑚 𝑑𝑚 𝑑𝑚 𝑑𝑝 𝑑𝑝 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 ⟹ = =𝐶 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑑𝑡 𝑑𝑡 𝑑𝑝 𝑑𝑡 𝑑𝑡 - Also 𝑑𝑚 𝑑𝑚 𝑑ℎ 𝑑ℎ 𝑑ℎ = = 𝜌𝐴 ⟹ 𝜌𝐴 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑑𝑡 𝑑ℎ 𝑑𝑡 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.21 Nguyen Tan Tien Fluid and Thermal Systems §2.Fluid Capacitance b Dynamic Model which is a nonlinear equation because of the product 𝑝𝑝 Obtain the model for the height by substituting ℎ = 𝑝/𝜌𝑔 𝑑ℎ 2𝜌𝐿𝑡𝑎𝑛𝜃 ℎ = 𝑞𝑚𝑖 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering 7.23 Nguyen Tan Tien Fluid and Thermal Systems §3.Fluid Resistance - The relation 𝑝 = 𝑓(𝑞𝑚 ) • is linear in a limited number of cases, such as pipe flow under certain conditions 𝑞𝑚 = 𝑝/𝑅 • is a square-root relation in some other applications 𝑞𝑚 = 7.20 Fluid and Thermal Systems §2.Fluid Capacitance - Example 7.2.2 Capacitance of a V-Shaped Trough a Derive the capacitance of the V-shaped b Derive the dynamic models for the bottom pressure 𝑝 and the height ℎ The mass inflow rate is 𝑞𝑚𝑖 (𝑡) Solution a The fluid mass 𝑚 = 𝜌𝑉 = 𝜌 ℎ𝐷 𝐿 = 𝜌𝐿𝑡𝑎𝑛𝜃 ℎ2 2 𝑝 𝐿𝑡𝑎𝑛𝜃 = 𝜌𝐿𝑡𝑎𝑛𝜃 = 𝑝 𝜌𝑔 𝜌𝑔2 From the definition of capacitance 𝑑𝑚 2𝐿𝑡𝑎𝑛𝜃 𝐶= = 𝑝 𝑑𝑝 𝜌𝑔2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.22 Nguyen Tan Tien Fluid and Thermal Systems §3.Fluid Resistance 𝑑𝑝 𝑑𝑚 𝐶 = = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑑𝑡 𝑑𝑡 with 𝑞𝑚𝑜 = 0, 𝐶 = 𝑝 = 𝑞𝑚𝑖 2𝐿𝑡𝑎𝑛𝜃 𝑑𝑝 𝑝 = 𝑞𝑚𝑖 𝜌𝑔2 𝑑𝑡 System Dynamics System Dynamics 𝑝 𝑅1 𝑚𝑟 the reference values of 𝑞𝑚𝑟 and 𝑝𝑟 depend on the particular application HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 𝑅𝑟 : the linearized resistance at the reference condition (𝑞𝑚𝑟 , 𝑝𝑟 ) Nguyen Tan Tien 7.24 Nguyen Tan Tien Fluid and Thermal Systems §3.Fluid Resistance - Deviation variable at the reference values of 𝑞𝑚𝑟 and 𝑝𝑟 𝛿𝑝 ≡ 𝑝 − 𝑝𝑟 𝛿𝑞𝑚 ≡ 𝑞𝑚 − 𝑞𝑚𝑟 ⟹ 𝛿𝑝 = 𝑅𝑟 𝛿𝑞𝑚 = 2𝑅1 ⟹ 𝛿𝑞𝑚 = • can be linearized the expression near a reference operating point (𝑞𝑚𝑟 , 𝑝𝑟 ) 𝑑𝑝 𝑝 = 𝑝𝑟 + 𝑞 − 𝑞𝑚𝑟 = 𝑝𝑟 + 𝑅𝑟 𝑞𝑚 − 𝑞𝑚𝑟 𝑑𝑞𝑚 𝑟 𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering - Fluid meets resistance when flowing through a conduit such as a pipe, through a component such as a valve, or even through a simple opening or orifice, such as a hole - The relation between mass flow rate 𝑞𝑚 and the pressure difference 𝑝 across the resistance 𝑝 = 𝑓(𝑞𝑚 ) is shown in the figure - Define the fluid resistance 𝑅 𝑑𝑝 𝑅≡ 𝑑𝑞𝑚 𝑞=𝑞 𝑅1𝑝𝑟 𝑝𝑟 𝛿𝑞 = 𝑅1 𝑝𝑟 𝛿𝑞𝑚 𝑅1 𝑚 𝛿𝑝 - The resistance symbol • series resistances • parallel resistances HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 7.25 Fluid and Thermal Systems §3.Fluid Resistance 1.Laminar Pipe Resistance - Fluid motion is generally divided into two types • Laminar flow: 𝑅𝑒 < 2300 • Turbulent flow: 𝑅𝑒 > 2300 for circular pipe, 𝑅𝑒 ≡ 𝜇 - The laminar resistance (Hagen-Poiseuille formula) 128𝜇𝐿 𝑅= 𝜋𝜌𝐷 𝑅: flow resistance, 𝜇: the fluid viscosity, 𝑁𝑠/𝑚2 𝐿: the length of pipe, 𝑚 𝜌: the fluid density, 𝑘𝑔/𝑚3 𝐷: the diameter of pipe, 𝑚 System Dynamics §3.Fluid Resistance - Example 7.3.1 7.27 Nguyen Tan Tien Fluid and Thermal Systems Liquid-Level System with a Flow Source The cylindrical tank shown in the figure has a bottom area 𝐴 The mass inflow rate is 𝑞𝑚𝑖 (𝑡) The outlet resistance is linear and the outlet discharges to atmospheric pressure 𝑝𝑎 Develop a model of the liquid height ℎ Slolution Total mass in the tank is 𝑚 = 𝜌𝐴ℎ, from conservation of mass 𝑑𝑚 𝑑ℎ 1 = 𝜌𝐴 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜, 𝑞𝑚𝑜 = 𝜌𝑔ℎ + 𝑝𝑎 − 𝑝𝑎 = 𝜌𝑔ℎ 𝑑𝑡 𝑑𝑡 𝑅 𝑅 The desired model 𝑑ℎ 𝜌𝑔 𝜌𝐴 = 𝑞𝑚𝑖 − ℎ 𝑑𝑡 𝑅 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.29 §3.Fluid Resistance 3.Torricelli’s Principle - An orifice can simply be a hole in the side of a tank or it can be a passage in a valve - The mass flow rate 𝑞𝑚 through the orifice 𝑞𝑚 = 𝐶𝑑 𝐴𝑜 2𝜌𝑝 = 𝐶𝑑 𝐴𝑜 2𝜌 𝑝 = Nguyen Tan Tien Fluid and Thermal Systems Fluid and Thermal Systems - In liquid-level systems such as shown in the figure, energy is stored in two ways • potential energy in the mass of liquid in the tank • kinetic energy in the mass of liquid flowing in the pipe - If the mass of liquid in a pipe is small enough or is flowing at a small enough velocity, the kinetic energy contained in it will be negligible compared to the potential energy stored in the liquid in the tank HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 𝑝/𝑅𝑜 ℎ: the height of fluid, 𝑚 Nguyen Tan Tien 7.28 Nguyen Tan Tien Fluid and Thermal Systems §3.Fluid Resistance - Consider the circuit model 𝑑𝑣 𝐶 = 𝑖𝑠 − 𝑣 𝑑𝑡 𝑅 - The fluid flow system is analogous to the electric circuit system • pressure difference, 𝜌𝑔ℎ ⟺ voltage difference, 𝑣 • mass flow rate, 𝑞𝑚𝑖 ⟺ current, 𝑖𝑠 • resistance resists flow ⟺ resistor resists current • capacitance stores fluid mass, 𝐴/𝑔 ⟺ capacitor stores charge, 𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics §3.Fluid Resistance - Example 7.3.2 𝐶𝑑 : factor, 𝐴0 : the area of the orifice, 𝑚2 𝑝: the pressure of fluid, 𝑁/𝑚2 𝜌: the fluid density, 𝑘𝑔/𝑚3 𝑅𝑜 : orifice resistance 𝑅𝑜 ≡ 2𝜌𝐶𝑑2 𝐴𝑜2 - The volume flow rate 𝑞𝑣 through the orifice 𝑞𝑣 = 𝐶𝑑 𝐴𝑜 2𝑔 ℎ0.5 HCM City Univ of Technology, Faculty of Mechanical Engineering 7.26 §3.Fluid Resistance 2.System Model 𝜌 𝑣𝐷 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.30 Nguyen Tan Tien Fluid and Thermal Systems Liquid-Level System with an Orifice The cylindrical tank shown in the figure has a circular bottom area 𝐴 The volume inflow rate from the flow source is 𝑞𝑣𝑖 (𝑡), a given function of time The orifice in the side wall has an area 𝐴𝑜 and discharges to atmospheric pressure 𝑝𝑎 Develop a model of the liquid height ℎ, assuming that ℎ1 > 𝐿 Solution From conservation of mass and the orifice flow relation 𝑑ℎ 𝜌𝐴 = 𝜌𝑞𝑣𝑖 − 𝐶𝑑 𝐴𝑜 2𝑝𝜌 𝑑𝑡 where 𝑝 = 𝜌𝑔ℎ Thus the model becomes 𝑑ℎ 𝐴 = 𝑞𝑣𝑖 − 𝐶𝑑 𝐴𝑜 2𝑔ℎ 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 7.31 Fluid and Thermal Systems §3.Fluid Resistance 4.Turbulance and Component Resistance - The practical importance of the difference between laminar and turbulent flow lies in the fact that • laminar flow can be described by the linear relation 𝑞𝑚 = 𝑝/𝑅 • turbulent flow is described by the nonlinear relation 𝑞𝑚 = 𝑝/𝑅1 - Components, such as valves, elbow bends, couplings, porous plugs, and changes in flow area resist flow and usually induce turbulent flow at typical pressures, and 𝑞𝑚 = 𝑝/𝑅1 is often used to model them - Experimentally determined values of 𝑅 are available for common types of components HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.33 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Because the outlet resistance is linear 𝜌𝑔ℎ 𝑞𝑚𝑜 = 𝜌𝑔ℎ + 𝑝𝑎 − 𝑝𝑎 = 𝑅2 𝑅2 The mass inflow rate 𝑞𝑚𝑖 = 𝑝 + 𝑝𝑎 − 𝜌𝑔ℎ + 𝑝𝑎 𝑅1 𝑠 = (𝑝𝑠 − 𝜌𝑔ℎ) 𝑅1 The desired model 𝑑ℎ 𝜌𝑔 𝜌𝐴 = 𝑝 − 𝜌𝑔ℎ − ℎ 𝑑𝑡 𝑅1 𝑠 𝑅2 which can be rearranged as 𝑑ℎ 1 1 𝑅1 + 𝑅2 𝜌𝐴 = 𝑝𝑠 − 𝜌𝑔 + ℎ = 𝑝𝑠 − 𝜌𝑔 ℎ 𝑑𝑡 𝑅1 𝑅1 𝑅2 𝑅1 𝑅1𝑅2 The time constant 7.32 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems - Example 7.4.1 Liquid-Level System with a Pressure Source Consider the system shown in the figure The linear resistance 𝑅 represents the pipe resistance lumped at the outlet of the pressure source The bottom of the water tank is a height 𝐿 above the pressure source Develop a model of the water height ℎ with the supply pressure 𝑝𝑠 and the flow rate 𝑞𝑚𝑜 (𝑡) as the inputs Solution The total mass in the tank 𝑚 = 𝜌𝐴ℎ Since 𝜌 and 𝐴 are constants, from conservation of mass 𝑑𝑚 𝑑ℎ = 𝜌𝐴 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.34 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems - Example 7.4.2 Water Tank Model Consider the system shown in the figure, the input was the specified flow rate 𝑞𝑚𝑖 The linear resistance 𝑅 represents at the outlet of the pressure source The bottom of the water tank is a height 𝐿 above the pressure source Develop a model of the water height ℎ with the supply pressure 𝑝𝑠 and the flow rate 𝑞𝑚𝑜 (𝑡) as the inputs 𝜏 = 𝑅1 𝑅2 𝐴/𝑔(𝑅1 + 𝑅2 ) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics System Dynamics 7.35 Nguyen Tan Tien Fluid and Thermal Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.36 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Solution The mass flow rate into the bottom of the tank 𝑞𝑚𝑖 = 𝑝 + 𝑝𝑎 − 𝜌𝑔 ℎ + 𝐿 + 𝑝𝑎 𝑅 𝑠 = 𝑝𝑠 − 𝜌𝑔 ℎ + 𝐿 𝑅 From conservation of mass, 𝑑 𝜌𝐴ℎ = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑡 𝑑𝑡 = 𝑝𝑠 − 𝜌𝑔 ℎ + 𝐿 − 𝑞𝑚𝑜(𝑡) 𝑅 Because 𝜌 and 𝐴 