ME 2121 yunus a cengel, michael a boles thermodynamics solution

1-1 Solutions Manual for Thermodynamics: An Engineering Approach Seventh Edition Yunus A Cengel, Michael A Boles McGraw-Hill, 2011 Chapter INTRODUCTION AND BASIC CONCEPTS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc (“McGraw-Hill”) and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-2 Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed There is no creation of energy, and thus no violation of the conservation of energy principle 1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics Therefore, this cannot happen Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill 1-3C There is no truth to his claim It violates the second law of thermodynamics PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-3 Mass, Force, and Units 1-4C The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system You should get into the habit of never writing the unit “lb”, but always use either “lbm” or “lbf” as appropriate since the two units have different dimensions 1-5C In this unit, the word light refers to the speed of light The light-year unit is then the product of a velocity and time Hence, this product forms a distance dimension and unit 1-6C There is no acceleration, thus the net force is zero in both cases 1-7E The weight of a man on earth is given His weight on the moon is to be determined Analysis Applying Newton's second law to the weight force gives W = mg ⎯ ⎯→ m = 210 lbf W = g 32.10 ft/s ⎛ 32.174 lbm ⋅ ft/s ⎜ ⎜ lbf ⎝ ⎞ ⎟ = 210.5 lbm ⎟ ⎠ Mass is invariant and the man will have the same mass on the moon Then, his weight on the moon will be lbf ⎞ ⎛ W = mg = (210.5 lbm)(5.47 ft/s )⎜ ⎟ = 35.8 lbf ⎝ 32.174 lbm ⋅ ft/s ⎠ 1-8 The interior dimensions of a room are given The mass and weight of the air in the room are to be determined Assumptions The density of air is constant throughout the room Properties The density of air is given to be ρ = 1.16 kg/m3 Analysis The mass of the air in the room is 3 m = ρV = (1.16 kg/m )(6 × × m ) = 334.1 kg ROOM AIR 6X6X8 m3 Thus, ⎛ 1N W = mg = (334.1 kg)(9.81 m/s )⎜ ⎜ kg ⋅ m/s ⎝ ⎞ ⎟ = 3277 N ⎟ ⎠ PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-4 1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude The height at which the weight of a body will decrease by 0.5% is to be determined z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 × 10−6 z ) In our case, W = 0.995W s = 0.995mg s = 0.995(m)(9.81) Substituting, 0.995(9.81) = (9.81 − 3.32 × 10 −6 z) ⎯ ⎯→ z = 14,774 m ≅ 14,770 m Sea level 1-10 The mass of an object is given Its weight is to be determined Analysis Applying Newton's second law, the weight is determined to be W = mg = (200 kg)(9.6 m/s ) = 1920 N 1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units Analysis Applying Newton's second law, the weight is determined in various units to be ⎛ kJ/kg ⋅ K ⎞ ⎟⎟ = 1.005 kJ/kg ⋅ K c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ kJ/kg ⋅ °C ⎠ ⎛ 1000 J ⎞⎛ kg ⎞ ⎟⎟ = 1.005 J/g ⋅ °C c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟⎜⎜ ⎝ kJ ⎠⎝ 1000 g ⎠ ⎛ kcal ⎞ c p = (1.005 kJ/kg ⋅ °C)⎜ ⎟ = 0.240 kcal/kg ⋅ °C ⎝ 4.1868 kJ ⎠ ⎛ Btu/lbm ⋅ °F ⎞ ⎟⎟ = 0.240 Btu/lbm ⋅ °F c p = (1.005 kJ/kg ⋅ °C)⎜⎜ ⎝ 4.1868 kJ/kg ⋅ °C ⎠ PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-5 1-12 A rock is thrown upward with a specified force The acceleration of the rock is to be determined Analysis The weight of the rock is ⎛ 1N W = mg = (3 kg)(9.79 m/s )⎜ ⎜ kg ⋅ m/s ⎝ ⎞ ⎟ = 29.37 N ⎟ ⎠ Then the net force that acts on the rock is Fnet = Fup − Fdown = 200 − 29.37 = 170.6 N Stone From the Newton's second law, the acceleration of the rock becomes a= F 170.6 N ⎛⎜ kg ⋅ m/s = m kg ⎜⎝ N ⎞ ⎟ = 56.9 m/s ⎟ ⎠ PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-6 1-13 Problem 1-12 is reconsidered The entire EES solution is to be printed out, including the numerical results with proper units Analysis The problem is solved using EES, and the solution is given below "The weight