1-1 Chapter INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles 1-2C 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 1-3C There is no truth to his claim It violates the second law of thermodynamics Mass, Force, and Units 1-4C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit One pound-force is the force required to accelerate a mass of 32.174 lbm by ft/s2 In other words, the weight of a 1-lbm mass at sea level is lbf 1-5C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit 1-kg-force is the force required to accelerate a 1-kg mass by 9.807 m/s2 In other words, the weight of 1-kg mass at sea level is kg-force 1-6C There is no acceleration, thus the net force is zero in both cases 1-7 A plastic tank is filled with water The weight of the combined system is to be determined Assumptions The density of water is constant throughout Properties The density of water is given to be ρ = 1000 kg/m3 Analysis The mass of the water in the tank and the total mass are mtank = kg V =0.2 m mw =ρV =(1000 kg/m )(0.2 m ) = 200 kg 3 mtotal = mw + mtank = 200 + = 203 kg Thus,  1N   W = mg = (203 kg)(9.81 m/s )   kg ⋅ m/s = 1991 N   H2O 1-2 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 ROOM AIR m = ρV = (1.16 kg/m )(6 × × m ) = 334.1 kg 6X6X8 m3 Thus,  1N W = mg = (334.1 kg)(9.81 m/s )  kg ⋅ m/s    = 3277 N   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 1% 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 = 0.99Ws = 0.99mgs = 0.99(m)(9.807) Substituting, 0.99(9.81) = (9.81 − 3.32 × 10 −6 z)  → z = 29,539 m Sea level 1-10E An astronaut took his scales with him to space It is to be determined how much he will weigh on the spring and beam scales in space Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body:  lbf W = mg = (150 lbm)(5.48 ft/s )  32.2 lbm ⋅ ft/s    = 25.5 lbf   (b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration The beam scale will read what it reads on earth, W = 150 lbf 1-11 The acceleration of an aircraft is given in g’s The net upward force acting on a man in the aircraft is to be determined Analysis From the Newton's second law, the force applied is  1N F = ma = m(6 g) = (90 kg)(6 × 9.81 m/s )  kg ⋅ m/s    = 5297 N   1-3 1-12 [Also solved by EES on enclosed CD] 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 = (5 kg)(9.79 m/s )  kg ⋅ m/s    = 48.95 N   Then the net force that acts on the rock is Fnet = Fup − Fdown = 150 − 48.95 = 101.05 N From the Newton's second law, the acceleration of the rock becomes a= F 101.05 N  kg ⋅ m/s  = m kg  1N  Stone   = 20.2 m/s   1-13 EES 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 W=m*g"[N]" m=5"[kg]" g=9.79"[m/s^2]" "The force balance on the rock yields the net force acting on the rock as" F_net = F_up - F_down"[N]" F_up=150"[N]" F_down=W"[N]" "The acceleration of the rock is determined from Newton's second law." F_net=a*m "To Run the program, press F2 or click on the calculator icon from the Calculate menu" SOLUTION a=20.21 [m/s^2] F_down=48.95 [N] F_net=101.1 [N] F_up=150 [N] g=9.79 [m/s^2] m=5 [kg] W=48.95 [N] 1-4 1-14 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an altitude of 13,000 m Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from ∆g 9.807 − 9.767 × 100 = × 100 = 0.41% %Reduction in weight = %Reduction in g = 9.807 g Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude Discussion Note that the weight loss at cruising altitudes is negligible Systems, Properties, State, and Processes 1-15C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system 1-16C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the system 1-17C Intensive properties not depend on the size (extent) of the system but extensive properties 1-18C 