are constants, the model can be written as 𝑑ℎ 𝐴 = 𝑞𝑚𝑖 − 𝑞𝑚𝑜 𝑡 = 𝑝𝑠 − 𝜌𝑔 ℎ + 𝐿 − 𝑞𝑚𝑜 𝑡 𝑑𝑡 𝑅 §4.Dynamic Models of Hydraulic Systems - Example 7.4.3 Two Connected Tanks The cylindrical tanks have bottom areas 𝐴1 and 𝐴2 The mass inflow rate 𝑞𝑚𝑖 (𝑡) from the flow source is a given function of time The resistances are linear and the outlet discharges with pressure 𝑝𝑎 a.Develop a model of the liquid heights ℎ1 and ℎ2 b.Suppose 𝑅1 = 𝑅2 = 𝑅, and 𝐴1 = 𝐴 and 𝐴2 = 3𝐴 Obtain the transfer function 𝐻1 (𝑠)/𝑄𝑚𝑖 (𝑠) c.Use the transfer function to solve for the steady state response for ℎ1 if the inflow rate 𝑞𝑚𝑖 is a unit-step function, and estimate how long it will take to reach steady state Is it possible for liquid heights to oscillate in the step response? HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 8/25/2013 System Dynamics 7.37 Fluid and Thermal Systems System Dynamics 7.38 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Solution a.Assume that ℎ1 > ℎ2 so that the mass flow rate 𝑞𝑚1 is positive if flowing from tank to tank Conservation of mass applied to tank gives 𝜌𝑔 𝜌𝐴1ℎ1 = −𝑞𝑚1 = − (ℎ1 − ℎ2) 𝑅1 For tank 𝜌𝑔 𝜌𝐴2 ℎ2 = 𝑞𝑚𝑖 + 𝑞𝑚1 − 𝑞𝑚𝑜 = 𝑞𝑚𝑖 + 𝑞𝑚1 − ℎ 𝑅2 Canceling 𝜌 where possible, we obtain the desired model 𝑔 𝐴1 ℎ1 = − (ℎ1 − ℎ2 ) 𝑅1 𝜌𝑔 𝜌𝑔 𝜌𝐴2 ℎ2 = 𝑞𝑚𝑖 + ℎ − ℎ2 − ℎ 𝑅1 𝑅2 §4.Dynamic Models of Hydraulic Systems b.Substituting 𝑅1 = 𝑅2 = 𝑅, and 𝐴1 = 𝐴 and 𝐴2 = 3𝐴 into the differential equations and dividing by 𝐴, and letting 𝐵 ≡ 𝑔/𝑅𝐴 we obtain ℎ1 = −𝐵 ℎ1 − ℎ2 and 𝑞𝑚𝑖 𝑞𝑚𝑖 3ℎ2 = + 𝐵 ℎ1 − ℎ2 − 𝐵ℎ2 = + 𝐵ℎ1 − 2𝐵ℎ2 𝜌𝐴 𝜌𝐴 Assuming zero initial conditions, apply the Laplace transform 𝑠 + 𝐵 𝐻1 𝑠 − 𝐵𝐻2 𝑠 = −𝐵𝐻1 𝑠 + (3𝑠 + 2𝐵)𝐻2 (𝑠) = 𝑄 (𝑠) 𝜌𝐴 𝑚𝑖 𝐻1 (𝑠) 𝑅𝐵 /𝜌𝑔 ⟹ = 𝑄𝑚𝑖 (𝑠) 3𝑠 + 5𝐵𝑠 + 𝐵 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.39 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems c.The characteristic equation is 3𝑠 + 5𝐵𝑠 + 𝐵 = 0, with roots 𝑠 = −5 ± 13 𝐵/6 = −1.43𝐵, −0.232𝐵 ⟹ the system is stable, and there will be a constant steadystate response to a step input The step response cannot oscillate because both roots are real The steady-state height can be obtained by applying the final value theorem with 𝑄𝑚𝑖 𝑠 = 1/𝑠 𝑅𝐵 2/𝜌𝑔 𝑅 ℎ1𝑠𝑠 = lim 𝑠𝐻1(𝑠) = lim = 𝑠→0 𝑠→0 3𝑠 + 5𝐵𝑠 + 𝐵 𝑠 𝜌𝑔 The time constants are 0.699 4.32 𝜏1 = = , 𝜏2 = = 1.43𝐵 𝐵 0.232𝐵 𝐵 The largest time constant is 𝜏2 and thus it will take a time equal to approximately 4𝜏2 = 17.2𝐵 to reach steady state HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.41 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.40 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems 5.Hydraulic Damper Dampers oppose a velocity difference across them, and thus they are used to limit velocities The most common application of dampers is in vehicle shock absorbers - Example 7.4.4 Linear Damper Consider a shock absorber: a piston of diameter 𝑊 and thickness 𝐿 has a cylindrical hole of diameter 𝐷 The piston rod extends out of the housing, which is sealed and filled with a viscous incompressible fluid Assuming that the flow through the hole is laminar and that the entrance length 𝐿𝑒 is small compared to 𝐿, develop a model of the relation between the applied force 𝑓 and 𝑥 , the relative velocity between the piston and the cylinder HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.42 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Solution Assume that the rod’s cross-sectional area and the hole area 𝜋(𝐷/2)2 are small compared to the piston area 𝐴 Let 𝑚 be the combined mass of the piston and rod Then the force 𝑓 acting on the piston rod creates a pressure difference (𝑝1 − 𝑝2) across the piston such that 𝑚𝑦 = 𝑓 − 𝐴(𝑝1 − 𝑝2 ) If 𝑚 or 𝑦 is small, then 𝑚𝑦 ≈ ⟹ 𝑓 = 𝐴(𝑝1 − 𝑝2 ) For laminar flow through the hole 1 𝑞𝑣 = 𝑞𝑚 = (𝑝 − 𝑝2 ) 𝜌 𝜌𝑅 The volume flow rate 𝑞𝑣 can be expressed as 𝑞𝑣 = 𝐴 𝑦 − 𝑧 = 𝐴𝑥 §4.Dynamic Models of Hydraulic Systems Combining the above equations, we obtain 𝑓 = 𝐴 𝜌𝑅𝐴𝑥 = 𝜌𝑅𝐴2 𝑥 = 𝑐 𝑥, 𝑐 ≡ 𝜌𝑅𝐴2 From the Hagen-Poiseuille formula, for a cylindrical conduit 128𝜇𝐿 128𝜇𝐿𝐴2 𝑅= ⟹𝑐= 𝜋𝜌𝐷 𝜋𝐷 The approximation 𝑚𝑦 ≈ is commonly used for hydraulic systems to simplify the resulting model To see the effect of this approximation, rewrite 𝑞𝑣 and 𝑓 1 𝑞𝑣 = 𝑝 − 𝑝2 = 𝑓 − 𝑚𝑦 = 𝐴𝑥 𝜌𝑅 𝜌𝑅𝐴 𝑓 = 𝑚𝑦 + 𝜌𝑅𝐴2 𝑥 If 𝑚𝑦 cannot be neglected, the damper force is a function of the absolute acceleration as well as the relative velocity HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 8/25/2013 System Dynamics 7.43 Fluid and Thermal Systems System Dynamics 7.44 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems 6.Hydraulic Actuators Hydraulic actuators are widely used with high pressures to obtain high forces for moving large loads or achieving high accelerations The working fluid may be liquid, as is commonly found with construction machinery, or it may be air, as with the air cylinder-piston units frequently used in manufacturing and parts-handling equipment - Example 7.4.5 Hydraulic Piston and Load The figure shows a double-acting piston and cylinder The device moves the load mass 𝑚 in response to the pressure sources 𝑝1 and 𝑝2 Assume the fluid is incompressible, the resistances are linear, and the piston mass is included in 𝑚 Derive the equation of motion for 𝑚 §4.Dynamic Models of Hydraulic Systems Solution Define the pressures 𝑝3 and 𝑝4 to be the pressures on the leftand right-hand sides of the piston The mass flow rates through the resistances are 𝑞𝑚1 = 𝑝 + 𝑝2 − 𝑝3 𝑅1 1 𝑞𝑚2 = 𝑝 − 𝑝2 − 𝑝𝑎 𝑅2 From conservation of mass 𝑞𝑚1 = 𝑞𝑚2 𝑞𝑚1 = 𝜌𝐴𝑥 Combining these equations we obtain 𝑝1 + 𝑝𝑎 − 𝑝3 = 𝑅1 𝜌𝐴𝑥 𝑝4 − 𝑝2 − 𝑝𝑎 = 𝑅2 𝜌𝐴𝑥 ⟹ 𝑝4 − 𝑝3 = 𝑝2 − 𝑝1 + (𝑅1 + 𝑅2 )𝜌𝐴𝑥 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.45 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.46 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems From Newton’s law 𝑚𝑥 = 𝐴(𝑝3 − 𝑝4 ) Rearrange to obtain the desired model 𝑚𝑥 + 𝑅1 + 𝑅2 𝜌𝐴2 𝑥 = 𝐴(𝑝1 − 𝑝2 ) Note that if the resistances are zero, the 𝑥 term disappears, and we obtain 𝑚𝑥 = 𝐴(𝑝1 − 𝑝2 ) which is identical to the model derived in part (a) of Example 7.1.3 §4.Dynamic Models of Hydraulic Systems - Example 7.4.6 Hydraulic Piston with Negligible Load Develop a model for the motion of the load mass 𝑚 in the figure, assuming that the product of the load mass 𝑚 and the load acceleration 𝑥 is very small Solution If 𝑚𝑥 is very small, from 𝑚𝑥 + 𝑅1 + 𝑅2 𝜌𝐴2 𝑥 = 𝐴(𝑝1 − 𝑝2 ), we obtain the model 𝑅1 + 𝑅2 𝜌𝐴2 𝑥 = 𝐴(𝑝1 − 𝑝2 ) which can be expressed as 𝑝1 − 𝑝2 𝑥= (𝑅1 + 𝑅2 )𝜌𝐴 if 𝑝1 − 𝑝2 is constant, the mass velocity 𝑥 will also be constant HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.47 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.48 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems The implications of the approximation 𝑚𝑥 = can be seen from Newton’s law 𝑚𝑥 = 𝐴(𝑝3 − 𝑝4 ) If 𝑚𝑥 = 0, the above equation implies that 𝑝3 = 𝑝4 ⟹ the pressure is the same on both sides of the piston From this we can see that the pressure difference across the piston is produced by a large load mass or a large load acceleration The modeling implication of this fact is that if we neglect the load mass or the load acceleration, we can develop a simpler model of a hydraulic system - a model based only on conservation of mass and not on Newton’s law The resulting model will be first order rather than second order §4.Dynamic Models of Hydraulic Systems - Example 7.4.A Hydraulic Actuator The pilot valve controls the flow rate of the hydraulic fluid from the supply to the cylinder When the pilot valve is moved to the right of its neutral position, the fluid enters the right-hand piston chamber and pushes the piston to the left The fluid displaced by this motion exits through the left-hand drain port The action is reversed for a pilot valve displacement to the left Both return lines are connected to a sump from which a pump draws fluid to deliver to the supply line Derive a model of the system assuming that 𝑚𝑥 = is negligible HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 8/25/2013 System Dynamics 7.49 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Solution The volume flow rate through the cylinder port is given by 1 𝑞𝑣 = 𝑞𝑚 = 𝐶𝑑 𝐴0 2∆𝑝𝜌 𝜌 𝜌 = 𝐶𝑑 𝐴0 2∆𝑝/𝜌 𝐴𝑜 :uncovered area of the port 𝐴𝑜 ≈ 𝑦𝐷, 𝐷: the port depth 𝐶𝑑 : the discharge coefficient 𝜌: mass density of the fluid If 𝐶𝑑 , 𝜌, 𝑝, and 𝐷 are taken to be constant 𝑞𝑣 = 𝐶𝑑 𝐷𝑦 2∆𝑝/𝜌 = 