# AME Homework Solutions 1 Fall 2011

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1 Homework 1 AME Homework Solutions 1 Fall CPIG air enters an isentropic nozzle at 1.30 atm and 24 Cwithavelocityof2.5m/s. The nozzle entrance diameter is 120 mm. The air exits the nozzle at 1.24 atm with a velocity of 90 m/s. Determine the temperature of the exiting air and the nozzle exit diameter. Given: CPIG air, 1 T 1 =24 C, P 1 =1.30 atm, v 1 =2.5 m/s,d 1 =120mm 2 P 2 =1.24 atm, v 2 =90m/s Assumptions: Find: T 2, d 2 For T 2,useenergyconservation: For d 2,usemassconservation: h 1 + v = h 2 + v (h 1 h 2 )+ v = v c p (T 1 T 2 )+ v = v c p (T 1 T 2 )=1/2(v 2 2 v 2 1 ) J kg K (T 1 T 2 )=1/2((90 m/s) 2 (2.5 m/s) 2 ) T 1 T 2 =4.029 C T 2 =19.97 C. m 1 = m 2 ρ 1 A 1 v 1 = ρ 2 A 2 v 2 P = ρrt ρ = P/RT P 1 A 1 v 1 = P 2A 2 v 2 T 1 T 2 A 2 = T 2P 1 v 1 T 1 P 2 v 2 A 1 πd = T 2P 1 v 1 πd 2 1 T 1 P 2 v 2 4 d 2 ( K)(1.30 atm)(2.5 m/s) 2 = (120 mm)2 ( K)(1.24 atm)(90 m/s) d 2 =20.34 mm Consider a steam turbine power plant operating near critical pressure. As a first approximation, it may be assumed that the turbine and the pump processes are reversible and adiabatic. Neglect any changes in kinetic and potential energies. 1 Solutions adapted from Borgnakke, Sonntag (2008) Solutions Manual, Fundamentals of Thermodynamics, 7th Edition and previous AME Homework Solutions documents. 1

2 Given: CPIG air, 1 T 1 = 750 C, P 1 = 20 MPa 2 P 2 = 15 kpa 3 T 3 = 40 C, P 3 = 15 kpa 4 P 4 = 20 MPa Assumptions: v =0, z =0,pumpwater is incompressible Find: (a) w T,turbineexitstate(b)w P, h 4 (c) η TH (a) h 1 = kj/kg, s 1 = kj kg K Thus s 2 = s 1 = kj kg K s 2 = kj kg K + x kj kj kg K kg K x 2 = = kj kg K h 2 =53.97 kj/kg + x 2 ( kj/kg) = kj/kg w T = h 1 h 2 = kj/kg kj/kg = kj/kg = w T. (b) h 3 = kj/kg, v 3 = m 3 /kg Turbine exit state: liquid-vapor mixture w P = v 3 (P 4 P 3 ) w P =( m 3 /kg)(20000 kpa 15 kpa) w P =20.14 kj/kg h 4 = h 3 + w P = kj/kg = h 4 (c) η TH = w NET = (h 1 h 2 ) (h 4 h 3 ) = w T w P q H h 1 h 4 q H kj/kg kj/kg η TH = kj/kg kj/kg η TH = Homework Athermalstorageismadewitharock(granite)bedof2m 3 which is heated to 390 K using solar energy. A heat engine receives a Q h from the bed and rejects heat to the ambient at 290 K. The rock bed therefore cools down and as it reaches 290 K the process 2