of the rock is" W=m*g m=3 [kg] g=9.79 [m/s2] "The force balance on the rock yields the net force acting on the rock as" F_up=200 [N] F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=m*a "To Run the program, press F2 or select Solve from the Calculate menu." SOLUTION a=56.88 [m/s^2] F_down=29.37 [N] F_net=170.6 [N] F_up=200 [N] g=9.79 [m/s2] m=3 [kg] W=29.37 [N] 200 160 a [m/s2] 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21 a [m/s ] m [kg] 10 120 80 40 m [kg] 10 PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-7 1-14 During an analysis, a relation with inconsistent units is obtained A correction is to be found, and the probable cause of the error is to be determined Analysis The two terms on the right-hand side of the equation E = 25 kJ + kJ/kg not have the same units, and therefore they cannot be added to obtain the total energy Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage 1-15 A resistance heater is used to heat water to desired temperature The amount of electric energy used in kWh and kJ are to be determined Analysis The resistance heater consumes electric energy at a rate of kW or kJ/s Then the total amount of electric energy used in hours becomes Total energy = (Energy per unit time)(Time interval) = (4 kW)(2 h) = kWh Noting that kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (8 kWh)(3600 kJ/kWh) = 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy 1-16 A gas tank is being filled with gasoline at a specified flow rate Based on unit considerations alone, a relation is to be obtained for the filling time Assumptions Gasoline is an incompressible substance and the flow rate is constant Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline Also, we know that the unit of time is ‘seconds’ Therefore, the independent quantities should be arranged such that we end up with the unit of seconds Putting the given information into perspective, we have t [s] ↔ V [L], and V& [L/s} It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate Therefore, the desired relation is t= V V& Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-8 1-17 A pool is to be filled with water using a hose Based on unit considerations, a relation is to be obtained for the volume of the pool Assumptions Water is an incompressible substance and the average flow velocity is constant Analysis The pool volume depends on the filling time, the cross-sectional area which depends on hose diameter, and flow velocity Also, we know that the unit of volume is m3 Therefore, the independent quantities should be arranged such that we end up with the unit of seconds Putting the given information into perspective, we have V [m3] is a function of t [s], D [m], and V [m/s} It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square of D Therefore, the desired relation is V = CD2Vt where the constant of proportionality is obtained for a round hose, namely, C =π/4 so that V = (πD2/4)Vt Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach 1-18 It is to be shown that the power needed to accelerate a car is proportional to the mass and the square of the velocity of the car, and inversely proportional to the time interval Assumptions The car is initially at rest Analysis The power needed for acceleration depends on the mass, velocity change, and time interval Also, the unit of power W& is watt, W, which is equivalent to W = J/s = N⋅m/s = (kg⋅m/s2)m/s = kg⋅m2/s3 Therefore, the independent quantities should be arranged such that we end up with the unit kg⋅m2/s3 for power Putting the given information into perspective, we have W& [ kg⋅m2/s3] is a function of m [kg], V [m/s], and t [s] It is obvious that the only way to end up with the unit “kg⋅m2/s3” for power is to multiply mass with the square of the velocity and divide by time Therefore, the desired relation is W& is proportional to mV / t or, W& = CmV / t where C is the dimensionless constant of proportionality (whose value is ½ in this case) Discussion Note that this approach