𝐵𝑦 where, 𝐵 = 𝐶𝑑 𝐷 2∆𝑝/𝜌 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.51 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.50 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems The rate at which the piston pushes fluid out of the cylinder is 𝐴𝑑𝑥/𝑑𝑡 From conservation of volume 𝑑𝑥 𝑞𝑣 = 𝐴 𝑑𝑡 Combining the last two equations gives the model for the servomotor 𝑑𝑥 𝐵 = 𝑦 𝑑𝑡 𝐴 This model predicts a constant piston velocity 𝑑𝑥/𝑑𝑡 if 𝑦 is held fixed The same pressure drop 𝑝 across both the inlet and outlet valves ∆𝑝 = 𝑝𝑠 + 𝑝𝑎 − 𝑝1 = 𝑝2 − 𝑝𝑎 ⟹ 𝑝1 − 𝑝2 = 𝑝𝑠 − 2∆𝑝 From Newton’s law 𝑚𝑥 = 𝐴(𝑝1 − 𝑝2 ), 𝑚𝑥 ≈ ⟹ 𝑝1 = 𝑝2 , and thus 𝑝 = 𝑝𝑠 /2 Therefore 𝐵 = 𝐶𝑑 𝐷 𝑝𝑠 /𝜌 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.52 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems 7.Pump Models - Pump behavior, especially dynamic response, can be quite complicated and difficult to model - Here, based on the steady-state performance curves to obtain linearized models for pump - Typical performance curves for a centrifugal pump which relates the mass flow rate 𝑞𝑚 through the pump to the pressure increase 𝑝 in going from the pump inlet to its outlet, for a given pump speed 𝑠𝑗 §4.Dynamic Models of Hydraulic Systems - For a given speed and given equilibrium values (𝑞𝑚 )𝑒 and (𝑝)𝑒 , we can obtain a linearized description of the figure 𝛿𝑞𝑚 = − 𝛿(∆𝑝) 𝑟 𝛿𝑞𝑚 : the deviations of 𝑞𝑚 𝛿𝑞𝑚 = 𝑞𝑚 − (𝑞𝑚 )𝑒 𝛿(𝑝): the deviations of 𝑝 𝛿 𝑝 = 𝑝 − (𝑝)𝑒 - Identification of the equilibrium values depends on the load connected downstream of the pump Once this load is known, the resulting equilibrium flow rate of the system can be found as a function of 𝑝 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.53 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.54 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems - Example 7.4.8 A Liquid-Level System with a Pump The figure shows a liquid-level system with a pump input and a drain whose linear resistance is 𝑅2 The inlet from the pump to the tank has a linear resistance 𝑅1 Obtain a linearized model of the liquid height ℎ Soution Let 𝑝 ≡ 𝑝1 − 𝑝2 Denote the mass flow rates through each resistance as 𝑞𝑚1 and 𝑞𝑚2 These flow rates are 1 𝑞𝑚1 = 𝑝 − 𝜌𝑔ℎ − 𝑝𝑎 = (∆𝑝 − 𝜌𝑔ℎ) (1) 𝑅1 𝑅1 1 𝑞𝑚2 = 𝜌𝑔ℎ + 𝑝𝑎 − 𝑝𝑎 = 𝜌𝑔ℎ (2) 𝑅2 𝑅2 §4.Dynamic Models of Hydraulic Systems From conservation of mass 𝑑ℎ 𝜌𝐴 = 𝑞𝑚1 − 𝑞𝑚2 𝑑𝑡 1 = ∆𝑝 − 𝜌𝑔ℎ − 𝜌𝑔ℎ (3) 𝑅1 𝑅2 At equilibrium, 𝑞𝑚1 = 𝑞𝑚2 , from eq.(3) 1 𝑅2 ∆𝑝 − 𝜌𝑔ℎ = 𝜌𝑔ℎ ⟹ 𝜌𝑔ℎ = ∆𝑝 (4) 𝑅1 𝑅2 𝑅1 + 𝑅2 Substituting eq.(4) into eq.(2) to obtain an expression for the equilibrium value of the flow rate 𝑞𝑚2 as a function of 𝑝 𝑞𝑚2 = ∆𝑝 (5) 𝑅 +𝑅 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien This is simply an expression of the series resistance law, which applies here because ℎ = at equilibrium and thus the same flow occurs through 𝑅1 and 𝑅2 Nguyen Tan Tien 8/25/2013 System Dynamics 7.55 Fluid and Thermal Systems System Dynamics 7.56 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Plotted the flow rate 𝑞𝑚2 on the same plot as the pump curve, the intersection gives the equilibrium values of 𝑞𝑚1 and 𝑝 A straight line tangent to the pump curve and having the slope −1/𝑟 then gives the linearized model (6) 𝛿𝑞𝑚1 = − 𝛿(∆𝑝) 𝑟 𝛿𝑞𝑚1 , 𝛿(𝑝): the deviations from the equilibrium values From eq.(4) and eq.(6) 𝑅1 + 𝑅2 𝑅1 + 𝑅2 ∆𝑝 = 𝜌𝑔ℎ, 𝛿 ∆𝑝 = 𝜌𝑔𝛿ℎ 𝑅2 𝑅2 1 𝑅1 + 𝑅2 𝛿𝑞𝑚1 = − 𝛿 ∆𝑝 = − 𝜌𝑔𝛿ℎ (7) 𝑟 𝑟 𝑅2 §4.Dynamic Models of Hydraulic Systems The linearized form of eq.(3) 𝜌𝐴𝑑(𝛿ℎ)/𝑑𝑡 = 𝛿𝑞𝑚1 − 𝛿𝑞𝑚2 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.57 Nguyen Tan Tien Fluid and Thermal Systems From eq.(2) and eq.(7) 𝑑 𝑅1 + 𝑅2 𝜌𝑔 𝜌𝐴 𝛿ℎ = − 𝜌𝑔𝛿ℎ − 𝛿ℎ 𝑑𝑡 𝑟 𝑅2 𝑅2 𝑑 𝑅1 + 𝑅2 ⟹ 𝐴 𝛿ℎ = − + 𝑔𝛿ℎ 𝑑𝑡 𝑟 𝑅2 𝑅2 This is the linearized model, and it is of the form 𝑑 𝑅1 + 𝑅2 𝑔 𝛿ℎ = −𝑏𝛿ℎ, 𝑏= + 𝑑𝑡 𝑟 𝑅2 𝑅2 𝐴 The equation has the solution 𝛿ℎ(𝑡) = 𝛿ℎ(0)𝑒 −𝑏𝑡 Thus if additional liquid is added to or taken from the tank so that 𝛿ℎ(0) = , the liquid height will eventually return to its equilibrium value The time to return is indicated by the time constant, which is 1/𝑏 System Dynamics 7.58 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems 8.Nonlinear System - Common causes of nonlinearities in hydraulic system models are a nonlinear resistance relation, such as due to orifice flow or turbulent flow, or a nonlinear capacitance relation, such as a tank with a variable cross section - If the liquid height is relatively constant, say because of a liquid-level controller, we can analyze the system by linearizing the model - In cases where the height varies considerably, we must solve the nonlinear equation numerically §4.Dynamic Models of Hydraulic Systems - Example 7.4.9 Liquid-Level System with an Orifice Consider the liquid-level system with an orifice as in the figure The model is 𝑑ℎ 𝐴 = 𝑞𝑣𝑖 − 𝑞𝑣𝑜 𝑑𝑡 = 𝑞𝑣𝑖 − 𝐶𝑑 𝐴𝑜 2𝑔ℎ HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.59 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Solution Substituting the given values, we obtain 𝑑ℎ = 𝑞𝑣𝑖 − 𝑞𝑣𝑜 = 𝑞𝑣𝑖 − ℎ (1) 𝑑𝑡 When the inflow rate 𝑞𝑣𝑒 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, the liquid height reaches an equilibrium value ℎ𝑒 that can be found by setting 𝑑ℎ/𝑑𝑡 = The two cases of interest to us are (i) ℎ𝑒 = 122 /36 = 4𝑓𝑡 and (ii) ℎ𝑒 = 242 /36 = 16𝑓𝑡 The graph is a plot of the flow rate ℎ through the orifice as a function of the height ℎ The two points corresponding to ℎ𝑒 = and ℎ𝑒 = 16 are indicated on the plot HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien where 𝐴 = 2𝑓𝑡 and 𝐶𝑑 𝐴𝑜 2𝑔 = Estimate the system’s time constant for two cases (i) the inflow rate is held constant at 𝑞𝑣𝑖 = 12𝑓𝑡 /𝑠𝑒𝑐 (ii)the inflow rate is held constant at 𝑞𝑣𝑖 = 24𝑓𝑡 /𝑠𝑒𝑐 System Dynamics 7.60 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems In the figure two straight lines are shown, each passing through one of the points of interest (ℎ𝑒 = and ℎ𝑒 = 16), and having a slope equal to the slope of the curve at that point The general equation for these lines is 𝑑𝑞𝑣𝑜 − 𝑞𝑣𝑜 = ℎ = ℎ𝑒 + ℎ − ℎ𝑒 = ℎ𝑒 + 3ℎ𝑒 2(ℎ − ℎ𝑒) 𝑑ℎ 𝑒 Substitute this into equation (1) to obtain 1 𝑑ℎ − − = 𝑞𝑣𝑖 − ℎ𝑒 − 3ℎ𝑒 ℎ − ℎ𝑒 = 𝑞𝑣𝑖 − ℎ𝑒 − 3ℎ𝑒 2ℎ 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10 8/25/2013 System Dynamics 7.61 Fluid and Thermal Systems System Dynamics 7.62 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems 𝑑ℎ − −1/2 = 𝑞𝑣𝑖 − ℎ𝑒 − 3ℎ𝑒 ℎ − ℎ𝑒 = 𝑞𝑣𝑖 − ℎ𝑒 − 3ℎ𝑒 ℎ 𝑑𝑡 The time constant of this linearized model is τ = ℎ𝑒 /3 𝜏 = , 𝜏 = 3 ℎ𝑒 =4 ℎ𝑒 =16 - If the input rate 𝑞𝑣𝑖 is changed slightly from its equilibrium value of 𝑞𝑣𝑖 = 12, the liquid height will take about 4(4/3) = 16/3sec to reach its new height - If the input rate 𝑞𝑣𝑖 is changed slightly from its value of 𝑞𝑣𝑖 = 24, the liquid height will take about 8(4/3) = 32/3sec to reach its new height §4.Dynamic Models of Hydraulic Systems 9.Fluid Inertance Fluid inertance 𝐼: the ratio of the pressure difference over the rate of change of the mass flow rate 𝑝 𝐼≡ 𝑑𝑞𝑚 /𝑑𝑡 - Example 7.4.10 Calculation of Inertance Consider fluid flow (either liquid or gas) in a nonaccelerating pipe Derive the expression for the inertance of a slug of fluid of length 𝐿 Solution The mass of the slug: 𝜌𝐴𝐿, 𝜌: the fluid mass density The net force acting on the slug due to the pressures 𝑝1 and 𝑝2 𝐴(𝑝2 − 𝑝1 ) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.63 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.64 Nguyen Tan Tien Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems Applying Newton’s law to the slug 𝑑𝑣 𝜌𝐴𝐿 = 𝐴(𝑝2 − 𝑝1 ) 𝑑𝑡 𝜌𝐴𝑣 = 𝑞𝑚 𝑣: the fluid velocity 𝑞𝑚 : the mass flow rate 𝑑𝑞𝑚 𝐿 𝑑𝑞𝑚 𝐿 = 𝐴(𝑝2 − 𝑝1 ) ⟹ = 𝑝2 − 𝑝1 𝑑𝑡 𝐴 𝑑𝑡 With 𝑝 = 𝑝2 − 𝑝1 , we obtain 𝐿 𝑝 = 𝐴 𝑑𝑞𝑚 /𝑑𝑡 From the definition of inertance 𝐼 𝐿 𝐼= 𝐴 Inertance is larger for longer pipes and for smaller cross section pipes §5.Pneumatic Systems - Working fluid: a compressible fluid, most commonly air - The response of pneumatic systems can be slower and more oscillatory than that of hydraulic systems because of the compressibility of working fluid - The inertance relation is not usually needed to develop a model because the kinetic energy of a gas is usually negligible Instead, capacitance and resistance elements form the basis of most pneumatic system models - The perfect gas law 𝑝𝑉 = 𝑚𝑅𝑔 𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.65 Nguyen Tan Tien Fluid and Thermal Systems 𝑝: the absolute