3 stops. Find the energy the rock bed can give out. What is the heat engine efficiency at the beginning of the process and what is it at the end of the process? Given: T i =390K,T f =290K,V =2m 3 Assumptions: Reversible, heat engine operates in a Carnot cycle Find: 1 Q 2, η i, η f 1q 2 = u 2 u 1 = C T =(0.89 kj/kg K)(390 K 290 K) = 89.0 kj/kg m = ρv =(2750kg/m 3 )(2 m 3 )=5500kg Q = m 1 q 2 =(5500kg)(89.0 kj/kg) = 489, 500 kj = 1 Q 2 To get the efficiencies, use the Carnot cycle: η i =1 T 0 /T i =1 290/390 = = η i η f =1 T 0 /T i =1 290/290 = 0=η f An insulated piston/cylinder setup contains carbon dioxide gas at 400 kpa, 300 K which is then compressed to 4 MPa in a reversible adiabatic process. Calculate the final temperature and the specific work using (a) ideal gas Table A.8 and (b) using constant specific heats Table A.5. Given: 1 T 1 =300K,P 1 =400kPa, u 1 = kj/kg K (from Table A.8) 2 P 2 =4 MPa Assumptions: CV CO 2,acontrolmassundergoingareversible,adiabatic(isentropic)process Find: T 2, 1 q 2 (a) From Table A.8 for CO 2 and Eq. (8.19), s 2 s 1 =0=s o T 2 s o T 1 R ln P 2 /P 1 s o T 2 = s o T 1 + R ln (P 2 /P 1 )= ln (4000/400) = kj kg K Now interpolate in Table A.8: u 2 = kj/kg K, T 2 =481.2 K 1w 2 = (u 2 u 1 )= ( ) = kj/kg = 1 q 2 3

4 (b) Table A.5: k =1.289, C v0 =0.653 kj/kg K. From Eq. (8.23): k 1 P2 k T 2 = T 1 P 1 = = K = T w 2 = C v0 (T 2 T 1 )= 0.653( ) = kj/kg = 1 q Ammonia is contained in a rigid sealed tank of unknown quality at 0 C. When heated in boiling water to 100 C its pressure reaches 1200 kpa. Find the initial quality, the heat transfer to the ammonia and the total entropy generation. Given: 1 T 1 =0 C, 2 P 2 =1200kPa,T 2 =100 C Assumptions: Control volume ammonia, which is a control mass of a control volume Find: x 1, 1 q 2, 1 s 2 From the tables: s 2 = kj kg K, v 2 = m 3 /kg, u 2 = kj/kg The volume is constant, v 1 = v 2,so x 1 = Also because the volume is constant, 1 w 2 =0,so = = x 1 1q 2 = u 2 u 1 = = kj/kg = 1 q 2 For the entropy generation, 1s 2 = s 2 s 1 1 q 2 /T = / s 2 =0.547 kj/kg K The air-standard Carnot cycle was not shown in the text; show the Ts diagram for this cycle. In an air-standard Carnot cycle the low temperature is 280 K and the efficiency is 60 %. If the pressure before compression and after heat rejection is 100 kpa, find the high temperature and the pressure just before heat addition. Given: T L =280K,η =0.6, P 1 =100kPa,air-standardCarnotcycle Assumptions: Find: the T s diagram, T H, P 2 From Eq. (7.5), η =0.6 =1 T H /T L T H = T L /0.4 =700K. State 2 is the state just before heat addition, and we already have the temperature (T L )and pressure (P 1 ) for State 1. As seen in the T s diagram, State 1 to State 2 is an isentropic compression, so using Eq. (8.23), P 2 = P 1 (T H /T L ) k 1.4 k 1 =100(700/280) 0.4 4

5 Homework 3 P 2 =2.47 MPa A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a leaves the boiler at a temperature of 85 C, and the condenser temperature is 35 C. Calculate the thermal efficiency of this cycle. Given: R-134a, ideal Rankine Cycle; 1 h 1 = kj/kg, P 1 =887.6 kj/kg 2 P 2 =kpa 3 h 3 = kj/kg 4 Assumptions: Find: η TH w P = h 2 h 1 = 2 1 vdp v 1 (P 2 P 1 )= ( ) w P =1.747 kj/kg h 2 = h 1 + w P = = kj/kg For the boiler: q H = h 3 h 2 = = kj/kg For the turbine: s 4 = s 3 = = x 4 (0.5465) x 4 =