cannot determine the numerical value of the dimensionless numbers involved PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-9 Systems, Properties, State, and Processes 1-19C This system is a region of space or open system in that mass such as air and food can cross its control boundary The system can also interact with the surroundings by exchanging heat and work across its control boundary By tracking these interactions, we can determine the energy conversion characteristics of this system 1-20C The system is taken as the air contained in the piston-cylinder device This system is a closed or fixed mass system since no mass enters or leaves it 1-21C Any portion of the atmosphere which contains the ozone layer will work as an open system to study this problem Once a portion of the atmosphere is selected, we must solve the practical problem of determining the interactions that occur at the control surfaces which surround the system's control volume 1-22C Intensive properties not depend on the size (extent) of the system but extensive properties 1-23C If we were to divide the system into smaller portions, the weight of each portion would also be smaller Hence, the weight is an extensive property 1-24C If we were to divide this system in half, both the volume and the number of moles contained in each half would be one-half that of the original system The molar specific volume of the original system is v = V N and the molar specific volume of one of the smaller systems is v = V/ V = N /2 N which is the same as that of the original system The molar specific volume is then an intensive property 1-25C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not However, there should be no unbalanced pressure forces present The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight 1-26C A process during which a system remains almost in equilibrium at all times is called a quasi-equilibrium process Many engineering processes can be approximated as being quasi-equilibrium The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-10 1-27C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric 1-28C The state of a simple compressible system is completely specified by two independent, intensive properties 1-29C The pressure and temperature of the water are normally used to describe the state Chemical composition, surface tension coefficient, and other properties may be required in some cases As the water cools, its pressure remains fixed This cooling process is then an isobaric process 1- 30C When analyzing the acceleration of gases as they flow through a nozzle, the proper choice for the system is the volume within the nozzle, bounded by the entire inner surface of the nozzle and the inlet and outlet cross-sections This is a control volume since mass crosses the boundary 1-31C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-105 17-147 Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-33 for methane Properties The specific heat ratio is given to be k = 1.3 for methane Analysis The normal shock relations listed below are expressed in EES and the results are tabulated Ma = P2 + kMa 12 2kMa 12 − k + = = P1 + kMa 22 k +1 (k − 1)Ma 12 + 2kMa 12 − k +1 (k + 1)Ma 12 V ρ P2 / P1 = = = , ρ T2 / T1 + (k − 1)Ma 12 V T2 + Ma 12 (k − 1) = T1 + Ma 22 (k − 1) k +1 P02 Ma ⎡1 + Ma 22 (k − 1) / ⎤ 2( k −1) = ⎢ ⎥ P01 Ma ⎢⎣1 + Ma 12 (k − 1) / ⎥⎦ P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 + kMa 22 Methane: k=1.3 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2) Ma1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 Ma2 1.0000 0.6942 0.5629 0.4929 0.4511 0.4241 0.4058 0.3927 0.3832 0.3760 0.3704 0.3660 0.3625 0.3596 0.3573 0.3553 0.3536 0.3522 0.3510 P2/P1 1.0000 2.4130 4.3913 6.9348 10.0435 13.7174 17.9565 22.7609 28.1304 34.0652 40.5652 47.6304 55.2609 63.4565 72.2174 81.5435 91.4348 101.8913 112.9130 ρ2/ρ1 1.0000 1.9346 2.8750 3.7097 4.4043 4.9648 5.4118 5.7678 6.0526 6.2822 6.4688 6.6218 6.7485 6.8543 6.9434 7.0190 