pressure, 𝑁/𝑚2 𝑉: gas volume, 𝑚3 𝑚: the mass, 𝑘𝑔 𝑇: absolute temperature, 𝐾 𝑅𝑔 : the gas constant, for air 𝑅𝑔 = 287𝑁𝑚/𝑘𝑔 0𝐾 System Dynamics 7.66 Nguyen Tan Tien Fluid and Thermal Systems §5.Pneumatic Systems - The perfect gas law enables us to solve for one of the variables 𝑝,𝑉,𝑚, or 𝑇 if the other three are given Additional information is usually available in the form of a pressurevolume or “process” relation - The following process models are commonly used • Constant-Pressure Process (𝑝1 = 𝑝2 ) 𝑉2 𝑇2 𝑝𝑉 = 𝑚𝑅𝑔 𝑇 ⟹ = 𝑉1 𝑇1 • Constant-Volume Process (𝑉1 = 𝑉2 ) 𝑝2 𝑇2 𝑝𝑉 = 𝑚𝑅𝑔 𝑇 ⟹ = 𝑝1 𝑇1 • Constant-Temperature (isothermal) Process (𝑇1 = 𝑇2) 𝑝2 𝑉1 𝑝𝑉 = 𝑚𝑅𝑔 𝑇 ⟹ = 𝑝1 𝑉2 §5.Pneumatic Systems • Reversible Adiabatic (Isentropic) Process 𝛾 𝛾 𝑝1 𝑉1 = 𝑝2 𝑉2 𝛾 = 𝑐𝑝 /𝑐𝑣 𝑐𝑝 : constant pressure 𝑐𝑣 : constant volume HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Adiabatic: no heat is transferred to or from the gas Reversible: the gas and its surroundings can be returned to their original thermodynamic conditions 𝑊 = 𝑚𝑐𝑣 (𝑇1 − 𝑇2 ) 𝑊:the external work • Polytropic Process A process can be more accurately modeled by properly choosing the exponent 𝑛 in the polytropic process 𝑉 𝑛 𝑝 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑚 If 𝑚 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, this reduces to the previous processes if 𝑛 is chosen as 0,∞,1, 𝛾, and if the perfect gas law is used Nguyen Tan Tien 11 8/25/2013 System Dynamics 7.67 Fluid and Thermal Systems System Dynamics 7.68 Fluid and Thermal Systems §5.Pneumatic Systems 1.Pneumatics Capacitance Fluid capacitance 𝐶 is the ratio of the change in stored mass, 𝑚, to the change in pressure, 𝑝 𝑑𝑚 𝐶≡ 𝑑𝑝 For a container of constant volume 𝑉 with a gas density 𝜌, 𝑚 = 𝜌𝑉 𝑑(𝜌𝑉) 𝑑𝜌 𝐶= =𝑉 𝑑𝑝 𝑑𝑝 If the gas undergoes a polytropic process 𝑉 𝑛 𝑝 𝑑𝜌 𝜌 𝑚 𝑝 = 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⟹ = = 𝑚 𝜌 𝑑𝑝 𝑛𝑝 𝑛𝑝𝑉 For a perfect gas, this shows the capacitance of the container 𝑚𝑉 𝑉 𝐶= = 𝑛𝑝𝑉 𝑛𝑅𝑔 𝑇 §5.Pneumatic Systems - Example 7.5.1 Capacitance of an Air Cylinder Obtain the capacitance of air in a rigid cylinder of volume 0.03𝑚3 , if the cylinder is filled by an isothermal process Assume the air is initially at room temperature, 293𝐾 Solution The filling of the cylinder can be modeled as an isothermal process if it occurs slowly enough to allow heat transfer to occur between the air and its surroundings In this case, 𝑛 = in the polytropic process equation, and we obtain 𝑉 0.03 𝐶= = = 3.57 × 10−7 𝑘𝑔𝑚2 /𝑁 𝑛𝑅𝑔 𝑇 × 287 × 293 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.69 Nguyen Tan Tien Fluid and Thermal Systems §5.Pneumatic Systems - Example 7.5.2 Pressurizing an Air Cylinder Air at temperature 𝑇 passes through a valve into a rigid cylinder of volume 𝑉, as shown in the figure The mass flow rate through the valve depends on the pressure difference ∆𝑝 = 𝑝𝑖 − 𝑝, and is given by an experimentally determined function 𝑞𝑚𝑖 = 𝑓(𝑝) Assume the filling process is isothermal Develop a dynamic model of the gage pressure 𝑝 in the container as a function of the input pressure 𝑝𝑖 System Dynamics 7.70 Nguyen Tan Tien Fluid and Thermal Systems §5.Pneumatic Systems Solution From conservation of mass, if 𝑝𝑖 = 𝑝 > 𝑑𝑚 𝑞𝑚𝑖 = 𝑓(𝑝) ⟹ = 𝑞𝑚𝑖 = 𝑓(∆𝑝) 𝑑𝑡 But 𝑑𝑚 𝑑𝑚 𝑑𝑝 𝑑𝑝 = =𝐶 𝑑𝑡 𝑑𝑝 𝑑𝑡 𝑑𝑡 𝑑𝑝 𝐶 = 𝑓 ∆𝑝 = 𝑓(𝑝 − 𝑝) and thus 𝑖 𝑑𝑡 where the capacitance 𝐶 is given with 𝑛 = 𝑉 𝐶= 𝑅𝑔 𝑇 If the function 𝑓 is nonlinear, then the dynamic model is nonlinear HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.71 Nguyen Tan Tien Fluid and Thermal Systems Part Thermal Systems HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.72 Nguyen Tan Tien Fluid and Thermal Systems - A thermal system is one in which energy is stored and transferred as thermal energy, commonly called heat - Thermal systems operate because of temperature differences, as heat energy flows from an object with the higher temperature to an object with the lower temperature - Thermal systems are analogous to electric circuits • conservation of charge ⟺ conservation of heat • voltage difference ⟺ temperature difference HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 12 8/25/2013 System Dynamics 7.73 Fluid and Thermal Systems §6.Thermal Capacitance - The amount of heat energy 𝐸 stored in the object at a temperature 𝑇 is 𝐸 = 𝑚𝑐𝑝 (𝑇 − 𝑇𝑟 ) 𝑚: mass, 𝑘𝑔 𝑐𝑝 : specific heat, 𝐽𝑘𝑔/𝐾 𝑇𝑟 : an arbitrarily selected reference temperature, 𝐾 - Thermal capacitance 𝑑𝐸 𝐶≡ 𝑑𝑇 𝐶: thermal capacitance, 𝐽/𝐾 𝐸: the stored heat energy If 𝑐𝑝 does not depend on temperature 𝐶 = 𝑚𝑐𝑝 = 𝜌𝑉𝑐𝑝 𝜌: the density, 𝑘𝑔/𝑚3 𝑚: the mass, 𝑘𝑔 7.74 Fluid and Thermal Systems §6.Thermal Capacitance - Example 7.6.1 Temperature Dynamics of a Mixing Process Liquid at a temperature 𝑇𝑖 is pumped into a mixing tank at a constant volume flow rate 𝑞𝑣 The container walls are perfectly insulated so that no heat escapes through them Container volume is 𝑉, and the liquid within is well mixed so that its temperature throughout is 𝑇 The liquid’s specific heat and mass density are 𝑐𝑝 and 𝜌 Develop a model for the temperature 𝑇 as a function of time, with 𝑇𝑖 as the input 𝑉: the volume, 𝑚3 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics System Dynamics 7.75 Nguyen Tan Tien Fluid and Thermal Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.76 Nguyen Tan Tien Fluid and Thermal Systems §6.Thermal Capacitance Solution The amount of heat energy in the tank liquid (1) 𝜌𝑐𝑝 𝑉(𝑇 − 𝑇𝑟 ) From conservation of energy 𝑑 𝜌𝑐𝑝𝑉(𝑇 − 𝑇𝑟) = ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 𝑖𝑛 − ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡 𝑑𝑡 𝑚 = 𝜌𝑞𝑣 ⟹ heat energy is flowing into the tank at the rate ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 𝑖𝑛 = 𝑚𝑐𝑝 𝑇𝑖 − 𝑇𝑟 = 𝜌𝑞𝑣 𝑐𝑝 (𝑇𝑖 − 𝑇𝑟 ) Similarly,ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡 = 𝜌𝑞𝑣 𝑐𝑝 (𝑇 − 𝑇𝑟 ) From eq.(1), since 𝜌, 𝑐𝑝 , 𝑉, and 𝑇𝑟 are constants 𝑑𝑇 𝜌𝑐𝑝𝑉 = 𝜌𝑞𝑣𝑐𝑝 𝑇𝑖 − 𝑇𝑟 − 𝜌𝑞𝑣𝑐𝑝 𝑇 − 𝑇𝑟 = 𝜌𝑞𝑣𝑐𝑝(𝑇𝑖 − 𝑇) 𝑑𝑡 𝑉 𝑑𝑇 ⟹ + 𝑇 = 𝑇𝑖 𝑞𝑣 𝑑𝑡 §7.Thermal Resistance 1.Conduction, convection, and Radiation - Temperature is a measure of the amount of heat energy in an object - Heat transfer can occur by one or more modes: conduction, convection, and radiation, as illustrated by the figure - Conduction: the transfer of heat energy by diffusion of heat through a substance - Convection: the transfer of heat energy by the movement of fluids - Radiation: the transfer of heat energy by radiation occurs through infrared waves HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.77 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance 2.Newton’s Law of Cooling - Newton’s law of cooling (for both convection and conduction) 𝑞ℎ = ∆𝑇 𝑅 𝑞ℎ : the heat flow rate, 𝐽/𝑠 = 𝑊 𝑅: the thermal resistance, 0𝐶/𝑊 𝑇: the temperature difference, 0𝐶 - For conduction through material of thickness 𝐿, an approximate formula for the conductive resistance is 𝐿 𝑘𝐴 𝑅= ⟹ 𝑞ℎ = ∆𝑇 𝑘𝐴 𝐿 𝑘: the thermal conductivity of the material 𝐴: the surface area, 𝑚2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 7.78 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance - The thermal resistance for convection occurring at the boundary of a fluid and a solid is given by 𝑅= ⟹ 𝑞ℎ = ℎ𝐴∆𝑇 ℎ𝐴 ℎ: the convection coefficient of the fluid-solid interface, 𝑊/𝑚2 0𝐶 𝐴: the involved surface area, 𝑚2 - When two bodies are in visual contact, radiation heat transfer occurs through a mutual exchange of heat energy by emission and absorption The net heat transfer rate 𝑞ℎ = 𝛽(𝑇14 − 𝑇24 ) 𝛽: factor incorporating the other effects 𝑇: absolute body temperatures, 𝐾 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 13 8/25/2013 System Dynamics 7.79 Fluid and Thermal Systems System Dynamics 7.80 §7.Thermal Resistance - The radiation model is nonlinear, however, we can use a linearized model if the temperature change is not too large Note that linear thermal resistance is a special case of the more general definition of thermal resistance 𝑅= 𝑑𝑞ℎ /𝑑𝑇 Suppose that 𝑇2 is constant, then 1 𝑅= = 𝑑𝑞ℎ /𝑑𝑇1 4𝛽𝑇13 When this is evaluated at a specific temperature 𝑇1, we can obtain a specific value for the linearized radiation resistance §7.Thermal Resistance 3.Heat Transfer Through a Plate - Consider a solid plate or wall of thickness 𝐿 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.81 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance - Under steady-state