6 h 4 = x 4 (168.42) = kj/kg w T = h 3 h 4 = = kj/kg w NET = w T w P = = kj/kg ηth = w NET /q H =19.78/ = = η TH The reheat pressure affects the operating variables and thus turbine performance. Repeat Problem twice, using 0.6 MPa and 1.0 MPa for the reheat pressure. Given: Q L =10, 000 kw 1x 1 =0, T 1 =45 C 2P 2 =3MPa 3P 3 = P 2, T 3 =600 C h 3 = kj/kg, s 3 = kj/kg 6x 6 = 1, T 6 = T 1 h 6 = kj/kg, s 6 = kj/kg K Assumptions: Pump: reversible and adiabatic and incompressible flow, 3 4 isentropic, 5 6 isentropic Find: T 5, Ẇ T,tot, Q H w P = v 1 (P 2 P 1 )= ( ) = 3.02 kj/kg h 2 = h 1 + w P = = kj/kg For P 4 =1MPa=P 5 : s 3 = s 4 state 4 is a superheated vapor, h 4 = kj/kg State 5: T 5 =655.6 C, P 4 = P 5 h 5 = kj/kg, s 5 = s 6 = kj/kg K For P 4 =0.6 MPa=P 5 : s 3 = s 4 state 4 is a superheated vapor, h 4 = kj/kg State 5: T 5 =561.3 C, P 4 = P 5 h 5 = kj/kg, s 5 = s 6 = kj/kg K Total condenser heat transfer: q L = h 6 h 1 = kj/kg Mass flow rate: ṁ = Q L q L = 10, =4.176 kg/s 6

7 Total turbine power: Ẇ T,tot = ṁ(w T,tot )=ṁ (h 3 h 4 + h 5 h 6 ) Total boiler heat rate: Q H = ṁ(h 3 h 2 + h 5 h 4 ) See the results for both iterations of the problem in the table below: P 4 = P 5 T 5 Ẇ T,tot Q H 1MPa C 6797 kw kw 0.6 MPa C 6568 kw kw ARankinecycleoperatingwithammoniaisheatedbysomelowtemperature source so the highest temperature is 130 C at a pressure of 5000 kpa. Its low pressure is 1003 kpa and it operates with one open feedwater heater at 2033 kpa. The total flow rate is 5 kg/s. Find the extraction flow rate to the feedwater heater assuming its outlet state is saturated liquid at 2033 kpa. Find the total power to the two pumps. Given: 1 x 1 =0,h 1 = kj/kg, v 1 = m 3 /kg 3 x 3 =0,h 3 = kj/kg, v 3 = m 3 /kg 5 h 5 =1621.8, s 5 = kj/kg K 6 s 6 = s 5 x 6 =(s 6 s f )/s fg = ( )/ = , h 6 = kj/kg Assumptions: Incompressible fluid in both pumps Find: Ẇ P For the pump: w P 1 = h 2 h 1 v 1 (P 2 P 1 )= ( ) = kj/kg h 2 = h 1 + w P 1 = = kj/kg For the feedwater heater, call ṁ 6 /ṁ tot = y (the extraction fraction) Energy equation: (1 y)h 2 +(y)h 6 = h 3 y = h 3 h 2 h 6 h 2 = =

8 For pump 2: Total pump work: ṁ extr = y(ṁ tot )=0.1054(5) = kg/s ṁ 1 =(1 y)ṁ tot =( )5 = kg/s w P 2 = h 4 h 3 = v 3 (P 4 P 3 )= ( ) = kj/kg Ẇ P = ṁ 1 w P 1 +ṁ tot w P 2 =(4.473)(1.708) + (5)(5.272) = 34.0 kw=ẇp From problem 11.30: The power plant in Problem is modified to have asuperheatersectionfollowingtheboilersothesteamleavesthesuperheaterat3.0mpa, 400 C. Calculate the thermal efficiency of the cycle and the moisture content of the steam leaving the turbine for turbine exhaust pressures of 5, 10, 50, and 100 kpa. Plot the thermal efficiency versus turbine exhaust pressure. Given: h 3 = kj/kg, s 3 = kj/kg K Assumptions: Find: Pump: Boiler: Isentropic turbine (s 4 = s 3 ): w P = 2 2 vdp v 1 (P 2 P 1 ) h 2 = h 1 + w P q H = h 3 h 2 x 4 = s 3 s f s fg h 4 = h f +(x 4 )h fg Finally, the cycle efficiency: w T,s = h 3 h 4 η CY CLE = w NET q H = w T,s w P q H 8

