The mass of argon in the tank.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

(b)

To determine

The final pressure.

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

(c)

To determine

The heat transfer.

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain h.

h=hideal−ZhRTcr (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

h2−h1=(h2−h1)ideal−(Zh2−Zh1)RTcr (VII)

The enthalpy difference at ideal gas state is expressed as follows.

(h2−h1)ideal=cp(T2−T1) (VIII)

Here, the specific heat at constant pressure is cp.

Write the energy balance equation for the system (piston-cylinder).

Ein−Eout=ΔEsystemQin−0=ΔuQin=m(u2−u1) (IX)

Here, the net energy in is Ein, the net energy out is Eout and the change in net energy of the system is ΔEsystem.

The internal energy is expressed as follows.

u=h−Pν

Here, the enthalpy is h, the pressure is P, and the volume is ν.

The change in internal energy is expressed as follows.

u2−u1=h2−h1−(P2ν2−P1ν1)=h2−h1−(Z2RT2−Z1RT1)=h2−h1−R(Z2T2−Z1T1) (X)

Substitute h2−h1−R(Z2T2−Z1T1) for u2−u1 in Equation (X).

Qin=m[h2−h1−R(Z2T2−Z1T1)] (XI)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) is 0.5203 kJ/kg⋅K.

Conclusion:

Substitute 0.5203 kJ/kg⋅K for cp, 0 °C for T2, and −100 °C in Equation (VIII).

(h2−h1)ideal=0.5203 kJ/kg⋅K[0 °C−(−100 °C)]=0.5203 kJ/kg⋅K[(0 +273) K−(−100 +273) K]=52.03 kJ/kg

Substitute 52.03 kJ/kg for (h2−h1)ideal, 0 for Zh2, 0.18 for Zh1, 0.2081 kPa⋅m3/kg⋅K for R, and 151 K for Tcr in Equation (VII).

h2−h1={52.03 kJ/kg−[(0−0.18)(0.2081 kPa⋅m3/kg⋅K)(151 K)]}=52.03 kJ/kg+5.6561 kJ/kg=57.6861 kJ/kg≃57.69 kJ/kg

Substitute 57.69 kJ/kg for h2−h1, 0.2081 kPa⋅m3/kg⋅K for R, 0.99 for Z2, 273 K for T2, 0.95 for Z1, and 173 K for T1 in Equation (X).

u2−u1={57.69 kJ/kg−(0.2081 kPa⋅m3/kg⋅K)[(0.99×273 K)−(0.95×173 K)]}=57.69 kJ/kg−22.0419 kJ/kg=35.648 kJ/kg

Substitute 35.1 kg for m and 35.648 kJ/kg for u2−u1 in Equation (XI).

Qin=35.1 kg(35.648 kJ/kg)=1251.2465 kJ≃1251 kJ

Thus, the heat transfer is 1251 kJ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

3.) 13.152* - A sphere (m = 2 kg) is connected to a fixed point O by an inextensible cord (L = 1.2 m). The sphere rests on a frictionless horizontal surface at a distance of x = 0.5 m from O when it is given a velocity (V.) in a direction perpendicular to OA. It moves freely until it reach A', when the cord becomes taut. Find the maximum possible velocity v. if the impulse of the force exerted on the cord is not to exceed F = 3Ns.

4.) 13.165* - Two billiard balls A & B (same mass m & diameter d = 2.37") are as shown. Ball A has velocity V₁ = 3 ft/s when it strikes B, which is at rest. B then moves in the x direction after impact. Find (a) angle 0, (b) the velocity of B after impact (VB'). Positional variables are x = 6" and y = 10", and e = 0.9.