conditions, the average temperature is at the center Consider the entire mass 𝑚 of the plate to be concentrated (“lumped”) at the plate centerline, and consider conductive heat transfer to occur over a path of length 𝐿/2 between temperature 𝑇1 and temperature 𝑇 The thermal resistance for this path is 𝐿/2 𝑅1 = = 𝑅2 𝑘𝐴 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.83 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance 5.Series and Parallel Thermal Resistances - Suppose the capacitance 𝐶 in the circuit is zero This is equivalent to removing the capacitance We can see immediately that the two resistances are in series Therefore they can be combined by the series law 𝑅 = 𝑅1 + 𝑅2 to obtain the equivalent circuit HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Fluid and Thermal Systems If 𝑇1 > 𝑇2 , heat will flow from the left side to the right side - Fourier’s law of heat conduction: the heat transfer rate per unit area within a homogeneous substance is directly proportional to the negative temperature gradient 𝑘𝐴(𝑇1 − 𝑇2 ) 𝑞ℎ = 𝐿 𝑘: the thermal conductivity, 𝑊/𝑚 0𝐶 𝐴: the plate area, 𝑚2 System Dynamics 7.82 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance - Applying conservation of heat energy with sssuming that 𝑇1 > 𝑇 > 𝑇2, we can derive the following model 𝑑𝑇 1 = 𝑞1 − 𝑞2 = 𝑇 − 𝑇 − (𝑇 − 𝑇2 ) 𝑑𝑡 𝑅1 𝑅2 The thermal capacitance is 𝐶 = 𝑚𝑐𝑝 𝑚𝑐𝑝 - This system is analogous to the circuit shown in the figure • the voltages 𝑣, 𝑣1 , 𝑣2 ⟺ the temperatures 𝑇, 𝑇1, 𝑇2 • the current 𝑖1, 𝑖2 ⟺ the heat flow rate • the current 𝑖3 ⟺ the net heat flow rate into the mass 𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.84 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance - If the plate mass 𝑚 is very small ⟹ its thermal capacitance 𝐶 is also very small: the mass absorbs a negligible amount of heat energy the heat flow rate 𝑞1 through the left-hand conductive path = the heat flow rate 𝑞2 through the right-hand path That is, if 𝐶 = 1 𝑞1 = 𝑇 − 𝑇 = 𝑞2 = (𝑇 − 𝑇2 ) 𝑅1 𝑅2 The solution of these equations is 𝑅2 𝑇1 + 𝑅1 𝑇2 𝑇= 𝑅1 + 𝑅2 𝑇1 − 𝑇2 𝑇1 − 𝑇2 𝑞1 = 𝑞2 = = 𝑅1 + 𝑅2 𝑅 the resistances 𝑅1, 𝑅2 are equivalent to the single resistance 𝑅 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 14 8/25/2013 System Dynamics 7.85 Fluid and Thermal Systems §7.Thermal Resistance - Thermal resistances are in series if they pass the same heat flow rate; if so, they are equivalent to a single resistance equal to the sum of the individual resistances 𝑅 = 𝑅1 + 𝑅2 - It can also be shown that thermal resistances are in parallel if they have the same temperature difference; if so, they are equivalent to a single resistance calculated by the reciprocal formula 1 = + +⋯ 𝑅 𝑅1 𝑅2 - If convection occurs on both sides of the plate, the convective resistances 𝑅𝑐1 and 𝑅𝑐2 are in series with the conductive resistance 𝑅, and the total resistance is given by 𝑅 + 𝑅𝑐1 + 𝑅𝑐2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.87 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance Solution a.The series resistance law 𝑅 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 = 0.036+ 4.01+ 0.408 + 0.038 = 4.492 0𝐶/𝑊 for 1𝑚2 of wall area The total heat loss 1 𝑞ℎ = 15 𝑇𝑖 − 𝑇𝑜 = 15 20 + 10 = 100.2𝑊 𝑅 4.492 This is the heat rate that must be supplied by the building’s heating system to maintain the inside temperature at 20 0𝐶, if the outside temperature is −10 0𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.89 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance - Example 7.7.2 Parallel Resistances A certain wall section is composed of a 15𝑐𝑚 by 15𝑐𝑚 glass block 8𝑐𝑚 thick Surrounding the block is a 50𝑐𝑚 × 50𝑐𝑚𝑚 brick section, which is also 8𝑐𝑚 thick The thermal conductivity of the glass is 𝑘 = 0.81𝑊/𝑚 0𝐶 For the brick, 𝑘 = 0.45𝑊/𝑚 0𝐶 a.Determine the thermal resistance of the wall section b.Compute the heat flow rate through (1) the glass, (2) the brick, and (3) the wall if the temperature difference across the wall is 30 0𝐶 System Dynamics 7.86 Fluid and Thermal Systems §7.Thermal Resistance - Example 7.7.1 Thermal Resistance of Building Wall The wall cross section shown in figure consists of four layers: 10𝑚𝑚 plaster/ lathe, 125𝑚𝑚 fiberglass insulation, 60𝑚𝑚 wood, and 50𝑚𝑚 brick For the given materials, the resistances for a wall area of 1𝑚2 are 𝑅1 = 0.036 0𝐶/𝑊, 𝑅2 = 4.01 0𝐶/𝑊, 𝑅3 = 0.408 0𝐶/ 𝑊 , and 𝑅4 = 0.038 0𝐶/𝑊 Suppose that 𝑇𝑖 = 20 0𝐶 , 𝑇𝑜 = − 10 0𝐶 a Compute the total wall resistance for 1𝑚2 of wall area, and compute the heat loss rate if the wall’s area is 3𝑚 × 5𝑚 b Find the temperatures 𝑇1, 𝑇2 , and 𝑇3 , assuming steadystate conditions HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 7.88 Fluid and Thermal Systems §7.Thermal Resistance b.If we assume that the inner and outer temperatures 𝑇𝑖 and 𝑇𝑜 have remained constant for some time, then the heat flow rate through each layer is the same, 𝑞ℎ Applying conservation of energy gives 1 1 𝑞ℎ = 𝑇𝑖 − 𝑇1 = 𝑇1 − 𝑇2 = 𝑇 −𝑇 = 𝑇 −𝑇 𝑅1 𝑅2 𝑅3 𝑅4 These equations can be rearranged as follows 𝑅1 + 𝑅2 𝑇1 − 𝑅1 𝑇2 = 𝑅2 𝑇𝑖 𝑅3 𝑇1 − 𝑅2 + 𝑅3 𝑇2 + 𝑅2 𝑇3 = −𝑅4 𝑇2 + 𝑅3 + 𝑅4 𝑇3 = 𝑅3 𝑇𝑜 Solution: 𝑇1 = 19.7596 0𝐶 , − 9.7462 0𝐶 𝑇2 = −7.0214 0𝐶 , HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 𝑇3 = Nguyen Tan Tien 7.90 Fluid and Thermal Systems §7.Thermal Resistance Solution a.The wall resistance 𝑅= 𝐿 𝑘𝐴 0.08 = 4.39 0.81 × 0.152 0.08 = 0.781 For the brick 𝑅2 = 0.45(0.52 − 0.152) Because the temperature difference is the same across both the glass and the brick, the resistances are in parallel, and thus their total resistance is given by 1 = + = 0.228 + 1.28 = 1.51 𝑅 𝑅1 𝑅2 For the glass 𝑅1 = or 𝑅 = 0.633 0𝐶/𝑊 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 15 8/25/2013 System Dynamics 7.91 Fluid and Thermal Systems System Dynamics 7.92 Fluid and Thermal Systems §7.Thermal Resistance Solution b.The heat flow through the glass is 1 𝑞1 = ∆𝑇 = 30 = 6.83𝑊 𝑅1 4.39 The heat flow through the brick is 1 𝑞2 = ∆𝑇 = 30 = 38.4𝑊 𝑅2 0.781 The total heat flow through the wall section is 𝑞ℎ = 𝑞1 + 𝑞2 = 45.2𝑊 This rate could also have been calculated from the total resistance as follows 1 𝑞ℎ = ∆𝑇 = 30 = 45.2𝑊 𝑅 0.663 §7.Thermal Resistance - Example 7.7.3 Radial Conductive Resistance Consider a cylindrical tube whose inner and outer radii are 𝑟𝑖 and 𝑟𝑜 Heat flow in the tube wall can occur in the axial direction along the length of the tube and in the radial direction If the tube surface is insulated, there will be no radial heat flow, and the heat flow in the axial direction is given by 𝑘𝐴 𝑞ℎ = ∆𝑇 𝐿 where 𝐿 is the length of the tube, 𝑇 is the temperature difference between the ends a distance 𝐿 apart, and 𝐴 is area of the solid cross section If only the ends of the tube are insulated, then the heat flow will be entirely radial Derive an expression for the conductive resistance in the radial direction HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.93 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.94 Fluid and Thermal Systems §7.Thermal Resistance Solution From Fourier’s law, the heat flow rate per unit area through an element of thickness 𝑑𝑟 is proportional to the negative of the temperature gradient 𝑑𝑇/𝑑𝑟 Assuming that the temperature inside the tube wall does not change with time, the heat flow rate 𝑞ℎ out of the section of thickness 𝑑𝑟 is the same as the heat flow into the section 𝑞ℎ 𝑑𝑇 = −𝑘 2𝜋𝑟𝐿 𝑑𝑟 𝑑𝑇 𝑑𝑇 ⟹ 𝑞ℎ = −𝑘 2𝜋𝑟𝐿 = −2𝜋𝐿𝑘 𝑑𝑟 𝑑𝑟/𝑟 𝑟𝑜 𝑇𝑜 𝑑𝑟 ⟹ 𝑞ℎ = −2𝜋𝐿𝑘 𝑑𝑇 𝑟 𝑟1 𝑇𝑖 §7.Thermal Resistance 𝑟𝑜 𝑑𝑟 𝑞ℎ = −2𝜋𝐿𝑘 𝑟 𝑟1 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics §7.Thermal Resistance - Example 7.7.4 7.95 Nguyen Tan Tien Fluid and Thermal Systems Heat Loss from Water in a Pipe Water at 120 0𝐹 flows in a copper pipe 6𝑓𝑡 long, whose inner and outer radii are 1/4𝑖𝑛 and 3/8𝑖𝑛 The temperature of the surrounding air is 70 0𝐹 Compute the heat loss rate from the water to the air in the radial direction Use the following values For copper, 𝑘 = 50𝑙𝑏/𝑠𝑒𝑐 0𝐹 The convection coefficient at the inner surface between the water and the copper is ℎ𝑖 = 16𝑙𝑏/𝑠𝑒𝑐 0𝐹 The convection coefficient at the outer surface between the air and the copper is ℎ𝑜 = 11𝑙𝑏/𝑠𝑒𝑐 0𝐹 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 𝑇𝑜 𝑑𝑇 𝑇𝑖 Because 𝑞ℎ is constant, the integration yields 𝑟𝑜 𝑞ℎ ln = −2𝜋𝐿𝑘(𝑇𝑜 − 𝑇𝑖 ) 𝑟𝑖 or 𝑟𝑜 𝑞ℎ ln = −2𝜋𝐿𝑘 𝑇𝑜 − 𝑇𝑖 𝑟𝑖 The radial resistance is thus given by 2𝜋𝐿𝑘 𝑞ℎ = 