7.0837 7.1393 7.1875 T2/T1 1.0000 1.2473 1.5274 1.8694 2.2804 2.7630 3.3181 3.9462 4.6476 5.4225 6.2710 7.1930 8.1886 9.2579 10.4009 11.6175 12.9079 14.2719 15.7096 P02/P01 0.9261 0.7006 0.461 0.2822 0.1677 0.09933 0.05939 0.03613 0.02243 0.01422 0.009218 0.006098 0.004114 0.002827 0.001977 0.001404 0.001012 0.000740 P02/P1 1.8324 3.2654 5.3700 8.0983 11.4409 15.3948 19.9589 25.1325 30.9155 37.3076 44.3087 51.9188 60.1379 68.9658 78.4027 88.4485 99.1032 110.367 122.239 PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-106 17-148 Air flowing at a supersonic velocity in a duct is accelerated by cooling For a specified exit Mach number, the rate of heat transfer is to be determined Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid Q& P01 = 240 kPa T01 = 350 K Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a) Ma2 = Ma1 = 1.2 Analysis Knowing stagnation properties, the static properties are determined to be ⎛ k −1 ⎞ Ma 12 ⎟ T1 = T01 ⎜1 + ⎝ ⎠ −1 ⎛ k −1 ⎞ Ma 12 ⎟ P1 = P01 ⎜1 + ⎝ ⎠ ρ1 = ⎛ 1.4 - ⎞ = (350 K)⎜1 + 1.2 ⎟ ⎝ ⎠ − k /( k −1) −1 = 271.7 K ⎛ 1.4 - ⎞ = (240 kPa)⎜1 + ⎟ ⎝ ⎠ − / = 98.97 kPa P1 98.97 kPa = = 1.269 kg/m RT1 (0.287 kJ/kgK)(271.7 K) Then the inlet velocity and the mass flow rate become ⎛ 1000 m / s c1 = kRT1 = (1.4 )(0.287 kJ/kg ⋅ K)(271.7 K)⎜⎜ ⎝ kJ/kg ⎞ ⎟ = 330.4 m/s ⎟ ⎠ V1 = Ma 1c1 = 1.2(330.4 m/s) = 396.5 m/s m& air = ρ1 Ac1V1 = (1.269 kg/m )[π (0.20 m) / 4](330.4 m/s) = 15.81 kg/s The Rayleigh flow functions T0/T0* corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 1.8: T01/T0* = 0.9787 Ma2 = 2: T02/T0* = 0.7934 Then the exit stagnation temperature is determined to be T0 T02 / T0* 0.7934 = = = 0.8107 → T02 = 0.8107T01 = 0.8107(350 K ) = 283.7 K T0 T01 / T0* 0.9787 Finally, the rate of heat transfer is Q& = m& air c p (T02 − T01 ) = (15.81 kg/s)(1.005 kJ/kg ⋅ K )(283.7 − 350) K = -1053 kW Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated Also, it can be shown that the thermodynamic temperature drops to 158 K at the exit, which is extremely low Therefore, the duct may need to be heavily insulated to maintain indicated flow conditions PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-107 17-149 Air flowing at a subsonic velocity in a duct is accelerated by heating The highest rate of heat transfer without affecting the inlet conditions is to be determined Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid Inlet conditions (and thus the mass flow rate) remain constant Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a) Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = The inlet density and stagnation temperature are ρ1 = Q& P1 = 550 kPa T1 = 450 K P1 550 kPa = = 4.259 kg/m RT1 (0.287 kJ/kgK)(450 K) Ma2 = Ma1 = 0.3 ⎛ k −1 ⎞ ⎛ 1.4 - ⎞ Ma 12 ⎟ = (450 K)⎜1 + T01 = T1 ⎜1 + 0.3 ⎟ = 458.1 K 2 ⎝ ⎠ ⎝ ⎠ Then the inlet velocity and the mass flow rate become ⎛ 1000 m / s c1 = kRT1 = (1.4)(0.287 kJ/kg ⋅ K)(450 K)⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 425.2 m/s ⎟ ⎠ V1 = Ma 1c1 = 0.3(425.2 m/s) = 127.6 m/s m& air = ρ1 Ac1V1 = (4.259 kg/m )(0.08 × 0.08 m )(127.6 m/s) = 3.477 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = (since Ma2 = 1) T01 T0* = (k + 1)Ma 12 [2 + (k − 1)Ma 12 ] (1 + kMa 12 ) = (1.4 + 1)0.3 [2 + (1.4 − 1)0.3 ] (1 + 1.4 × 0.3 ) = 0.3469 Ther efore, T0 T02 / T0* = = → T02 = T01 / 0.3469 = (458.1 K ) / 0.3469 = 1320.7 K T0 T01 / T0* 0.3469 Then the rate of heat transfer becomes Q& = m& air c p (T02 − T01 ) = (3.477 kg/s)(1.005 kJ/kg ⋅ K)(1320.7 − 458.1) K = 3014 kW Discussion It can also be shown that T2 = 1101 K, which is the highest thermodynamic temperature that can be attained under stated conditions If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease We can also solve this problem using the Rayleigh function values listed in Table A-34 PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-108 17-150 Helium flowing at a subsonic velocity in a duct is accelerated by heating The highest rate of heat transfer without affecting the inlet