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 1-19C A process during which a system remains almost in equilibrium at all times is called a quasiequilibrium 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 1-20C 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 1-21C The state of a simple compressible system is completely specified by two independent, intensive properties 1-22C Yes, because temperature and pressure are two independent properties and the air in an isolated room is a simple compressible system 1-23C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries 1-24C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ρH2O = 1000 kg/m3) That is, SG = ρ / ρ H2O When specific gravity is known, density is determined from ρ = SG × ρ H2O 1-5 1-25 EES The variation of density of atmospheric air with elevation is given in tabular form A relation for the variation of density with elevation is to be obtained, the density at km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated Assumptions Atmospheric air behaves as an ideal gas The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km Properties The density data are given in tabular form as ρ, kg/m3 1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008 1.4 1.2 z, km 10 15 20 25 ρ, kg/m r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402 0.8 0.6 0.4 0.2 0 10 15 20 z, km 25 Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window Then specify 2nd order polynomial and enter/edit equation The results are: ρ(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2 for the unit of kg/m3, (or, ρ(z) = (1.20252 – 0.101674z + 0.0022375z2)×109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level At z = km, the equation would give ρ = 0.60 kg/m3 (b) The mass of atmosphere can be evaluated by integration to be m= ∫ V ρdV = ∫ h z =0 (a + bz + cz )4π (r0 + z ) dz = 4π [ ∫ h z =0 (a + bz + cz )(r02 + 2r0 z + z )dz = 4π ar02 h + r0 (2a + br0 )h / + (a + 2br0 + cr02 )h / + (b + 2cr0 )h / + ch / ] where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function Substituting and multiplying by the factor 109 for the density unity kg/km3, the mass of the atmosphere is determined to be m = 5.092×1018 kg Discussion Performing the analysis with excel would yield exactly the same results EES Solution for final result: a=1.2025166 b=-0.10167 c=0.0022375 r=6377 h=25 m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9 1-6 Temperature 1-26C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact 1-27C They are celsius(°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system 1-28C Probably, but not necessarily The operation of these two thermometers is based on the thermal expansion of a fluid If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings Otherwise, the two readings may deviate 1-29 A temperature is given in °C It is to be expressed in K Analysis The Kelvin scale is related to Celsius scale by T(K] = T(°C) + 273 Thus, T(K] = 37°C + 273 = 310 K 1-30E A temperature is given in °C It is to be expressed in °F, K, and R Analysis Using the conversion relations between the various temperature scales, T(K] = T(°C) + 273 = 18°C + 273 = 291 K T(°F] = 1.8T(°C) + 32 = (1.8)(18) + 32 = 64.4°F T(R] = T(°F) + 460 = 64.4 + 460 = 524.4 R 1-31 A temperature change is given in °C It is to be expressed in K Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales Thus, ∆T(K] = ∆T(°C) = 15 K 1-32E A temperature change is given in °F It is to be expressed in °C, K, and R Analysis This problem deals with temperature changes, which are identical in Rankine and Fahrenheit scales Thus, ∆T(R) = ∆T(°F) = 45 R The temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in Fahrenheit and Rankine scales by ∆T(K) = ∆T(R)/1.8 = 45/1.8 = 25 K and ∆T(°C) = ∆T(K) = 25°C 1-33 Two systems having different temperatures and energy contents are brought in contact The direction of heat transfer is to be determined Analysis Heat transfer occurs from warmer to cooler objects Therefore, heat will be transferred from system B to system A until both systems reach the same temperature 1-7 Pressure, Manometer, and Barometer 1-34C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure 1-35C The atmospheric pressure, which is the external pressure exerted on the skin, decreases with increasing elevation Therefore, the pressure is lower at higher elevations As a result, the difference between the blood pressure in the veins and the air pressure outside increases This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding The shortness of breath is caused by the lower air density at higher elevations, and thus lower amount of oxygen per unit volume 1-36C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled It is the gage pressure that doubles when the depth is doubled 1-37C If the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube will be the same 1-38C Pascal’s principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount This is a consequence of the pressure in a fluid remaining constant in the horizontal direction An example of Pascal’s principle is the operation of the hydraulic car jack 1-39C The density of air at sea level is higher than the density of air on top of a high mountain Therefore, the volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will be higher 1-40 The pressure in a vacuum chamber is measured by a vacuum gage The absolute pressure in the chamber is to be determined Analysis The absolute pressure in the chamber is determined from Pabs = Patm − Pvac = 92 − 35 = 57 kPa Pabs Patm = 92 kPa 35 kPa 1-8 1-41E The pressure in a tank is measured with a manometer by measuring the differential height of the manometer fluid The absolute pressure in the tank is to be determined for the cases of the manometer arm with the higher and lower fluid level being attached to the tank Assumptions incompressible The fluid in the manometer is Properties The specific gravity of the fluid is given to be SG = 1.25 The density of water at 32°F is 62.4 lbm/ft3 (Table A-3E) Air 28 in Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water, ρ = SG × ρ H O = (1.25)(62.4 lbm/ft ) = 78.0 lbm/ft SG = 1.25 Patm = 12.7 psia The pressure difference corresponding to a differential height of 28 in between the two arms of the manometer is   1ft  lbf   ∆P = ρgh = (78 lbm/ft3 )(32.174 ft/s )(28/12 ft)  32.174 lbm ⋅ ft/s  144 in  = 1.26 psia    Then the absolute pressures in the tank for the two cases become: (a) The fluid level in the arm attached to the tank is higher (vacuum): Pabs = Patm − Pvac = 12.7 − 1.26 = 11.44 psia (b) The fluid level in the arm attached to the tank is lower: Pabs = Pgage + Patm = 12.7 + 1.26 = 13.96 psia Discussion Note that we can determine whether the pressure in a tank is above or below atmospheric pressure by simply observing the side of the manometer arm with the higher fluid level 1-9 1-42 The pressure in a pressurized water tank is measured by a multi-fluid manometer The gage pressure of air in the tank is to be determined Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m3, respectively Analysis Starting with the pressure at point at the air-water interface, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives P1 + ρ water gh1 + ρ oil gh2 − ρ mercury gh3 = Patm Air Solving for P1, P1 = Patm − ρ water gh1 − ρ oil gh2 + ρ mercury gh3 h1 or, h3 P1 − Patm = g ( ρ mercury h3 − ρ water h1 − ρ oil h2 ) Water Noting that P1,gage = P1 - Patm and substituting, h2 