We will consider a linear system and a nonlinear system under uncertainty, each expressed in the form of a set of stochastic differential equation (SDE) as follows: = da (Ax+ Bu)dt + Gdw, dx = f(x,u,t)dt+Gdw, (1) (2) where x is the state, u is the control, and dw is a differential increment of standard Brownian motion, i.e., E[dw] = 0 and E [dw(t)dw(t)] = dt-1. In this problem, we consider the linear SDE, Eq. (1), with a very simple model where x = R², u = [0,0] (no control), and dw R². The matrices A, B, and G are given as follows: A=02x2, B=02x2; G = [000] (3) where σp Є R represents the degree of the uncertainty, and let us take σ₁ = 2 and σ2 = 3. Assume that the initial state is deterministic and e(t = 0) = [0,0]. Take the following steps to simulate the given SDE for 1 € [0, 1]: (a): Consider the increments of w between each time interval [+1). Derive the analytical expression of Aw using w~N(02, 12), where Aw, w(tk+1) — w(tk). (c): Derive the approximate continuous-time EoM from…

Answer 2

Question

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

(b)

To determine

The final pressure.

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

(c)

To determine

The heat transfer.

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain h.

h=hideal−ZhRTcr (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

h2−h1=(h2−h1)ideal−(Zh2−Zh1)RTcr (VII)

The enthalpy difference at ideal gas state is expressed as follows.

(h2−h1)ideal=cp(T2−T1) (VIII)

Here, the specific heat at constant pressure is cp.

Write the energy balance equation for the system (piston-cylinder).

Ein−Eout=ΔEsystemQin−0=ΔuQin=m(u2−u1) (IX)

Here, the net energy in is Ein, the net energy out is Eout and the change in net energy of the system is ΔEsystem.

The internal energy is expressed as follows.

u=h−Pν

Here, the enthalpy is h, the pressure is P, and the volume is ν.

The change in internal energy is expressed as follows.

u2−u1=h2−h1−(P2ν2−P1ν1)=h2−h1−(Z2RT2−Z1RT1)=h2−h1−R(Z2T2−Z1T1) (X)

Substitute h2−h1−R(Z2T2−Z1T1) for u2−u1 in Equation (X).

Qin=m[h2−h1−R(Z2T2−Z1T1)] (XI)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) is 0.5203 kJ/kg⋅K.

Conclusion:

Substitute 0.5203 kJ/kg⋅K for cp, 0 °C for T2, and −100 °C in Equation (VIII).

(h2−h1)ideal=0.5203 kJ/kg⋅K[0 °C−(−100 °C)]=0.5203 kJ/kg⋅K[(0 +273) K−(−100 +273) K]=52.03 kJ/kg

Substitute 52.03 kJ/kg for (h2−h1)ideal, 0 for Zh2, 0.18 for Zh1, 0.2081 kPa⋅m3/kg⋅K for R, and 151 K for Tcr in Equation (VII).

h2−h1={52.03 kJ/kg−[(0−0.18)(0.2081 kPa⋅m3/kg⋅K)(151 K)]}=52.03 kJ/kg+5.6561 kJ/kg=57.6861 kJ/kg≃57.69 kJ/kg

Substitute 57.69 kJ/kg for h2−h1, 0.2081 kPa⋅m3/kg⋅K for R, 0.99 for Z2, 273 K for T2, 0.95 for Z1, and 173 K for T1 in Equation (X).

u2−u1={57.69 kJ/kg−(0.2081 kPa⋅m3/kg⋅K)[(0.99×273 K)−(0.95×173 K)]}=57.69 kJ/kg−22.0419 kJ/kg=35.648 kJ/kg

Substitute 35.1 kg for m and 35.648 kJ/kg for u2−u1 in Equation (XI).

Qin=35.1 kg(35.648 kJ/kg)=1251.2465 kJ≃1251 kJ

Thus, the heat transfer is 1251 kJ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

(b)

To determine

The final pressure.

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

(c)

To determine

The heat transfer.

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain h.

h=hideal−ZhRTcr (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

h2−h1=(h2−h1)ideal−(Zh2−Zh1)RTcr (VII)

The enthalpy difference at ideal gas state is expressed as follows.

(h2−h1)ideal=cp(T2−T1) (VIII)

Here, the specific heat at constant pressure is cp.

Write the energy balance equation for the system (piston-cylinder).

Ein−Eout=ΔEsystemQin−0=ΔuQin=m(u2−u1) (IX)

Here, the net energy in is Ein, the net energy out is Eout and the change in net energy of the system is ΔEsystem.

The internal energy is expressed as follows.

u=h−Pν

Here, the enthalpy is h, the pressure is P, and the volume is ν.