conditions is to be determined Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid Inlet conditions (and thus the mass flow rate) remain constant Properties We take the properties of helium to be k = 1.667, cp = 5.193 kJ/kg⋅K, and R = 2.077 kJ/kg⋅K (Table A-2a) Q& Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = The inlet density and stagnation temperature are P 550 kPa ρ1 = = = 0.5885 kg/m RT1 (2.077 kJ/kgK)(450 K) P1 = 550 kPa T1 = 450 K Ma2 = Ma1 = 0.3 ⎛ k −1 ⎞ ⎛ 1.667 - ⎞ Ma 12 ⎟ = (450 K)⎜1 + T01 = T1 ⎜1 + 0.3 ⎟ = 463.5 K 2 ⎝ ⎠ ⎝ ⎠ Then the inlet velocity and the mass flow rate become ⎛ 1000 m / s c1 = kRT1 = (1.667)(2.077 kJ/kg ⋅ K)(450 K)⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 1248 m/s ⎟ ⎠ V1 = Ma 1c1 = 0.3(1248 m/s) = 374.5 m/s m& air = ρ1 Ac1V1 = (0.5885 kg/m )(0.08 × 0.08 m )(374.5 m/s) = 1.410 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = (since Ma2 = 1) T01 T0* Therefore, = (k + 1)Ma 12 [2 + (k − 1)Ma 12 ] (1 + kMa 12 ) = (1.667 + 1)0.33 [2 + (1.667 − 1)0.3 ] (1 + 1.667 × 0.3 ) = 0.3739 T0 T02 / T0* = = → T02 = T01 / 0.3739 = (463.5 K ) / 0.3739 = 1239.8 K T0 T01 / T0* 0.3739 Then the rate of heat transfer becomes Q& = m& air c p (T02 − T01 ) = (1.410 kg/s)(5.193 kJ/kg ⋅ K)(1239.8 − 463.5) K = 5685 kW Discussion It can also be shown that T2 = 930 K, which is the highest thermodynamic temperature that can be attained under stated conditions If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease Also, in the solution of this problem, we cannot use the values of Table A-34 since they are based on k = 1.4 PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-109 17-151 Air flowing at a subsonic velocity in a duct is accelerated by heating For a specified exit Mach number, the heat transfer for a specified exit Mach number as well as the maximum heat transfer are to be determined Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid Inlet conditions (and thus the mass flow rate) remain constant Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a) Analysis The inlet Mach number and stagnation temperature are ⎛ 1000 m / s c1 = kRT1 = (1.4 )(0.287 kJ/kg ⋅ K)(400 K)⎜⎜ ⎝ kJ/kg Ma = ⎞ ⎟ = 400.9 m/s ⎟ ⎠ V1 100 m/s = = 0.2494 c1 400.9 m/s k −1 ⎞ ⎛ Ma12 ⎟ T01 = T1⎜1 + ⎠ ⎝ ⎞ ⎛ 1.4 - 0.2494 ⎟ = (400 K)⎜1 + ⎠ ⎝ = 405.0 K q P1 = 35 kPa T1 = 400 K Ma2 = 0.8 V1 = 100 m/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 0.2494: T01/T* = 0.2559 Ma2 = 0.8: T02/T* = 0.9639 Then the exit stagnation temperature and the heat transfer are determined to be T0 T02 / T * 0.9639 = = = 3.7667 → T0 = 3.7667T01 = 3.7667(405.0 K ) = 1526 K T0 T01 / T * 0.2559 q = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(1526 − 405) K = 1126 kJ/kg Maximum heat transfer will occur when the flow is choked, and thus Ma2 = and thus T02/T* = Then, T0 T02 / T * = = → T0 = T01 / 0.2559 = 405.0 K ) / 0.2559 = 1583 K T0 T01 / T * 0.2559 q max = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(1583 − 405) K = 1184 kJ/kg Discussion This is the maximum heat that can be transferred to the gas without affecting the mass flow rate If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-110 17-152 Air flowing at sonic conditions in a duct is accelerated by cooling For a specified exit Mach number, the amount of heat transfer per unit mass is to be determined Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a) Analysis Noting that Ma1 = 1, the inlet stagnation temperature is q ⎛ k −1 ⎞ T01 = T1 ⎜1 + Ma 12 ⎟ ⎝ ⎠ ⎛ 1.4 - ⎞ = (500 K)⎜1 + ⎟ = 600 K ⎝ ⎠ P01 = 420 kPa T01 = 500 K The Rayleigh flow functions T0/T0* corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 1: T01/T0* = Ma2 = 1.6: T02/T0* = 0.8842 Ma2 = 1.6 Ma1 = Then the exit stagnation temperature and heat transfer are determined to be T0 T02 / T0* 0.8842 = = = 0.8842 T0 T01 / T0* → T02 = 0.8842T01 = 0.8842(600 K ) = 530.5 K q = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(530.5 − 600) K = - 69.8 kJ/kg Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated Also, it can be shown that the thermodynamic temperature drops to 351 K at the exit PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-111 17-153 Saturated steam enters a converging-diverging nozzle with a low velocity The throat area, exit velocity, mass flow rate, and exit Mach number are to be determined for isentropic and 90 percent efficient nozzle cases Assumptions Flow through the nozzle is steady and one-dimensional The nozzle is adiabatic Analysis (a) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible Thus h10 = h1 At the inlet, Steam Vi ≈ h1 = (h f + x1h fg ) @1.75 MPa = 878.16 + 0.90 × 1917.1 = 2603.5 kJ/kg s1 = ( s f + x1 s fg ) @1.75 MPa = 2.3844 + 0.90 × 4.0033 = 5.9874 kJ/kg ⋅ K t a) ηN = 100% b) ηN = 92% At the exit, P2 = 1.2 MPa and s2 = s2s = s1 = 5.9874 kJ/kg·K Thus, s = s f + x s fg → 5.9874 = 2.2159 + x (4.3058) → x = 0.8759 h2 = h f + x h fg = 798.33 + 0.8759 × 1985.4 = 2537.4 kJ/kg v = v f + x 2v fg = 0.001138 + 0.8759 × (0.16326 − 0.001138) = 0.14314 m / kg Then the exit velocity is determined from the steady-flow energy balance to be h1 + V12 V2 V − V12 = h2 + → = h2 − h1 + 2 2 Solving for V2, ⎛ 1000 m / s V2 = 2(h1 − h2 ) = 2(2603.5 − 2537.4)kJ/kg⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 363.7 m/s ⎟ ⎠ The mass flow rate is determined from m& = v2 A2V2 = 0.14314 m / kg (25 × 10 − m )(363.7 m/s) = 6.35 kg/s The velocity of sound at the exit of the nozzle is determined from 1/ ⎛ ∂P ⎞ c=⎜ ⎟ ⎝ ∂r ⎠ s 1/ ⎛ ∆P ⎞ ⎟⎟ ≅ ⎜⎜ ⎝ ∆(1 / v ) ⎠ s The specific volume of steam at s2 = 5.9874 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1547 and 0.1333 m3/kg Substituting, c2 = (1300 − 1100) kPa ⎛ 1000 m / s ⎜ ⎜ ⎞ ⎛ ⎝ kPa ⋅ m − ⎜ ⎟ kg/m ⎝ 0.1333 0.1547 ⎠ ⎞ ⎟ = 438.9 m/s ⎟ ⎠ Then the exit Mach number becomes Ma = V2 363.7 m/s = = 0.829 c 438.9 m/s The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576 × P01 = 0.576 × 1.75 = 1.008 MPa Then at the throat, Pt = 1.008 MPa and s t = s1 = 5.9874 kJ/kg ⋅ K Thus, PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-112 ht = 2507.7 kJ/kg v t = 0.1672 m / kg Then the throat velocity is determined from the steady-flow energy balance, Ê0 h1 + V12 = ht + Vt V2 → = ht − h1 + t 2 Solving for Vt, ⎛ 1000 m / s Vt = 2(h1 − ht ) = 2(2603.5 − 2507.7)kJ/kg⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 437.7 m/s ⎟ ⎠ Thus the throat area is m& v t (6.35 kg/s)(0.1672 m / kg) = = 24.26 × 10 − m = 24.26 cm 437.7 m/s Vt At = (b) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible Thus h10 = h1 At the inlet, h1 = (h f + x1 h fg ) @1.75 MPa = 878.16 + 0.90 × 1917.1 = 2603.5 kJ/kg s1 = ( s f + x1 s fg ) @1.75 MPa = 2.3844 + 0.90 × 4.0033 = 5.9874 kJ/kg ⋅ K Vi ≈ At state 2s, P2 = 1.2 MPa and s2 = s2s = s1 = 5.9874 kJ/kg·K Thus, s s = s f + x s fg → 5.9874 = 2.2159 + x (4.3058) → x s = 0.8759 Steam t a) ηN = 100% b) ηN = 92% h2 s = h f + x h fg = 798.33 + 0.8759 × 1985.4 = 2537.4 kJ/kg The enthalpy of steam at the actual exit state is determined from ηN = h01 − h2 2603.5 − h2 ⎯ ⎯→ 0.92 = ⎯ ⎯→ h2 = 2542.7 kJ/kg 2603.4 − 2537.4 h01 − h2 s Therefore at the exit, P2 = 1.2 MPa and h2 = 2542.7 kJ/kg·K Thus, h2 = h f + x h fg ⎯ ⎯→ 2542.7 = 798.33 + x (1985.4) ⎯ ⎯→ x = 0.8786 s = s f + x s fg = 2.2159 + 0.8786 × 4.3058 = 5.9989 v = v f + x 2v fg = 0.001138 + 0.8786 × (0.16326 − 0.001138) = 0.1436 kJ / kg Then the exit velocity is determined from the steady-flow energy balance to be h1 + V12 V2 V − V12 = h2 + → = h2 − h1 + 2 2 Solving for V2, ⎛ 1000 m / s V2 = 2(h1 − h2 ) = 2(2603.5 − 2542.7)kJ/kg⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 348.9 m/s ⎟ ⎠ The mass flow rate is determined from m& = v2 A2V2 = 0.1436 m / kg (25 × 10 − m )(348.9 m/s) = 6.07 kg/s The velocity of sound at the exit of the nozzle is determined from 1/ ⎛ ∂P ⎞ c = ⎜⎜ ⎟⎟ ⎝ ∂ρ ⎠ s 1/ ⎛ ∆P ⎞ ⎟⎟ ≅ ⎜⎜ ⎝ ∆(1 / v ) ⎠ s PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-113 The specific volume of steam at s2 = 5.9989 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1551 