𝑇 − 𝑇𝑜 ln(𝑟𝑜 /𝑟𝑖 ) 𝑖 System Dynamics 7.96 Nguyen Tan Tien Fluid and Thermal Systems §7.Thermal Resistance Solution Assuming constant temperature inside the pipe wall, then the same heat flow rate occurs in the inner and outer convection layers and in the pipe wall ⟹ the three resistances are in series The inner and outer surface areas are 1 𝐴𝑖 = 2𝜋𝑟𝑖 𝐿 = 2𝜋 × × × = 0.785𝑓𝑡 12 𝐴𝑜 = 2𝜋𝑟𝑜 𝐿 = 2𝜋 × × × = 1.178𝑓𝑡 12 The inner convective resistance is 1 𝑠𝑒𝑐 0𝐹 𝑅𝑖 = = = 0.08 ℎ𝑖 𝐴𝑖 16 × 0.785 𝑓𝑡𝑙𝑏 1 𝑠𝑒𝑐 0𝐹 𝑅𝑜 = = = 0.77 ℎ𝑜 𝐴𝑜 1.1 × 1.178 𝑓𝑡𝑙𝑏 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 16 8/25/2013 System Dynamics 7.97 Fluid and Thermal Systems §7.Thermal Resistance The conductive resistance of the pipe wall is ln(𝑟𝑜/𝑟𝑖) ln (3 8)/(1 4) 𝑠𝑒𝑐0𝐹 𝑅𝑐 = = = 2.15 × 10−4 2𝜋𝐿𝑘 2𝜋 × × 50 𝑓𝑡𝑙𝑏 Thus the total resistance is 𝑠𝑒𝑐0𝐹 𝑓𝑡𝑙𝑏 Assuming that the water temperature is a constant 120 0𝐹 along the length of the pipe, the heat loss from the pipe is 1 𝑓𝑡𝑙𝑏 𝑞ℎ = ∆𝑇 = 120 − 70 = 59 𝑅 0.85 𝑠𝑒𝑐 𝑅 = 𝑅𝑓 + 𝑅𝑐 + 𝑅𝑜 = 0.08 + 2.15 × 10−4 + 0.77 = 0.85 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.99 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems 1.The Biot Criterion - For solid bodies immersed in a fluid, a useful criterion for determining the validity of the uniform-temperature assumption is based on the Biot number, defined as ℎ𝐿 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 𝑁𝐵 = , 𝐿= 𝑘 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 ℎ: convection coefficient, 𝑊/𝑚2 0𝐶 𝐿: a representative dimension of the object, 𝑚 𝑘: thermal conductivity, 𝑊/𝑚 0𝐶 - If the shape of the body resembles a plate, cylinder, or sphere, it is common practice to consider the object to have a single uniform temperature if 𝑁𝐵 is small - Often, if 𝑁𝐵 < 0.1, the temperature is taken to be uniform The accuracy of this approximation improves if the inputs vary slowly HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.101 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.98 Fluid and Thermal Systems §7.Thermal Resistance To investigate the assumption that the water temperature is constant, compute the thermal energy 𝐸 of the water in the pipe, using the mass density 𝜌 = 1.94𝑠𝑙𝑢𝑔/𝑓𝑡 and 𝑐𝑝 = 25,000𝑓𝑡 − 𝑙𝑏/𝑠𝑙𝑢𝑔 0𝐹 𝐸 = 𝑚𝑐𝑝 𝑇𝑖 = 𝜋𝑟𝑖2 𝐿𝜌 𝑐𝑝 𝑇𝑖 = 47,624𝑓𝑡𝑙𝑏 Assuming that the water flows at 1𝑓𝑡/𝑠𝑒𝑐, a slug of water will be in the pipe for 6𝑠𝑒𝑐 During that time it will lose 59 × = 354𝑓𝑡𝑙𝑏 of heat Because this amount is very small compared to 𝐸 , our assumption that the water temperature is constant is confirmed HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.100 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems - The Biot number is the ratio of the convective heat transfer rate to the conductive rate This can be seen by expressing 𝑁𝐵 for a plate of thickness 𝐿 as follows 𝑞𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 ℎ𝐴∆𝑇 ℎ𝐿 𝑁𝐵 = = = 𝑞𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑘𝐴∆𝑇/𝐿 𝑘 - The Biot criterion reflects the fact that if the conductive heat transfer rate is large relative to the convective rate, any temperature changes due to conduction within the object will occur relatively rapidly, and thus the object’s temperature will become uniform relatively quickly - Calculation of the ratio 𝐿 depends on the surface area that is exposed to convection For example, a cube of side length 𝑑 has a value of 𝐿 = 𝑑 /(6𝑑 ) = 𝑑/6 if all six sides are exposed to convection, whereas if four sides are insulated, the value is 𝐿 = 𝑑3 /(2𝑑2 ) = 𝑑/2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.102 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems - Example 7.8.1 Quenching with Constant Bath Temperature Consider a lead cube with a side length of 𝑑 = 20𝑚𝑚 The cube is immersed in an oil bath for which ℎ = 200𝑊/𝑚2 0𝐶 The oil temperature is 𝑇𝑏 Thermal conductivity varies as function of temperature, but for lead the variation is relatively small (𝑘 for lead varies from 35.5𝑊/𝑚 0𝐶 at 0𝐶 to 31.2𝑊/𝑚 0𝐶 at 327 0𝐶 The density of lead is 1.134 × 104 𝑘𝑔/𝑚3 Take the specific heat of lead to be 129𝐽/𝑘𝑔 0𝐶 a.Show that temperature of the cube can be considered uniform b.Develop a model of the cube’s temperature as a function of the liquid temperature 𝑇𝑏 , which is assumed to be known §8.Dynamic Model of Thermal Systems Solution a.The ratio of volume of the cube to its surface area is 𝑑3 𝑑 0.02 𝐿= 2= = 6𝑑 6 Using an average value of 33.35𝑊/𝑚 0𝐶 for 𝑘, compute the Biot number ℎ𝐿 200 × 0.02 𝑁𝐵 = = = 0.02 𝑘 33.35 × 𝑁𝐵 < 0.1, according to the Biot criterion, the cube can be treated as a lumped-parameter system with a single uniform temperature, denoted 𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 17 8/25/2013 System Dynamics 7.103 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems b.Assume 𝑇 > 𝑇𝑏 , the heat flows from the cube to the liquid, and from conservation of energy we obtain 𝑑𝑇 𝐶 = − (𝑇 − 𝑇𝑏 ) 𝑑𝑡 𝑅 The thermal capacitance of the cube 𝐶 = 𝑚𝑐𝑝 = 𝜌𝑉𝑐𝑝 = 1.134 × 104 × 0.023 × 129 = 11.7𝐽/0𝐶 The thermal resistance 𝑅 is due to convection 1 𝑅= = 2.08 0𝐶𝑠/𝐽 ℎ𝐴 200 × × 0.022 Thus the model is 𝑑𝑇 𝑑𝑇 11.7 =− (𝑇 − 𝑇𝑏 ) ⟹ 24.4 + 𝑇 = 𝑇𝑏 𝑑𝑡 2.08 𝑑𝑡 The time constant is 𝜏 = 𝑅𝐶 = 24.4𝑠 The cube’s temperature will reach the temperature 𝑇𝑏 in approximately 4𝜏 = 98𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.105 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.104 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems 𝑑𝑇 24.4 + 𝑇 = 𝑇𝑏 𝑑𝑡 The thermal model of the quenching process is analogous to a circuit shown on the figure The voltages 𝑣 and 𝑣𝑏 are analogous to the temperatures 𝑇 and 𝑇𝑏 The circuit model is 𝑑𝑣 𝐶 = (𝑣 − 𝑣) 𝑑𝑡 𝑅 𝑏 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.106 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems 2.Multiple Thermal Capacitance - When it is not possible to identify one representative temperature for a system, we must identify a representative temperature for each distinct thermal capacitance - After identifying the resistance paths between each capacitance, apply conservation of heat energy to each capacitance - Arbitrarily but consistently assume that some temperatures are greater than others, to assign directions to the resulting heat flows The order of the resulting model equals the number of representative temperatures §8.Dynamic Model of Thermal Systems - Example 7.8.2 Quenching with Variable Bath Temperature Consider again the quenching process, if the thermal capacitance of the liquid bath is not large, the heat energy transferred from the cube will change the bath temperature, and we will need a model to describe its dynamics The temperature outside the bath is 𝑇0 The convective resistance between the cube and the bath is 𝑅1, and the combined convective/conductive resistance of the container wall and the liquid surface is 𝑅2 The capacitances of the cube and the liquid bath are 𝐶 and 𝐶𝑏 a.Derive a model of the cube temperature and the bath temperature assuming that the bath loses no heat to the surroundings (that is, 𝑅2 = ∞) b.Obtain the model’s characteristic roots & the form of the response HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.107 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.108 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems Solution a.Assume that 𝑇 > 𝑇𝑏 , then the heat flow is out of the cube and into the bath From conservation of energy for the cube 𝑑𝑇 (1) 𝐶 = − (𝑇 − 𝑇𝑏 ) 𝑑𝑡 𝑅1 and for the bath 𝑑𝑇𝑏 𝐶𝑏 = (𝑇 − 𝑇𝑏 ) (2) 𝑑𝑡 𝑅1 Equations (1) and (2) are the desired model Note that the heat flow rate in eq.(2) must have a sign opposite to that in eq.(1) because the heat flow out of the cube must be the same as the heat flow into the bath §8.Dynamic Model of Thermal Systems b.Applying the Laplace transform to equations (1) and (2) with zero initial conditions, we obtain 𝑅1 𝐶𝑠 + 𝑇 𝑠 − 𝑇𝑏 𝑠 = (3) 𝑅1 𝐶𝑏 𝑠 + 𝑇𝑏 𝑠 − 𝑇 𝑠 = (4) Solving eq.(3) for 𝑇𝑏 𝑠 and substituting into eq.