The change in internal energy is expressed as follows.

u2−u1=h2−h1−(P2ν2−P1ν1)=h2−h1−(Z2RT2−Z1RT1)=h2−h1−R(Z2T2−Z1T1) (X)

Substitute h2−h1−R(Z2T2−Z1T1) for u2−u1 in Equation (X).

Qin=m[h2−h1−R(Z2T2−Z1T1)] (XI)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) is 0.5203 kJ/kg⋅K.

Conclusion:

Substitute 0.5203 kJ/kg⋅K for cp, 0 °C for T2, and −100 °C in Equation (VIII).

(h2−h1)ideal=0.5203 kJ/kg⋅K[0 °C−(−100 °C)]=0.5203 kJ/kg⋅K[(0 +273) K−(−100 +273) K]=52.03 kJ/kg

Substitute 52.03 kJ/kg for (h2−h1)ideal, 0 for Zh2, 0.18 for Zh1, 0.2081 kPa⋅m3/kg⋅K for R, and 151 K for Tcr in Equation (VII).

h2−h1={52.03 kJ/kg−[(0−0.18)(0.2081 kPa⋅m3/kg⋅K)(151 K)]}=52.03 kJ/kg+5.6561 kJ/kg=57.6861 kJ/kg≃57.69 kJ/kg

Substitute 57.69 kJ/kg for h2−h1, 0.2081 kPa⋅m3/kg⋅K for R, 0.99 for Z2, 273 K for T2, 0.95 for Z1, and 173 K for T1 in Equation (X).

u2−u1={57.69 kJ/kg−(0.2081 kPa⋅m3/kg⋅K)[(0.99×273 K)−(0.95×173 K)]}=57.69 kJ/kg−22.0419 kJ/kg=35.648 kJ/kg

Substitute 35.1 kg for m and 35.648 kJ/kg for u2−u1 in Equation (XI).

Qin=35.1 kg(35.648 kJ/kg)=1251.2465 kJ≃1251 kJ

Thus, the heat transfer is 1251 kJ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

(b)

To determine

The final pressure.

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

(c)

To determine

The heat transfer.

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain h.

h=hideal−ZhRTcr (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

h2−h1=(h2−h1)ideal−(Zh2−Zh1)RTcr (VII)

The enthalpy difference at ideal gas state is expressed as follows.

(h2−h1)ideal=cp(T2−T1) (VIII)

Here, the specific heat at constant pressure is cp.

Write the energy balance equation for the system (piston-cylinder).

Ein−Eout=ΔEsystemQin−0=ΔuQin=m(u2−u1) (IX)

Here, the net energy in is Ein, the net energy out is Eout and the change in net energy of the system is ΔEsystem.

The internal energy is expressed as follows.

u=h−Pν

Here, the enthalpy is h, the pressure is P, and the volume is ν.

The change in internal energy is expressed as follows.

u2−u1=h2−h1−(P2ν2−P1ν1)=h2−h1−(Z2RT2−Z1RT1)=h2−h1−R(Z2T2−Z1T1) (X)

Substitute h2−h1−R(Z2T2−Z1T1) for u2−u1 in Equation (X).

Qin=m[h2−h1−R(Z2T2−Z1T1)] (XI)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) is 0.5203 kJ/kg⋅K.

Conclusion:

Substitute 0.5203 kJ/kg⋅K for cp, 0 °C for T2, and −100 °C in Equation (VIII).

(h2−h1)ideal=0.5203 kJ/kg⋅K[0 °C−(−100 °C)]=0.5203 kJ/kg⋅K[(0 +273) K−(−100 +273) K]=52.03 kJ/kg

Substitute 52.03 kJ/kg for (h2−h1)ideal, 0 for Zh2, 0.18 for Zh1, 0.2081 kPa⋅m3/kg⋅K for R, and 151 K for Tcr in Equation (VII).

h2−h1={52.03 kJ/kg−[(0−0.18)(0.2081 kPa⋅m3/kg⋅K)(151 K)]}=52.03 kJ/kg+5.6561 kJ/kg=57.6861 kJ/kg≃57.69 kJ/kg

Substitute 57.69 kJ/kg for h2−h1, 0.2081 kPa⋅m3/kg⋅K for R, 0.99 for Z2, 273 K for T2, 0.95 for Z1, and 173 K for T1 in Equation (X).

u2−u1={57.69 kJ/kg−(0.2081 kPa⋅m3/kg⋅K)[(0.99×273 K)−(0.95×173 K)]}=57.69 kJ/kg−22.0419 kJ/kg=35.648 kJ/kg

Substitute 35.1 kg for m and 35.648 kJ/kg for u2−u1 in Equation (XI).