and 0.1337 m3/kg Substituting, c2 = (1300 − 1100) kPa ⎛ 1000 m / s ⎜ ⎜ ⎞ ⎛ ⎝ kPa ⋅ m kg/m − ⎜ ⎟ ⎝ 0.1337 0.1551 ⎠ ⎞ ⎟ = 439.7 m/s ⎟ ⎠ Then the exit Mach number becomes Ma = V2 348.9 m/s = = 0.793 c 439.7 m/s The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576 × P01 = 0.576 × 1.75 = 1.008 MPa At state 2ts, Pts = 1.008 MPa and sts = s1 = 5.9874 kJ/kg·K Thus, hts = 2507.7 kJ/kg The actual enthalpy of steam at the throat is ηN = h01 − ht 2603.5 − ht ⎯ ⎯→ ht = 2515.4 kJ/kg ⎯ ⎯→ 0.92 = h01 − hts 2603.5 − 2507.7 Therefore at the throat, P2 = 1.008 MPa and ht = 2515.4 kJ/kg Thus, vt = 0.1679 m3/kg Then the throat velocity is determined from the steady-flow energy balance, Ê0 h1 + V12 = ht + Vt V2 → = ht − h1 + t 2 Solving for Vt, ⎛ 1000 m / s Vt = 2(h1 − ht ) = 2(2603.5 − 2515.4)kJ/kg⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 419.9 m/s ⎟ ⎠ Thus the throat area is At = m& v t (6.07 kg/s)(0.1679 m / kg) = = 24.30 × 10 − m = 24.30 cm 419.9 m/s Vt PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-114 Fundamentals of Engineering (FE) Exam Problems 17-154 An aircraft is cruising in still air at 5°C at a velocity of 400 m/s The air temperature at the nose of the aircraft where stagnation occurs is (a) 5°C (b) 25°C (c) 55°C (d) 80°C (e) 85°C Answer (e) 85°C Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.4 Cp=1.005 "kJ/kg.K" T1=5 "C" Vel1= 400 "m/s" T1_stag=T1+Vel1^2/(2*Cp*1000) "Some Wrong Solutions with Common Mistakes:" W1_Tstag=T1 "Assuming temperature rise" W2_Tstag=Vel1^2/(2*Cp*1000) "Using just the dynamic temperature" W3_Tstag=T1+Vel1^2/(Cp*1000) "Not using the factor 2" 17-155 Air is flowing in a wind tunnel at 25°C, 80 kPa, and 250 m/s The stagnation pressure at a probe inserted into the flow stream is (a) 87 kPa (b) 93 kPa (c) 113 kPa (d) 119 kPa (e) 125 kPa Answer (c) 113 kPa Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.4 Cp=1.005 "kJ/kg.K" T1=25 "K" P1=80 "kPa" Vel1= 250 "m/s" T1_stag=(T1+273)+Vel1^2/(2*Cp*1000) "C" T1_stag/(T1+273)=(P1_stag/P1)^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" T11_stag/T1=(W1_P1stag/P1)^((k-1)/k); T11_stag=T1+Vel1^2/(2*Cp*1000) "Using deg C for temperatures" T12_stag/(T1+273)=(W2_P1stag/P1)^((k-1)/k); T12_stag=(T1+273)+Vel1^2/(Cp*1000) "Not using the factor 2" T13_stag/(T1+273)=(W3_P1stag/P1)^(k-1); T13_stag=(T1+273)+Vel1^2/(2*Cp*1000) "Using wrong isentropic relation" PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-115 17-156 An aircraft is reported to be cruising in still air at -20°C and 40 kPa at a Mach number of 0.86 The velocity of the aircraft is (a) 91 m/s (b) 220 m/s (c) 186 m/s (d) 280 m/s (e) 378 m/s Answer (d) 280 m/s Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" T1=-20+273 "K" P1=40 "kPa" Mach=0.86 VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_vel=Mach*VS2; VS2=SQRT(k*R*T1) "Not using the factor 1000" W2_vel=VS1/Mach "Using Mach number relation backwards" W3_vel=Mach*VS3; VS3=k*R*T1 "Using wrong relation" 17-157 Air is flowing in a wind tunnel at 12°C and 66 kPa at a velocity of 230 m/s The Mach number of the flow is (a) 0.54 (b) 0.87 (c) 3.3 (d) 0.36 (e) 0.68 Answer (e) 0.68 Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" T1=12+273 "K" P1=66 "kPa" Vel1=230 "m/s" VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_Mach=Vel1/VS2; VS2=SQRT(k*R*(T1-273)*1000) "Using C for temperature" W2_Mach=VS1/Vel1 "Using Mach number relation backwards" W3_Mach=Vel1/VS3; VS3=k*R*T1 "Using wrong relation" PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-116 17-158 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane Now the nozzle exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same The nozzle exit velocity will (a) remain the same (b) double (c) quadruple (d) go down by half (e) go down to one-fourth Answer (a) remain the same 17-159 Air is approaching a converging-diverging nozzle with a low velocity at 12°C and 200 kPa, and it leaves the nozzle at a supersonic velocity The velocity of air at the