P1,gage = (9.81 m/s )[(13,600 kg/m )(0.46 m) − (1000 kg/m )(0.2 m)  1N − (850 kg/m )(0.3 m)]  kg ⋅ m/s  = 56.9 kPa  kPa    1000 N/m    Discussion 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 1-43 The barometric reading at a location is given in height of mercury column The atmospheric pressure is to be determined Properties The density of mercury is given to be 13,600 kg/m3 Analysis The atmospheric pressure is determined directly from Patm = ρgh  1N = (13,600 kg/m )(9.81 m/s )(0.750 m)  kg ⋅ m/s  = 100.1 kPa  kPa    1000 N/m    1-10 1-44 The gage pressure in a liquid at a certain depth is given The gage pressure in the same liquid at a different depth is to be determined Assumptions The variation of the density of the liquid with depth is negligible Analysis The gage pressure at two different depths of a liquid can be expressed as P1 = ρgh1 and P2 = ρgh2 Taking their ratio, h1 P2 ρgh2 h2 = = P1 ρgh1 h1 h2 Solving for P2 and substituting gives P2 = h2 9m P1 = (28 kPa) = 84 kPa h1 3m Discussion Note that the gage pressure in a given fluid is proportional to depth 1-45 The absolute pressure in water at a specified depth is given The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined Assumptions The liquid and water are incompressible Properties The specific gravity of the fluid is given to be SG = 0.85 We take the density of water to be 1000 kg/m3 Then density of the liquid is obtained by multiplying its specific gravity by the density of water, ρ = SG × ρ H 2O = (0.85)(1000 kg/m ) = 850 kg/m Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from Patm = P − ρgh  kPa = (145 kPa) − (1000 kg/m )(9.81 m/s )(5 m)  1000 N/m  = 96.0 kPa     Patm h P (b) The absolute pressure at a depth of m in the other liquid is P = Patm + ρgh  kPa = (96.0 kPa) + (850 kg/m )(9.81 m/s )(5 m)  1000 N/m  = 137.7 kPa     Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected 17-86 17-143 EES Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-33 for air Properties The specific heat ratio is given to be k = 1.4 for air Analysis The normal shock relations listed below are expressed in EES and the results are tabulated Ma = (k − 1)Ma + 2 P2 + kMa 2kMa − k + = = P1 + kMa k +1 2 2kMa − k + ρ P2 / P1 (k + 1)Ma V = = = , ρ T2 / T1 + (k − 1)Ma V 2 T2 + Ma (k − 1) = T1 + Ma (k − 1) k +1 P02 Ma ⎡1 + Ma (k − 1) / ⎤ 2( k −1) = ⎢ ⎥ P01 Ma ⎢1 + Ma (k − 1) / ⎥ ⎣ ⎦ P02 (1 + kMa )[1 + Ma (k − 1) / 2] k /( k −1) = P1 + kMa 2 Air: k=1.4 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 0.7011 0.5774 0.5130 0.4752 0.4512 0.4350 0.4236 0.4152 0.4090 0.4042 0.4004 0.3974 0.3949 0.3929 0.3912 0.3898 0.3886 0.3876 P2/P1 1.0000 2.4583 4.5000 7.1250 10.3333 14.1250 18.5000 23.4583 29.0000 35.1250 41.8333 49.1250 57.0000 65.4583 74.5000 84.1250 94.3333 105.1250 116.5000 ρ2/ρ1 1.0000 1.8621 2.6667 3.3333 3.8571 4.2609 4.5714 4.8119 5.0000 5.1489 5.2683 5.3651 5.4444 5.5102 5.5652 5.6117 5.6512 5.6850 5.7143 T2/T1 1.0000 1.3202 1.6875 2.1375 2.6790 3.3151 4.0469 4.8751 5.8000 6.8218 7.9406 9.1564 10.4694 11.8795 13.3867 14.9911 16.6927 18.4915 20.3875 P02/P01 0.9298 0.7209 0.499 0.3283 0.2129 0.1388 0.0917 0.06172 0.04236 0.02965 0.02115 0.01535 0.01133 0.008488 0.006449 0.004964 0.003866 0.003045 P02/P1 1.8929 3.4133 5.6404 8.5261 12.0610 16.2420 21.0681 26.5387 32.6535 39.4124 46.8152 54.8620 63.5526 72.8871 82.8655 93.4876 104.7536 116.6634 129.2170 PROPRIETARY MATERIAL © 2006 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 17-87 17-144 EES Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-14 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 = (k − 1)Ma + 2 P2 + kMa 2kMa − k + = = P1 + kMa k +1 2 2kMa − k + ρ P2 / P1 (k + 1)Ma V = = = , ρ T2 / T1 + (k − 1)Ma V 2 T2 + Ma (k − 1) = T1 + Ma (k − 1) k +1 P02 Ma ⎡1 + Ma (k − 1) / ⎤ 2( k −1) = ⎢ ⎥ P01 Ma ⎢1 + Ma (k − 1) / ⎥ ⎣ ⎦ P02 (1 + kMa )[1 + Ma (k − 1) / 2] k /( k −1) = P1 + kMa 2 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 © 2006 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 17-88 17-145 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 Q flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant crossP01 = 240 kPa sectional area duct with negligible frictional effects) are Ma2 = T01 = 350 K valid Ma1 = 1.2 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 Knowing stagnation properties, the static properties are determined to be ⎛ k −1 2⎞ T1 = T01 ⎜1 + Ma ⎟ ⎝ ⎠ −1 ⎛ k −1 2⎞ P1 = P01 ⎜1 + Ma ⎟ ⎝ ⎠ − k /( k −1) ρ1 = ⎛ 1.4 - ⎞ 1.2 ⎟ = (350 K)⎜1 + ⎠ ⎝ −1 = 271.7 K ⎛ 1.4 - ⎞ 1.2 ⎟ = (240 kPa)⎜1 + ⎠ ⎝ −1.4 / 0.4 = 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 T0 T01 / T0* 0.9787 → T02 = 0.8107T01 = 0.8107(350 K ) = 283.7 K 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 © 2006 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 17-89 17-146 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) & Q Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = The inlet density and stagnation temperature are P1 = 400 kPa T1 = 360 K ρ1 = P1 400 kPa = = 3.872 kg/m RT1 (0.287 kJ/kgK)(360 K) Ma2 = Ma1 = 0.4 ⎛ k −1 ⎛ 1.4 - ⎞ 2⎞ T01 = T1 ⎜1 + Ma ⎟ = (360 K)⎜1 + 0.4 ⎟ = 371.5 K 2 ⎝ ⎠ ⎝ ⎠ Then the inlet velocity and the mass flow rate become ⎛ 1000 m / s c1 = kRT1 = (1.4 )(0.287 kJ/kg ⋅ K)(360 K)⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 380.3 m/s ⎟ ⎠ V1 = Ma 1c1 = 0.4(380.3 m/s ) = 152.1 m/s & m air = ρ Ac1V1 = (3.872 kg/m )(0.1× 0.1 m )(152.1 m/s) = 5.890 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = (since Ma2 = 1) T01 T0* = 2 (k + 1)Ma [2 + (k − 1)Ma ] (1 + kMa ) = (1.4 + 1)0.4 [2 + (1.4 − 1)0.4 ] (1 + 1.4 × 0.4 ) = 0.5290 Therefore, T0 T02 / T0* = = * 0.5290 T0 T01 / T0 → T02 = T01 / 0.5290 = (371.5 K ) / 0.5290 = 702.3 K Then the rate of heat transfer becomes & & Q = m air c p (T02 − T01 ) = (5.890 kg/s )(1.005 kJ/kg ⋅ K )(702.3 − 371.5) K = 1958 kW Discussion It can also be shown that T2 = 585 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 © 2006 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 17-90 17-147 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) Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = The inlet density and stagnation temperature are P 400 kPa ρ1 = = = 0.5350 kg/m RT1 (2.077 kJ/kgK)(360 K) ⎛ k −1 ⎛ 1.667 - ⎞ 2⎞ T01 = T1 ⎜1 + Ma ⎟ = (360 K)⎜1 + 0.4 ⎟ = 379.2 K 2 ⎝ ⎠ ⎝ ⎠ & Q P1 = 400 kPa T1 = 360 K Ma2 = Ma1 = 0.4 Then the inlet velocity and the mass flow rate become ⎛ 1000 m / s c1 = kRT1 = (1.667)(2.077 kJ/kg ⋅ K)(360 K)⎜ ⎜ kJ/kg ⎝ ⎞ ⎟ = 1116 m/s ⎟ ⎠ V1 = Ma 1c1 = 0.4(1116 m/s ) = 446.6 m/s & m air = ρ1 Ac1V1 = (0.5350 kg/m )(0.1× 0.1 m )(446.6 m/s) = 2.389 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = (since Ma2 = 1) T01 T0* = 2 (k + 1)Ma [2 + (k − 1)Ma ] (1 + kMa ) = (1.667 + 1)0.4 [2 + (1.667 − 1)0.4 ] (1 + 1.667 × 0.4 ) = 0.5603 Therefore, T0 T02 / T0* = = * 0.5603 T0 T01 / T0 → T02 = T01 / 0.5603 = (379.2 K ) / 0.5603 = 676.8 K Then the rate of heat transfer becomes & & Q = m air c p (T02 − T01 ) = (2.389 kg/s)(5.193 kJ/kg ⋅ K )(676.8 − 379.2) K = 3693 kW Discussion It can also be shown that T2 = 508 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 A34 since they are based on k = 1.4 PROPRIETARY MATERIAL © 2006 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 17-91 17-148 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 A2a) 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 = V1 100 m/s = = 0.2494 c1 400.9 m/s ⎛ k −1 2⎞ T01 = T1⎜1 + Ma1 ⎟ ⎝ ⎠ 1.4 - ⎛ ⎞ 0.2494 ⎟ = (400 K)⎜1 + ⎝ ⎠ = 405.0 K ⎞ ⎟ = 400.9 m/s ⎟ ⎠ 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 * 0.2559 T0 T01 / T 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 © 2006 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 17-92 17-149 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 A2a) Analysis Noting that Ma1 = 1, the inlet stagnation temperature is T01 ⎛ k −1 ⎛ 1.4 - ⎞ 2⎞ = T1 ⎜1 + Ma ⎟ = (500 K)⎜1 + ⎟ = 600 K 2 ⎝ ⎠ ⎝ ⎠ 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: P01 = 420 kPa T01 = 500 K q Ma2 = 1.6 Ma1 = T02/T0* = 0.8842 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 © 2006 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 17-93 17-150 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, h1 = (h f + x1 h fg ) @ MPa = 1008.3 + 0.95 × 1794.9 = 2713.4 kJ/kg s1 = ( s f + x1 s fg ) @ MPa = 2.6454 + 0.95 × 3.5402 = 6.0086 kJ/kg ⋅ K At the exit, P2 = 1.2 MPa and s2 = s2s = s1 = 6.0086 kJ/kg·K Thus, s2 = s f + x2 s fg → 6.0086 = 2.2159 + x2 (4.3058) → x2 = 0.8808 Vi ≈ Steam h2 = h f + x2 h fg = 798.33 + 0.8808 × 1985.4 = 2547.2 kJ/kg t a) ηN = 100% b) ηN = 90% v = v f + x2v fg = 0.001138 + 0.8808 × (0.16326 − 0.001138) = 0.1439 m3 / 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 ⎞ ⎟ = 576.7 m/s V2 = 2(h1 − h2 ) = 2(2713.4 - 2547.2)kJ/kg⎜ ⎜ kJ/kg ⎟ ⎝ ⎠ The mass flow rate is determined from & m= v2 A2V2 = (16 × 10− m )(576.7 m/s) = 6.41 kg/s 0.1439 m / kg 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 = 6.0086 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1555 and 0.1340 m3/kg Substituting, c2 = (1300 − 1100) kPa ⎛ 1000 m / s ⎜ ⎜ ⎞ ⎛ ⎝ kPa ⋅ m − ⎟ kg/m ⎜ ⎝ 0.1340 0.1555 ⎠ ⎞ ⎟ = 440.3 m/s ⎟ ⎠ Then the exit Mach number becomes Ma = V2 576.7 m/s = = 1.310 c2 440.3 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 × = 1.728 MPa Then at the throat, Pt = 1.728 MPa and s t = s1 = 6.0086 kJ/kg ⋅ K Thus, PROPRIETARY MATERIAL © 2006 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 17-94 ht = 2611.4 kJ/kg v t = 0.1040 m3 / kg Then the throat velocity is determined from the steady-flow energy balance, h1 + V12 = ht + Vt V2 → = ht − h1 + t 2 Solving for Vt, ⎛ 1000 m / s ⎞ ⎟ = 451.7 m/s Vt = 2(h1 − ht ) = 2(2713.4 − 2611.4)kJ/kg⎜ ⎜ kJ/kg ⎟ ⎝ ⎠ Thus the throat area is & mv t (6.41 kg/s)(0.1040 m3 / kg) = = 14.75 × 10− m = 14.75 cm (451.7 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 ) @ MPa = 1008.3 + 0.95 × 1794.9 = 2713.4 kJ/kg s1 = ( s f + x1 s fg ) @ MPa = 2.6454 + 0.95 × 3.5402 = 6.0086 kJ/kg ⋅ K At state 2s, P2 = 1.2 MPa and s2 = s2s = s1 = 6.0086 kJ/kg·K Thus, Steam Vi ≈ s s = s f + x s s fg ⎯ ⎯→ 6.0086 = 2.2159 + x s (4.3058) ⎯ ⎯→ x s = 0.8808 h2 s = h f + x s h fg = 798.33 + 0.8808 × 1985.4 = 2547.2 kJ/kg t a) ηN = 100% b) ηN = 90% The enthalpy of steam at the actual exit state is determined from h01 − h2 2713.4 − h2 ⎯ 0.90 = ⎯→ ⎯ h2 = 2563.8 kJ/kg ⎯→ h01 − h2 s 2713.4 − 2547.2 ηN = Therefore at the exit, P2 = 1.2 MPa and h2 = 2563.8 kJ/kg·K Thus, h2 = h f + x2 h fg ⎯ ⎯→ 2563.8 = 798.33 + x2 (1985.4) ⎯ ⎯→ x2 = 0.8892 s2 = s f + x2 s fg = 2.2159 + 0.8892 × 4.3058 = 6.0447 v = v f + x2v fg = 0.001138 + 0.8892 × (0.16326 − 0.001138) = 0.1453 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 ⎞ ⎟ = 547.1 m/s V2 = 2(h1 − h2 ) = 2(2713.4 − 2563.8)kJ/kg⎜ ⎜ kJ/kg ⎟ ⎝ ⎠ The mass flow rate is determined from & m= v2 A2V2 = 0.1453 m / kg (16 × 10− m )(547.1 m/s) = 6.02 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 © 2006 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 17-95 The specific volume of steam at s2 = 6.0447 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1570 and 0.1353 m3/kg Substituting, (1300 − 1100) kPa ⎛ 1000 m / s ⎞ ⎟ = 442.6 m/s ⎜ ⎟ ⎜ ⎞ ⎛ ⎝ kPa ⋅ m ⎠ − ⎜ ⎟ kg/m ⎝ 0.1353 0.1570 ⎠ c2 = Then the exit Mach number becomes Ma = V2 547.1 m/s = = 1.236 c2 442.6 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 × = 1.728 MPa At state 2ts, Pts = 1.728 MPa and sts = s1 = 6.0086 kJ/kg·K Thus, hts = 2611.4 kJ/kg The actual enthalpy of steam at the throat is ηN = h01 − ht 2713.4 − ht ⎯ ht = 2621.6 kJ/kg ⎯→ ⎯ 0.90 = ⎯→ h01 − hts 2713.4 − 2611.4 Therefore at the throat, P2 = 1.728 MPa and ht = 2621.6 kJ/kg Thus, vt = 0.1046 m3/kg Then the throat velocity is determined from the steady-flow energy balance, V2 h1 + = ht + Vt V2 → = ht − h1 + t 2 Solving for Vt, ⎛ 1000 m / s ⎞ ⎟ = 428.5 m/s Vt = 2(h1 − ht ) = 2(2713.4 − 2621.6)kJ/kg⎜ ⎜ kJ/kg ⎟ ⎠ ⎝ Thus the throat area is At = & mv t (6.02 kg/s)(0.1046 m3 / kg) = = 14.70 × 10− m = 14.70 cm Vt (428.5 m/s) PROPRIETARY MATERIAL © 2006 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 17-96 Fundamentals of Engineering (FE) Exam Problems 17-151 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" 17-152 Air is flowing in a wind tunnel at 15°C, 80 kPa, and 200 m/s The stagnation pressure at a probe inserted into the flow stream is (a) 82 kPa (b) 91 kPa (c) 96 kPa (d) 101 kPa (e) 114 kPa Answer (d) 101 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=15 "K" P1=80 "kPa" Vel1= 200 "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 © 2006 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 17-97 17-153 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=-10+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" 17-154 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 (Problem changed, 2/2001) (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 © 2006 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 17-98 17-155 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 17-156 Air is approaching a converging-diverging nozzle with a low velocity at 20°C and 300 kPa, and it leaves the nozzle at a supersonic velocity The velocity of air at the throat of the nozzle is (a) 290 m/s (b) 98 m/s (c) 313 m/s (d) 343 m/s (e) 412 m/s Answer (c) 313 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=20+273 "K" P1=300 "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" 17-157 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" PROPRIETARY MATERIAL © 2006 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 17-99 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*(To273)/(k+1) "Using C for temp" W2_mass=rho2_throat*A*V_throat; rho2_throat=P1/(R*T1) "Using density at inlet" 17-158 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" PROPRIETARY MATERIAL © 2006 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 17-100 17-159 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 17-160 Combustion gases with k = 1.33 enter a converging nozzle at stagnation temperature and pressure of 400°C and 800 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) 26 kPa (b) 100 kPa (c) 321 kPa (d) 432 kPa (e) 272 kPa Answer (d) 432 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=800 "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" 17-161 ··· 17-163 Design and Essay Problems PROPRIETARY MATERIAL © 2006 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 ... State, and Processes 1-15C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system 1-16C A can of soft drink should be analyzed as a closed system... The temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in Fahrenheit and Rankine scales by ∆T(K) = ∆T(R)/1.8 = 45/1.8 = 25 K and ∆T(°C) = ∆T(K)... pressure exerted on the ground (and on the feet) is reduced by half when the person stands on both feet 1-48 The mass of a woman is given The minimum imprint area per shoe needed to enable her