Qin=35.1 kg(35.648 kJ/kg)=1251.2465 kJ≃1251 kJ

Thus, the heat transfer is 1251 kJ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

(b)

To determine

The final pressure.

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

(c)

To determine

The heat transfer.

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain h.

h=hideal−ZhRTcr (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

h2−h1=(h2−h1)ideal−(Zh2−Zh1)RTcr (VII)

The enthalpy difference at ideal gas state is expressed as follows.

(h2−h1)ideal=cp(T2−T1) (VIII)

Here, the specific heat at constant pressure is cp.

Write the energy balance equation for the system (piston-cylinder).

Ein−Eout=ΔEsystemQin−0=ΔuQin=m(u2−u1) (IX)

Here, the net energy in is Ein, the net energy out is Eout and the change in net energy of the system is ΔEsystem.

The internal energy is expressed as follows.

u=h−Pν

Here, the enthalpy is h, the pressure is P, and the volume is ν.

The change in internal energy is expressed as follows.

u2−u1=h2−h1−(P2ν2−P1ν1)=h2−h1−(Z2RT2−Z1RT1)=h2−h1−R(Z2T2−Z1T1) (X)

Substitute h2−h1−R(Z2T2−Z1T1) for u2−u1 in Equation (X).

Qin=m[h2−h1−R(Z2T2−Z1T1)] (XI)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) is 0.5203 kJ/kg⋅K.

Conclusion:

Substitute 0.5203 kJ/kg⋅K for cp, 0 °C for T2, and −100 °C in Equation (VIII).

(h2−h1)ideal=0.5203 kJ/kg⋅K[0 °C−(−100 °C)]=0.5203 kJ/kg⋅K[(0 +273) K−(−100 +273) K]=52.03 kJ/kg

Substitute 52.03 kJ/kg for (h2−h1)ideal, 0 for Zh2, 0.18 for Zh1, 0.2081 kPa⋅m3/kg⋅K for R, and 151 K for Tcr in Equation (VII).

h2−h1={52.03 kJ/kg−[(0−0.18)(0.2081 kPa⋅m3/kg⋅K)(151 K)]}=52.03 kJ/kg+5.6561 kJ/kg=57.6861 kJ/kg≃57.69 kJ/kg

Substitute 57.69 kJ/kg for h2−h1, 0.2081 kPa⋅m3/kg⋅K for R, 0.99 for Z2, 273 K for T2, 0.95 for Z1, and 173 K for T1 in Equation (X).

u2−u1={57.69 kJ/kg−(0.2081 kPa⋅m3/kg⋅K)[(0.99×273 K)−(0.95×173 K)]}=57.69 kJ/kg−22.0419 kJ/kg=35.648 kJ/kg

Substitute 35.1 kg for m and 35.648 kJ/kg for u2−u1 in Equation (XI).

Qin=35.1 kg(35.648 kJ/kg)=1251.2465 kJ≃1251 kJ

Thus, the heat transfer is 1251 kJ.

Answer 6

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

Answer 7

Chapter 12.6, Problem 99RP

(a)

To determine

The mass of argon in the tank.

Answer 8

(a)

Expert Solution

Answer to Problem 99RP

The mass of argon in the tank is 35.1 kg.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write formula for specific volume (v) in terms of compressibility factor (Z).

v=ZRTP (I)

Here, the gas constant of argon is R, the temperature is T, and the pressure is P.

Write formula for mass of the argon present in the tank.

m=νv (II)

Here, the volume of argon in the tank is ν.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of propane gas is as follows.

Tcr=151 KPcr=4.86 MPa

Refer Table A-2(a), “Ideal-gas specific heats of various common gases”.