9 See the results for all iterations of the problem in the table below: P 4 v 1 w P h 2 q H x 4 h 4 w T,s η CY CLE ηcy CLE Turbine exhaust pressure [kpa] Figure 1: Plot for problem Homework Repeat Problem 12.17, but assume variable specific heat for the air, table A.7. Consider an ideal air-standard Brayton cycle in which the air into the compressor is at 100 kpa, 20 C, and the pressure ratio across the compressor is 12:1. The maximum temperature in the cycle is 1100 C, and the air flow rate is 10 kg/s. Assume constant specific heat for the air, value from Table A.5. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. Given: h 1 =293.6 kj/kg, s T 1 = kj/kg K Assumptions: Constant c pair Find: w C, w T, η TH The compression is isentropic: s T 2 = s T 1 + R ln P 2 P 1 = (0.287) ln(12) = T 2 =590K,h 2 =597.2 kj/kg 9

10 Energy equation with compressor work in: w C = 1 w 2 = h 2 h 1 = =303.6 kj/kg Like the compression, the expansion is isentropic: s 4 = s 3 s T 4 = s T 3 + R ln P 4 P 3 = (0.287) ln 1 12 = Energy equation with turbine work out: T 4 =734.8 K,h 4 =751.1 kj/kg w T = h 3 h 4 = =732kJ/kg Find energy rates by multiplying by mass flow rate: Ẇ C = ṁw C = 3036 kw = ẆC, Ẇ T = ṁw T = 7320 kw = ẆT Energy added by the combustion process: q H = h 3 h 2 = =885.9 kj/kg w NET = w T W C = =428.4 kj/kg η TH = w NET /q H =428.4/885.9 = = η TH The gas-turbine cycle shown in Fig. P12.31 is used as an automotive engine. In the first turbine, the gas expands to pressure P 5,justlowenoughforthisturbinetodrive the compressor. The gas is then expanded through the second turbine connected to the drive wheels. The data for the engine are shown in the figure and assume that all processes are ideal. Determine the intermediate pressure P 5,thenetspecificworkoutputoftheengine, and the mass flow rate through the engine. Find also the air temperature entering the burner T 3,andthethermalefficiencyoftheengine. Given: Assumptions: Ideal generator Find: P 5, w NET, ṁ, T 3, η TH Consider the compressor: Now consider the turbine: k 1 P2 k s 2 = s 1 T 2 = T 1 P 1 =300(6) =500.8 K w C = 1 w 2 = c p0 (T 2 T 1 )=1.004( ) = kj/kg w T 1 = w C =201.6 =c p0 (T 4 T 5 )=1.004(1600 T 5 ) T 5 = K k T5 s 5 = s 4 P 5 = P 4 T 4 k 1 P6 k s 6 = s 5 T 6 = T P 5 k =600 = 375 kpa = P =958.8 K

11 The second turbine gives the net work out: w T 2 = c p0 (T 5 T 6 )=1.004( ) = kj/kg Ideal generator: T 6 = T 3 =958.8 K ṁ = ẆNET/w T 2 =150/442.2 = kg/s =ṁ q H = c p0 (T 5 T 6 )=1.004( ) = kj/kg η TH = w NET /q H =442.2/643.8 = = η TH Consider an ideal air-standard Ericsson cycle that has an ideal regenerator as shown in Fig. P The high pressure is 1 MPa and the cycle efficiency is 70 %. Heat is rejected in the cycle at a temperature of 350 K, and the cycle pressure at the beginning of the isothermal compression process is 150 kpa. Determine the high temperature, the compressor work, and the turbine work per kilogram of air. Given: P 2 = P 3 =1MPa,T 1 = T 2 =350K,P 1 =150kPa,r P = P 2 /P 1 =10 Assumptions: Ideal regenerator Find: T H, w C, w T Ideal regenerator: 2q 3 = 4 q 1 q H = 3 q 4, w T = q H η TH = η CARNOT =1 T L /T H =0.7 T 3 = T 4 = T H =1167K P2 q L = w C = vdp= RT 1 ln =(0.287)(350) ln 1000 P = kj/kg = q L P4 w T = q H = vdp= RT 3 ln = kj/kg = w T An afterburner in a jet engine adds fuel after the turbine thus raising the pressure and temperature due to the energy of combustion. Assume a standard condition of 800 K, 250 kpa after the turbine into the nozzle that exhausts at 95 kpa. Assume the afterburner adds 475 kj/kg to that state with a rise in pressure for same specific volume, and neglect any upstream effects on the turbine. Find the nozzle exit velocity before and after the afterburner is turned on. Given: 1 T 1 =800K,P 1 =250kPa 2 P 2 = 95 kpa; After afterburner is turned on: 3 v 3 = v 1, 4 P 4 =95kPa Assumptions: Isentropic nozzle flow Find: For the nozzle: T 2 = T 1 (P 2 /P 1 ) k 1 k =(800)(95/250) =606.8 K Energy equation: (1/2)v2 2 = c p (T 1 T 2 ) v 2 = 2c p (T 1 T 2 )= (2)(1004)( ) = m/s =v 2 Now with the q AB at assumed constant volume gives the energy equation as T 3 = T 1 + q AB /c v = /0.717 = K P 3 11