throat of the nozzle is (a) 338 m/s (b) 309 m/s (c) 280 m/s (d) 256 m/s (e) 95 m/s Answer (b) 309 m/s Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" "Properties at the inlet" T1=12+273 "K" P1=200 "kPa" Vel1=0 "m/s" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(To-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation" PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-117 17-160 Argon gas is approaching a converging-diverging nozzle with a low velocity at 20°C and 120 kPa, and it leaves the nozzle at a supersonic velocity If the cross-sectional area of the throat is 0.015 m2, the mass flow rate of argon through the nozzle is (a) 0.41 kg/s (b) 3.4 kg/s (c) 5.3 kg/s (d) 17 kg/s (e) 22 kg/s Answer (c) 5.3 kg/s Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.667 Cp=0.5203 "kJ/kg.K" R=0.2081 "kJ/kg.K" A=0.015 "m^2" "Properties at the inlet" T1=20+273 "K" P1=120 "kPa" Vel1=0 "m/s" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) rho_throat=P_throat/(R*T_throat) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) m=rho_throat*A*V_throat "Some Wrong Solutions with Common Mistakes:" W1_mass=rho_throat*A*V1_throat; V1_throat=SQRT(k*R*T1_throat*1000); T1_throat=2*(To-273)/(k+1) "Using C for temp" W2_mass=rho2_throat*A*V_throat; rho2_throat=P1/(R*T1) "Using density at inlet" PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-118 17-161 Carbon dioxide enters a converging-diverging nozzle at 60 m/s, 310°C, and 300 kPa, and it leaves the nozzle at a supersonic velocity The velocity of carbon dioxide at the throat of the nozzle is (a) 125 m/s (b) 225 m/s (c) 312 m/s (d) 353 m/s (e) 377 m/s Answer (d) 353 m/s Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" "Properties at the inlet" T1=310+273 "K" P1=300 "kPa" Vel1=60 "m/s" To=T1+Vel1^2/(2*Cp*1000) To/T1=(Po/P1)^((k-1)/k) "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(T_throat-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation" 17-162 Consider gas flow through a converging-diverging nozzle Of the five statements below, select the one that is incorrect: (a) The fluid velocity at the throat can never exceed the speed of sound (b) If the fluid velocity at the throat is below the speed of sound, the diversion section will act like a diffuser (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic (d) There will be no flow through the nozzle if the back pressure equals the stagnation pressure (e) The fluid velocity decreases, the entropy increases, and stagnation enthalpy remains constant during flow through a normal shock Answer (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 17-119 17-163 Combustion gases with k = 1.33 enter a converging nozzle at stagnation temperature and pressure of 350°C and 400 kPa, and are discharged into the atmospheric air at 20°C and 100 kPa The lowest pressure that will occur within the nozzle is (a) 13 kPa (b) 100 kPa (c) 216 kPa (d) 290 kPa (e) 315 kPa Answer (c) 216 kPa Solution Solved by EES Software Solutions can be verified by copying-and-pasting the following lines on a blank EES screen (Similar problems and their solutions can be obtained easily by modifying numerical values) k=1.33 Po=400 "kPa" "The critical pressure is" P_throat=Po*(2/(k+1))^(k/(k-1)) "The lowest pressure that will occur in the nozzle is the higher of the critical or atmospheric pressure." "Some Wrong Solutions with Common Mistakes:" W2_Pthroat=Po*(1/(k+1))^(k/(k-1)) "Using wrong relation" W3_Pthroat=100 "Assuming atmospheric pressure" 17-164 ··· 17-166 Design and Essay Problems KJ PROPRIETARY MATERIAL © 2011 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission ... teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 1-24 1-69 Both a gage and a manometer are attached to a gas to measure... Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly 1-54 The barometric reading at a location... density of air is given to be ρ = 1.20 kg/m3 h=? Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain Wair / A = Pbottom