Mechanical Engineering Systems 2008 Part 1 pdf

[...]... Example 2.2 .1 A perfect gas in an engine cylinder at the start of compression has a volume of 0. 01 m3, a temperature of 20°C and a pressure of 2 bar The piston rises to compress the gas, and at top dead centre the volume is 0.004 m3 and the pressure is 15 bar Find the temperature See Figure 2.2 .1 p1 = 2 bar p2 = 15 bar V1 = 0. 01 m 3 V2 = 0.004 m3 T1 = 20 + 273 = 293 K p1 V1 = T1 2 × 0. 01 293 T2 = T2... check on all formulas you use Example 2 .1. 1 The boiler in a canteen contains 6 kg of water at 20°C How much heat energy is required to raise the temperature of the water to 10 0°C? Specific heat of water = 419 0 J/kgK Q = m.c ␦ T Q = 6 × 419 0 × (10 0 – 20) Q = 2 011 200 J = 2 011 .2 kJ Thermodynamics 9 Example 2 .1. 2 How many kilograms of copper can be raised from 15 °C to 60°C by the absorption of 80 kJ... the ice into water, to raise the water to 10 0°C, and to change the water at 10 0°C into steam To heat the ice, Q1 = m.cice ␦T Q1 = 4 × 2.04 × 10 = 81. 6 kJ To change the ice at 0°C into water at 0°C, Q2 = m × latent heat of fusionice Q2 = 4 × 335 = 13 40 kJ To heat the water to 10 0°C, Q3 = m.cwater ␦T Q3 = 4 × 4.2 × 10 0 = 16 80 kJ To change the water into steam at 10 0°C, Q4 = m × latent heat of vaporizationwater... 16 80 kJ To change the water into steam at 10 0°C, Q4 = m × latent heat of vaporizationwater Q4 = 4 × 2256.7 = 9026.8 kJ Total heat energy = Q1 + Q2 + Q3 + Q4 = 81. 6 + 13 40 + 16 80 + 9026.8 = 12 12 8.4 kJ Thermodynamics 13 In this example, we could provide the 12 12 8.4 kJ very quickly with a large kW heater, or much more slowly with a small kW heater Neglecting losses, the result would be the same, i.e... value of its volume divided by its temperature will always be the same Hence, V = constant T or, V1 T1 = V2 T2 = V3 T3 , etc These laws can be remembered separately, but from an engineer’s point of view they are better combined to give a single very useful expression This is, p1 V1 T1 = constant or, p1 V1 T1 = p2 V2 T2 = p3 V3 T3 , etc It is important to remember that this expression is always valid,... = 390 kJ/kgK Q = m.c ␦ T 80 000 = m × 390 × (60 – 15 ) m = 80 000 390 × 45 = 4.56 kg Problems 2 .1. 1 (1) (2) Calculate the heat energy required to raise the temperature of 30 kg of copper from 12 °C to 70°C Assume the specific heat of copper to be 390 J/kgK A body of mass 10 00 kg absorbs 90 000 kJ of heat energy If the temperature of the body rises by 18 0°C, calculate the specific heat of the material... specific heat at constant pressure, cp , and the specific heat at constant volume, cv See Figure 2 .1. 1 Figure 2 .1. 1 gases Specific heats of Specific heat at constant pressure, cp This is the quantity of heat energy supplied to raise 1 kg of the gas through 1 C or K, while the gas is at constant pressure Think of 1 kg of gas trapped in a cylinder As heat energy is added, the pressure will rise If the piston... this is an example of the equivalence of heat energy and work energy Example 2 .1. 3 1. 5 kg of gas at 20°C is contained in a cylinder and heated to 75°C while the volume remains constant Calculate the heat energy supplied if cv = 700 J/kgK Q = m.cv.␦T Q = 1. 5 × 700 × (75 – 20) = 57 750 J = 57.75 kJ Thermodynamics 11 Example 2 .1. 4 A gas with a specific heat at constant pressure, cp = 900 J/kgK, is supplied... (Pa), is tiny It is equivalent to the weight of one average eating apple spread out over an engineering drawing board For practical purposes it is necessary to come up with a large number of pascals under a single name The unit often employed is the bar, which is equal to one hundred thousand pascals (1 bar = 1 × 10 5 Pa) The name this time comes from the same word that is used as the basis of ‘barometer’... say, 2 kW, it means that in 1 second it provides 2000 joules of heat energy Remember that power is the rate at which the energy is delivered, i.e work, or heat energy delivered, divided by time taken Let us say the kettle contains 2 kg of water and is at a room temperature of 18 °C, and the kettle is 2 kW Specific heat of water = 4.2 kJ/kgK Q = m.c ␦T Q = 2 × 4.2 × (10 0 – 18 ) = 688.8 kJ This is the heat . 89 2.8 Heat transfer 10 1 3 Fluid mechanics 11 2 3 .1 Hydrostatics – fluids at rest 11 3 3.2 Hydrodynamics – fluids in motion 13 5 4 Dynamics 16 9 4 .1 Introduction to kinematics 17 0 4.2 Dynamics – analysis. 7506 5 213 6 Composition by Genesis Typesetting, Laser Quay, Rochester, Kent Printed and bound in Great Britain Contents Series Preface vii 1 Introduction: the basis of engineering 1 1 .1 Real engineering.