12 v 3 = v 1 P 3 = P 1 (T 3 /T 1 )=250(1462.5/800) = kpa The expansion is isentropic, so from Eq. 8.23: T 4 = T 3 (P 4 /P 3 ) (k 1)/k =1462.5(95/457) =933.7 K v 2 = 2c p (T 1 T 2 )= (2)(1004)( ) = m/s =v 2 12

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16 Homework A 2 kg mixture of 25 % N 2,50%O 2 and 25 % CO 2 by mass is at 150 kpa and 300 K. Find the mixture gas constant and the total volume. Given: m =2kg,c N2 =0.25, c O2 =0.5, c CO2 =0.25 Assumptions: Find: R mix,v From Eq : R mix = c i R i =(0.25)(0.2968) + (0.5)(0.2598) + (0.25)(0.1889) = kj/kg K = R mix Ideal gas law: PV = mr mix T V = mr mixt P = (2 kg)( kj/kg K)(300 K) 150 kpa = m 3 = V 16

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19 Homework A new high-efficiency home heating system includes an air-to-air heat exchanger which uses energy from outgoing stale air to heat the fresh incoming air. If the outside ambient temperature is -10 Candtherelativehumidityis40%,howmuchwaterwillhave to be added to the incoming air, if it flows in at the rate of 1 m 3 /sandmusteventuallybe conditioned to 20 Cand40%relativehumidity? Given: φ 1 =0.4, T 1 = 10 C, T 2 =20 C, P g1 = kpa (from Table B.1.5) Assumptions: P 1 = P 2 =100kPa Find: ṁ liq,in Outside ambient air: P v1 =(φ 1 )(P g1 )=(0.40)(0.2602) = kpa P 1 = P 2 =100kPa,soP a1 = = kpa. ṁ a = P a1v 1 = (99.896)(1) = kg/s R a T 1 (0.287)(263.2) From Eq : w 1 =0.622 = On the outside: Eq : w 2 = P v2 =(φ 2 )(P g2 )=(0.4)(2.339) = kpa = Now applying the continuity equation to the water: ṁ liq,in = ṁ a (w 2 w 1 )=1.3228( ) = kg/s =24.9 kg/h =ṁ liq,in 19

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23 Homework Develop general expressions for u(t,v), h(t,v), s(t,v). That is, do not restrict your analysis to isothermal changes. In this problem, you may have to admit that c v = c v (T,v). Given: Assumptions: Find: u(t,v), h(t,v), s(t,v) Redlich-Kwong equation of state: From Eq. (14.30): = 2 1 P = RT v b P = RT T v v b a v(v + b)t 1/2 a v(v + b)t 3/2 u 2 u 1 = 2 C 1 v(t,v) dt + 2 3a 1 2v(v + b)t 1/2 C v ( ˆT )d ˆT 3a v2 + b v1 ln = u 2bT 1/2 2 u 1 v 1 + b We find the change in h from the change in u (dh = du + Pdv): h 2 h 1 = 2 1 C p ( ˆT )d ˆT 1 3a ln 2bT 1/2 Entropy follows from Eq. (14.35): 2 s 2 s 1 = C v (T,v) dt 2 T + s 2 s 1 = 2 1 v 2 v2 + b 1 v 2 v1 v 1 + b R v b + a/2 v(v + b)t 3/2 + P 2 v 2 P 1 v 1 dv d ˆT C v ( ˆT,v) ˆT + R ln v 2 b v 1 b a v2 + b v1 ln 2bt3/2 v 1 + b v 2 23