(4) gives 𝑅1 𝐶𝑏 𝑠 + 𝑅1 𝐶𝑠 + − 𝑇 𝑠 = from which we obtain 𝑅12 𝐶𝑏 𝐶𝑠 + 𝑅1 𝐶 + 𝐶𝑏 𝑠 = So the characteristic roots are 𝑠=0 𝐶 + 𝐶𝑏 𝑠=− 𝑅1 𝐶𝐶𝑏 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 18 8/25/2013 System Dynamics 7.109 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems Eq.s (3) and (4) are homogeneous, the form of the response 𝑡 𝑡 𝑅1𝐶𝐶𝑏 𝑇 𝑡 = 𝐴1 𝑒 0𝑡 + 𝐵1 𝑒 − 𝜏 = 𝐴1 + 𝐵1 𝑒 − 𝜏 , 𝜏= 𝐶 + 𝐶𝑏 𝑡 𝑡 𝑇𝑏 𝑡 = 𝐴2 𝑒 0𝑡 + 𝐵2 𝑒 − 𝜏 = 𝐴2 + 𝐵2 𝑒 − 𝜏 the constants 𝐴1 , 𝐴2 , 𝐵1 , 𝐵2 depend on the initial conditions The two temperatures become constant after approximately 4𝜏, note that lim 𝑇 = 𝐴1 , lim 𝑇𝑏 = 𝐴2 ⟹ 𝑡 → ∞ : 𝐴1 = 𝐴2 𝑡→∞ 𝑡→∞ Conservation of energy: the initial energy = the final energy 𝐶𝑇 + 𝐶𝑏𝑇𝑏(0) 𝐶𝑇 + 𝐶𝑏𝑇𝑏 = 𝐶𝐴1 + 𝐶𝑏𝐴1 ⟹ 𝐴1 = = 𝐴2 𝐶 + 𝐶𝑏 and 𝑇 = 𝐴1 + 𝐵1 ⟹ 𝐵1 = 𝑇 − 𝐴1 𝑇𝑏 = 𝐴2 + 𝐵2 ⟹ 𝐵2 = 𝑇𝑏 − 𝐴2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.111 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.110 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems 𝑑𝑇 𝐶 = − (𝑇 − 𝑇𝑏 ) 𝑑𝑡 𝑅1 𝑑𝑇𝑏 𝐶𝑏 = (𝑇 − 𝑇𝑏 ) 𝑑𝑡 𝑅1 The thermal model of the quenching with a variable bath temperature and infinite container resistance is analogous to a circuit shown on the figure The voltages 𝑣 and 𝑣𝑏 are analogous to the temperatures 𝑇 and 𝑇𝑏 (𝑅2 → ∞) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.112 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems - Example 7.8.3 Quenching with Heat Loss to the Surroundings Consider the quenching process treated in the previous example a.Derive a model of the cube temperature and the bath temperature assuming 𝑅2 is finite b.Obtain the model’s characteristic roots and the form of the response of 𝑇(𝑡) , assuming that the surrounding temperature 𝑇𝑜 is constant Solution a.Derive a model If 𝑅2 is finite, then we must now account for the heat flow into or out of the container Assume that 𝑇 > 𝑇𝑏 > 𝑇𝑜 : the heat flows from the cube into the bath and then into the surroundings §8.Dynamic Model of Thermal Systems From conservation of energy, we have the desired model 𝑑𝑇 𝐶 = − (𝑇 − 𝑇𝑏 ) (1) 𝑑𝑡 𝑅1 𝑑𝑇𝑏 1 (2) 𝐶𝑏 = 𝑇 − 𝑇𝑏 − (𝑇𝑏 − 𝑇𝑜 ) 𝑑𝑡 𝑅1 𝑅2 b.The model’s characteristic roots and the form of the response of 𝑇(𝑡) Applying the Laplace transform with zero initial conditions 𝑅1 𝐶𝑠 + 𝑇 𝑠 − 𝑇 𝑠 = (3) 𝑅1 𝑅2𝐶𝑏 𝑠 + 𝑅1 + 𝑅2 𝑇𝑏 𝑠 − 𝑅2 𝑇 𝑠 = 𝑅1 𝑇𝑜 (4) Solving eq.(3) for 𝑇𝑏 𝑠 and substituting into eq.(4) 𝑇(𝑠) (5) = 𝑇𝑜 (𝑠) 𝑅1 𝑅2 𝐶𝑏 𝐶𝑠 + [ 𝑅1 + 𝑅2 𝐶 + 𝑅2 𝐶𝑏 ]𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.113 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems The denominator gives the characteristic equation 𝑅1𝑅2 𝐶𝑏 𝐶𝑠 + 𝑅1 + 𝑅2 𝐶 + 𝑅2 𝐶𝑏 𝑠 + = So there will be two nonzero characteristic roots If these roots are real, say 𝑠 = −1/𝜏1 and 𝑠 = −1/𝜏2, and if 𝑇𝑜 is constant, the response will have the form 𝑇 𝑡 = 𝐴𝑒 −𝑡/𝜏1 + 𝐵𝑒 −𝑡/𝜏2 + 𝐷 System Dynamics 7.114 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems 𝑑𝑇 𝐶 = − (𝑇 − 𝑇𝑏 ) 𝑑𝑡 𝑅1 𝑑𝑇𝑏 1 𝐶𝑏 = 𝑇 − 𝑇𝑏 − (𝑇𝑏 − 𝑇𝑜 ) 𝑑𝑡 𝑅1 𝑅2 the constants 𝐴 and 𝐵 depend on the initial conditions Note that lim 𝑇(𝑡) = 𝐷 𝑡→∞ Applying the final value theorem to eq.(5) gives lim 𝑇 = 𝑇𝑜 𝑡→∞ and thus 𝐷 = 𝑇𝑜 We could have also obtained this result through physical reasoning HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien The thermal model of the quenching with a variable bath temperature and finite container resistance is analogous to a circuit shown on the figure The voltages 𝑣 and 𝑣𝑏 are analogous to the temperatures 𝑇 and 𝑇𝑏 (𝑅2 is finite) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 19 8/25/2013 System Dynamics 7.115 Fluid and Thermal Systems System Dynamics 7.116 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems 3.Experimental Determination of Thermal Resistance - Thermal parameters can get from the properties of materials • the mass density 𝜌 • the thermal capacitance 𝐶 • the specific heat 𝑐𝑝 ⟹ • the conductive resistance 𝐿/𝑘𝐴 • the thermal conductivity 𝑘 - However, determination of the convective resistance is difficult to analytically, and we must usually resort to experimentally determined values - In some cases, we may not be able to distinguish between the effects of conduction, convection, and radiation heat transfer, and the resulting model will contain a thermal resistance that expresses the aggregated effects §8.Dynamic Model of Thermal Systems - Example 7.8.4 Temperature Dynamics of a Cooling Object Consider the experiment with a cooling cup of water Water of volume 250𝑚𝑙 in a glass measuring cup was allowed to cool after being heated to 204 0𝐹 The surrounding air temperature was 70 0𝐹 The measured water temperature at various times is given in the table HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 7.117 Fluid and Thermal Systems From that data we derived the following model of the water temperature as a function of time 𝑇 = 129𝑒 −0.0007𝑡 + 70 (1) where 𝑇 is in 0𝐹 and time 𝑡 is in seconds Estimate the thermal resistance of this system System Dynamics 7.118 §8.Dynamic Model of Thermal Systems Solution Model the cup and water as the object shown in the figure Assume that • the convection has mixed the water well so that the water has the same temperature throughout • the air temperature 𝑇𝑜 is constant and select it as the reference temperature Let 𝑅 be the aggregated thermal resistance due to the combined effects of • conduction through the sides and bottom of the cup • convection from the water surface and from the sides of the cup into the air • radiation from the water to the surroundings §8.Dynamic Model of Thermal Systems Solution The heat energy in the water is 𝐸 = 𝜌𝑉𝑐𝑝 (𝑇 − 𝑇𝑜 ) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 7.119 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems The model’s complete response is 𝑡 𝑇 𝑡 = 𝑇 𝑒 −𝑡/𝑅𝐶 + − 𝑒 −𝑅𝐶 𝑇𝑜 = 𝑇 − 𝑇𝑜 𝑒 −𝑡/𝑅𝐶 + 𝑇𝑜 Comparing this with equation (1), we see that 𝑅𝐶 = = 1429𝑠𝑒𝑐 0.0007 1429 0𝐹 ⟹𝑅= 𝐶 𝑓𝑡𝑙𝑏 where 𝐶 = 𝜌𝑉𝑐𝑝 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Fluid and Thermal Systems From conservation of heat energy 𝑑𝐸 = − (𝑇 − 𝑇𝑜 ) 𝑑𝑡 𝑅 or, since 𝜌, 𝑉, 𝑐𝑝 , and 𝑇𝑜 are constant 𝑑𝑇 𝜌𝑉𝑐𝑝 = − (𝑇 − 𝑇𝑜 ) 𝑑𝑡 𝑅 The water’s thermal capacitance is 𝐶 = 𝜌𝑉𝑐𝑝 and the model can be expressed as 𝑑𝑇 𝑅𝐶 + 𝑇 = 𝑇𝑜 (2) 𝑑𝑡 System Dynamics 7.120 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems - Example 7.8.5 Temperature Sensor Response A thermocouple can be used to measure temperature The electrical resistance of the device is a function of the temperature of the surrounding fluid By calibrating the thermocouple and measuring its resistance, we can determine the temperature Because the thermocouple has mass, it has thermal capacitance, and thus its temperature change (and electrical resistance change) will lag behind any change in the fluid temperature Estimate the response time of a thermocouple suddenly immersed in a fluid Model the device as a sphere of copper constantin alloy, whose diameter is 2𝑚𝑚 , and whose properties are 𝜌 = 8920𝑘𝑔/𝑚3 , 𝑘 = 19𝑊/𝑚 0𝐶 , and 𝑐𝑝 = 362𝐽/𝑘𝑔 0𝐶 Take the convection coefficient to be ℎ = 200𝑊/𝑚3 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 20 8/25/2013 System Dynamics 7.121 Fluid and Thermal Systems System Dynamics 7.122 Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems Solution The Biot number 𝑉 (4/3)𝜋𝑟 𝑟 0.001 𝐿= = = = = 3.33 × 10−4 𝐴 4𝜋𝑟 3 ℎ𝐿 200 × 3.33 × 10−4 ⟹ 𝑁𝐵 = = 0.0035 𝑘 19 𝑁𝐵 < 0.1: so we can use a lumped-parameter model Applying conservation of heat energy to the sphere 𝑑𝑇 𝑐𝑝 𝜌𝑉 = ℎ𝐴(𝑇𝑜 − 𝑇) 𝑇: the temperature of the sphere 𝑑𝑡 𝑇0 : the fluid temperature The time constant of this model is 𝑐𝑝 𝜌 𝑉 362 × 8920 𝜏= = 3.33 × 10−4 = 5.38𝑠 ℎ 𝐴 200 System will reach 98% of the fluid temperature within 4𝜏 = 21.5𝑠 §8.Dynamic Model of Thermal Systems - Example 7.8.6 State-Variable Model of Wall Temperature Dynamics Consider the wall shown in cross section in the figure In that example the thermal capacitances of the layers were neglected We now want to develop a model that includes their effects Neglect any convective resistance on the inside and outside surfaces HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.123 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems Solution Lump each thermal mass at the centerline of its respective layer and assign half of the layer’s thermal resistance to the heat flow path on the left and half to the path on the right side of the lumped mass as shown 𝑅1 𝑅1 𝑅2 𝑅2 𝑅3 𝑅3 𝑅4 𝑅4 ,𝑅 = + , 𝑅𝑐 = + , 𝑅𝑑 = + , 𝑅𝑒 = 𝑏 2 2 2 An equivalent electrical circuit is shown 𝑅𝑎 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.125 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems These four equations may be put into state variable form 𝑑𝑻 = 𝑨𝑻 + 𝑩𝒖 𝑑𝑡 where 𝑎11 𝑎12 0 𝑇1 𝑏11 𝑎 𝑎22 𝑎23 𝑇 𝑇 0 𝑻 = , 𝒖 = 𝑖 , 𝑨 = 21 ,𝑩 = 𝑇3 𝑇𝑜 𝑎32 𝑎33 𝑎34 0 𝑇4 𝑏42 0 𝑎43 𝑎44 𝑅𝑎 + 𝑅𝑏 1 𝑅𝑏 + 𝑅𝑐 𝑎11 = − ,𝑎 = ,𝑎 = ,𝑎 = − 𝐶1 𝑅𝑎 𝑅𝑏 12 𝐶1 𝑅𝑏 21 𝐶2 𝑅𝑏 22 𝐶2 𝑅𝑏 𝑅𝑐 1 𝑅𝑐 + 𝑅𝑑 𝑎23 = , 𝑎32 = , 𝑎33 = − , 𝑎34 = 𝐶2 𝑅𝑐 𝐶3 𝑅𝑐 𝐶3 𝑅𝑐 𝑅𝑑 𝐶3 𝑅𝑑 𝑅𝑑 + 𝑅𝑒 1 𝑎43 = , 𝑎44 = − , 𝑏11 = , 𝑏42 = 𝐶4 𝑅𝑑 𝐶4 𝑅𝑑 𝑅𝑒 𝐶1 𝑅𝑎 𝐶4 𝑅𝑒 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 7.124 Nguyen Tan Tien Fluid and Thermal Systems §8.Dynamic Model of Thermal Systems For thermal capacitance 𝐶1 , conservation of energy gives 𝑑𝑇1 𝑇𝑖 − 𝑇1 𝑇1 − 𝑇2 𝐶1 = − 𝑑𝑡 𝑅𝑎 𝑅𝑏 For thermal capacitance 𝐶2 𝑑𝑇2 𝑇1 − 𝑇2 𝑇2 − 𝑇3 𝐶2 = − 𝑑𝑡 𝑅𝑏 𝑅𝑐 For thermal capacitance 𝐶3 𝑑𝑇3 𝑇2 − 𝑇3 𝑇3 − 𝑇4 𝐶3 = − 𝑑𝑡 𝑅𝑐 𝑅𝑑 For thermal capacitance 𝐶4 𝑑𝑇4 𝑇3 − 𝑇4 𝑇4 − 𝑇0 𝐶4 = − 𝑑𝑡 𝑅𝑑 𝑅𝑒 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.126 Nguyen Tan Tien Fluid and Thermal Systems Part Matlab and Simulink Applications HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 21 8/25/2013 System Dynamics 7.127 Fluid and Thermal Systems System Dynamics 7.128 Fluid and Thermal Systems §9.Matlab Applications - Exampe 7.9.1 Liquid Height in a Spherical Tank The figure shows a spherical tank for storing water The tank is filled through a hole in the top and drained through a hole in the bottom The following model for the liquid height ℎ 𝑑ℎ 𝜋 2𝑅ℎ − ℎ2 = −𝐶𝑑 𝐴𝑜 2𝑔ℎ (1) 𝑑𝑡 For water, 𝐶𝑑 = 0.6 is a common value Use Matlab to solve this equation to determine how long it will take for the tank to empty if the initial height is 9𝑓𝑡 The tank has a radius of 𝑅 = 5𝑓𝑡 and has a 1𝑖𝑛 diameter hole in the bottom Use 𝑔 = 32.2𝑓𝑡/𝑠𝑒𝑐 Discuss how to check the solution §9.Matlab Applications Solution With 𝐶𝑑 = 0.6 , 𝑅 = , 𝑔 = 32.2 , and 𝐴𝑜 = 𝜋 × (1/24) × 2, eq.(1) becomes 𝑑ℎ 0.0334 ℎ =− 𝑑𝑡 10ℎ − ℎ2 We can use our physical insight to guard against grossly incorrect results We can also check the preceding expression for 𝑑ℎ/𝑑𝑡 for singularities The denominator does not become zero unless ℎ = or ℎ = 10, which correspond to a completely empty and a completely full tank So we will avoid singularities if < ℎ(0) < 10 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.129 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.130 §9.Matlab Applications Matab function hdot = height(t,h) hdot = -(0.0334*sqrt(h))/(10*h-h^2); [t, h] = ode45(@height, [0, 2475], 9); plot(t,h),xlabel('Time (sec)'),ylabel('Height (ft)') §9.Matlab Applications - Exampe 7.9.2 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.131 Nguyen Tan Tien Fluid and Thermal Systems §9.Matlab Applications Solution The model was developed in Example 7.8.6 The given information shows that the outside temperature is described by 𝑇0 𝑡 = − 15𝑡, ≤ 𝑡 ≤ 3600𝑠 Matlab % htwall.m Heat transfer thru a multilayer wall % Resistance and capacitance data Ra = 0.018; Rb = 2.023; Rc = 2.204; Rd = 0.223; Re = 0.019; C1 = 8720; C2 = 6210; C3 = 6637; C4 = 20800; % Compute the matrix coefficients a11 = -(Ra+Rb)/(C1*Ra*Rb); a12 = 1/(C1*Rb); a21 = 1/(C2*Rb); a22 = -(Rb+Rc)/(C2*Rb*Rc); a23 = 1/(C2*Rc); a32 = 1/(C3*Rc); a33 = -(Rc+Rd)/(C3*Rc*Rd); a34 = 1/(C3*Rd); a43 = 1/(C4*Rd); a44 = -(Rd+Re)/(C4*Rd*Re); b11 = 1/(C1*Ra); b42 = 1/(C4*Re); % Define the A and B matrices A = [a11,a12,0,0; a21,a22,a23,0; 0,a32,a33,a34; 0,0,a43,a44]; B = [b11,0; 0,0; 0,0; 0,b42]; HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Fluid and Thermal Systems Heat Transfer Through a Wall Consider the wall cross section shown in the figure The temperature model was developed in Ex 7.8.6 Use the following values and plot the temperatures versus time for the case where the inside temperature is constant at 𝑇𝑖 = 20 0𝐶 and the outside temperature 𝑇𝑜 decreases linearly from 0𝐶 to −10 0𝐶 in 1ℎ The initial wall temperatures are 10 0𝐶 The resistances and capacitances are 𝑅𝑎 = 0.0180𝐶/𝑊, 𝑅𝑏 = 2.023 0𝐶/𝑊, 𝑅𝑐 = 2.204 0𝐶/𝑊 𝑅𝑑 = 0.2230𝐶/𝑊, 𝑅𝑒 = 0.019 0𝐶/𝑊 𝐶1 = 8720𝐽/ 0𝐶, 𝐶2 = 6210𝐽/ 0𝐶 𝐶3 = 6637𝐽/ 0𝐶, 𝐶2 = 2.08 × 104 𝐽/ 0𝐶 System Dynamics 7.132 Nguyen Tan Tien Fluid and Thermal Systems §9.Matlab Applications Matlab % Define the C and D matrices % The outputs are the four wall temperatures C = eye(4); D = zeros(size(B)); % Create the LTI model sys = ss(A,B,C,D); % Create the time vector for hour (3600 seconds) t = (0:1:3600); % Create the input vector u = [20*ones(size(t));(5-15*ones(size(t)).*t/3600)]; % Compute the forced response [yforced,t] = lsim(sys,u,t); % Compute the free response [yfree,t] = initial(sys,[10,10,10,10],t); % Plot the response along with the outside temperature plot(t,yforced+yfree,t,u(2,:)) % Compute the time constants tau =(-1./real(eig(A)))/60 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 22 8/25/2013 System Dynamics 7.133 Fluid and Thermal Systems §9.Matlab Applications The results System Dynamics 7.134 Fluid and Thermal Systems §10 Simulink Applications - One potential disadvantage of a graphical interface such as Simulink is that to simulate a complex system, the diagram can become rather large, and therefore somewhat cumbersome - Simulink, however, provides for the creation of subsystem blocks, which play a role analogous to subprograms in a programming language - A subsystem block is actually a Simulink program represented by a single block A subsystem block, once created, can be used in other Simulink programs HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.135 Nguyen Tan Tien Fluid and Thermal Systems §10 Simulink Applications 1.Subsystem Blocks HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.136 Nguyen Tan Tien Fluid and Thermal Systems §10 Simulink Applications - Create the following simulink model - The resistances are nonlinear and obey the following signed-squareroot relation 𝑢/𝑅 𝑢 ≥ 𝑞 = 𝑆𝑆𝑅(∆𝑝) = 𝑅 − 𝑢/𝑅 𝑢 < 𝑞: the mass flow rate 𝑅: the resistance 𝑝: the pressure difference across the resistance - The model of the system in the figure is the following 𝑑ℎ 1 𝜌𝐴 = 𝑞 + 𝑆𝑆𝑅 𝑝𝑙 − 𝑝 − 𝑆𝑆𝑅(𝑝 − 𝑝𝑟 ) 𝑑𝑡 𝑅𝑙 𝑅𝑟 𝑝𝑙 , 𝑝𝑟 : the gage pressures at the left and right-hand sides 𝐴: the bottom area 𝑞: a mass flow rate 𝑝 = 𝜌𝑔ℎ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.137 Nguyen Tan Tien Fluid and Thermal Systems §10 Simulink Applications - Suppose we want to create a simulation of the system shown in the figure, where the mass inflow rate 𝑞1 is a step function Create the Simulink model Run the model with 𝐴1 = 2, 𝐴2 = 𝜌 = 1.94, 𝑔 = 32.2 𝑅1 = 20, 𝑅2 = 50 𝑞1 = 20, ℎ10 = 1, ℎ20 = 10 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien - Create Subsystem from the Edit menu HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 7.138 Nguyen Tan Tien Fluid and Thermal Systems §10 Simulink Applications 2.Simulation of Thermal Systems - Example 7.10.1 Thermostatic Control of Temperature a.Develop a Simulink model of a thermostatic control system in which the temperature model is 𝑑𝑇 𝑅𝐶 + 𝑇 = 𝑅𝑞 + 𝑇𝑎 (𝑡) 𝑑𝑡 𝑇: the room air temperature in 0𝐹; 𝑇𝑎 : the ambient (outside) air temperature in 0𝐹; 𝑡: time, is measured in hours; 𝑞: the input from the heating system in 𝑓𝑡𝑙𝑏/ℎ𝑟; 𝑅: the thermal resistance; 𝐶: the thermal capacitance The thermostat switches 𝑞 on at the value 𝑞𝑚𝑎𝑥 whenever the temperature drops below 690 , and switches 𝑞 to 𝑞 = whenever the temperature is above 710 The value of 𝑞𝑚𝑎𝑥 indicates the heat output of the heating system HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 23 8/25/2013 System Dynamics 7.139 Fluid and Thermal Systems System Dynamics 7.140 §10 Simulink Applications Run the simulation for the case where 𝑇(0) = 700 and 𝑇𝑎 𝑡 = 50 + 10sin(𝜋𝑡/12) Use the values 𝑅 = × 10−5 0𝐹ℎ𝑟/𝑙𝑏𝑓𝑡 , 𝐶 = × 104 𝑙𝑏𝑓𝑡/ 0𝐹 Plot the temperatures 𝑇 and 𝑇𝑎 versus 𝑡 on the same graph, for ≤ 𝑡 24ℎ𝑟 Do this for two cases • 𝑞𝑚𝑎𝑥 = × 105𝑙𝑏𝑓𝑡/ℎ𝑟 • 𝑞𝑚𝑎𝑥 = × 105𝑙𝑏𝑓𝑡/ℎ𝑟 Investigate the effectiveness of each case b.The integral of 𝑞 over time is the energy used Plot 𝑞 𝑑𝑡 versus 𝑡 and determine how much energy is used in 24ℎ𝑟 for the case where 𝑞𝑚𝑎𝑥 = × 105𝑙𝑏𝑓𝑡/ℎ𝑟 §10 Simulink Applications Solution The model can be arranged as follows 𝑑𝑇 = 𝑅𝑞 + 𝑇𝑎 𝑡 − 𝑇 𝑑𝑡 𝑅𝐶 The Simulink model is shown in the figure HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Fluid and Thermal Systems Nguyen Tan Tien 24 ... 7.51 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 7.50 Fluid and Thermal Systems §4.Dynamic Models of Hydraulic Systems The rate at which the piston pushes fluid out of the cylinder... System Dynamics 7.67 Fluid and Thermal Systems System Dynamics 7.68 Fluid and Thermal Systems §5.Pneumatic Systems 1.Pneumatics Capacitance Fluid capacitance