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25 3. Consider a thermodynamic system in which there are two reversible work modes: compression and electrical. So take the version of the texts Eq. (4.16) giving dw to be dw = PdV EdZ, where E is the electrical potential difference and dz is the amount of charge that flows into the system. Extend the Gibbs equation to account for electrical work. Find the Legendre transformation which renders the independent variables to be P, E, and T and show how the other variables can be determined as functions of these independent variables. Find all Maxwell relations associated with this Legendre transformation. Gibbs equation: Knowing δq = Tds, Legendre transformation: du = δq δw du = TdS PdV + EdZ (1) ψ 1 = T,ψ 2 = P, ψ 3 = E.χ 1 = S, χ 2 = V,χ 3 = Z F 1 = U ψ 1 χ 1 = U TS F 2 = U ψ 2 χ 2 = U PV F 3 = U ψ 3 χ 3 = U EZ F 1,2,3 = U TS + PV EZ df 1,2,3 = du TdS SdT + PdV + VdP EdZ ZdE Re-arranging this equation and recalling Eq. (1), df 1,2,3 + TdS + SdT PdV VdP + EdZ + ZdE = du = TdS PdV + EdZ Canceling like terms and solving for df 1,2,3 shows that and that F 1,2,3 = F 1,2,3 (T,P,E). From Eq. (2): df 1,2,3 = SdT + VdP ZdE, (2) S = V = Z = F1,2,3 T F1,2,3 P F1,2,3 25 E P,E T,E T,P

26 Taking mixed partials of the right-hand sides of the three equations above results in the Maxwell relations: 2 F 1,2,3 S T P = 2 F 1,2,3 V S V, P P,E P T = = T T,E P P,E T T,E 2 F 1,2,3 S T E =, E P,E 2 F 1,2,3 V P E =, E T,E 2 F 1,2,3 E T = 2 F 1,2,3 E T = Z T P,T Z P P,T S = E P,E V = E T,E Z T P,T Z P P,T 26

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33 Homework Consider an isochoric, isothermal reaction in which n H2 =1kmole,n H =0kmoleat t =0,andforwhichT =5200KandP =1500kPa. Consider a Hydrogen dissociation reaction, H 2 + H 2 2H + H 2 For this reaction a = cm3 K 1/2 cal, β =0.5, and Ē =92600 (a) Formulate the mole s mole reaction kinetics in the form d ρ h2 = f( ρ H2 ) dt N d ρ h2 = ν 1 at β e Ē RT ρ ν k 1 1 N ρ ν k dt k c Therefore: ( ρ H 2 ) t=0 = P 0 RT = k=1 k= kpa kmol = K m kj kmol K V = (n H 2 ) t=0 ( ρ H2 ) t=0 =28.82 m 3 dn H2 =1/2n H (n H2 (n H2 ) t=0 )=1/2(n H2 (n H2 ) t=0 ) d ρ h2 dt k c = d ρ h 2 dt = ν H2 at β e Ē RT ( ρh2 ) ν H 2 ( ρ H ) ν H 1 1 ( ρ H2 ) ν H 2 ( ρh ) ν H k c νh 2 =2,νh =0,ν H2 = 1,ν H =2 d ρ h2 = at β e Ē RT ( ρh2 ) ( ρ H2 ) 2 dt k c ( ρ H ) k(t )=at β e Ē ĒT = N P0 i ν i e G ĒT, G = kj RT kmol k c = = ( ρ H2 ) [2( ρ H 2 )] 2 ρ H2 (b) Find all equilibria. Equilibria are located at the points where d ρ h 2 dt ρ H2 = {0, , } =0. (c) Ascertain the stability of each equilibrium point. For stability, d2 ρ h2 Therefore, dt 2 ρ =0 unstable ρ = stable ρ = unstable and non-physical must be negative. 33

### SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition

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