The gas constant (R) of argon is 0.2081 kPa⋅m3/kg⋅K.

The reduced pressure (PR1) and temperature (TR1) at initial state is expressed as follows.

TR1=T1Tcr=(−100+273) K151 K=1.146

PR1=P1Pcr=1 MPa4.86 MPa=0.206

At initial:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.95.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z1) corresponding the reduced pressure (PR1) and reduced temperature (TR1) is 0.18.

Conclusion:

Substitute 0.95 for Z, 0.2081 kPa⋅m3/kg⋅K for R, −100 °C for T, and 1 MPa for P in Equation (I).

v=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 °C)1 MPa=(0.95)(0.2081 kPa⋅m3/kg⋅K)(−100 +273) K1 MPa×103 kPa1 MPa=34.2505 kPa⋅m3/kg1000 kPa=0.0342 m3/kg

Substitute 1.2 m3 for ν and 0.0342 m3/kg for v in Equation (II).

m= 1.2 m30.0342 m3/kg=35.0877 kg≃35.1 kg

Thus, the mass of argon in the tank is 35.1 kg.

Answer 13

(b)

To determine

The final pressure.

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

Answer 14

(b)

To determine

The final pressure.

Answer 15

(b)

Expert Solution

Answer to Problem 99RP

The final pressure is 1531 kPa.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The reduced pressure (PR2) and temperature (TR2) at final state is expressed as follows.

TR2=T2Tcr=(0+273) K151 K=1.808

PR2=P2Pcr=P24.86 MPa×1000 kPa1 MPa=P24860 kPa (III)

Write the formula for reduced specific volume.

vR2=v2RTcr/Pcr (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

v2=v1=0.0342 m3/kg.

Substitute 0.0342 m3/kg for v2, 0.2081 kPa⋅m3/kg⋅K for R, 151 K for Pcr, and 4.86 MPa for Pcr in Equation (IV).

vR2=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4.86 MPa=0.0342 m3/kg(0.2081 kPa⋅m3/kg⋅K)151 K4860 kPa=0.0342 m3/kg0.00646 m3/kg=5.29

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor (Z2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.99.

The reduced pressure (PR2) corresponding the reduced volume (vR2) and reduced temperature (TR2) is 0.315.

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor (Zh2) corresponding the reduced pressure (PR2) and reduced temperature (TR2) is 0.00.

Substitute 0.315 for PR2 in Equation (III).

0.315=P24860 kPaP2=(0.315)4860 kPaP2=1530.9 kPaP2=1531 kPa

Thus, the final pressure is 1531 kPa.

Answer 20

(c)

To determine

The heat transfer.

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain h.

h=hideal−ZhRTcr (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

h2−h1=(h2−h1)ideal−(Zh2−Zh1)RTcr (VII)

The enthalpy difference at ideal gas state is expressed as follows.

(h2−h1)ideal=cp(T2−T1) (VIII)

Here, the specific heat at constant pressure is cp.

Write the energy balance equation for the system (piston-cylinder).

Ein−Eout=ΔEsystemQin−0=ΔuQin=m(u2−u1) (IX)

Here, the net energy in is Ein, the net energy out is Eout and the change in net energy of the system is ΔEsystem.

The internal energy is expressed as follows.

u=h−Pν

Here, the enthalpy is h, the pressure is P, and the volume is ν.

The change in internal energy is expressed as follows.

u2−u1=h2−h1−(P2ν2−P1ν1)=h2−h1−(Z2RT2−Z1RT1)=h2−h1−R(Z2T2−Z1T1) (X)

Substitute h2−h1−R(Z2T2−Z1T1) for u2−u1 in Equation (X).

Qin=m[h2−h1−R(Z2T2−Z1T1)] (XI)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) is 0.5203 kJ/kg⋅K.

Conclusion:

Substitute 0.5203 kJ/kg⋅K for cp, 0 °C for T2, and −100 °C in Equation (VIII).

(h2−h1)ideal=0.5203 kJ/kg⋅K[0 °C−(−100 °C)]=0.5203 kJ/kg⋅K[(0 +273) K−(−100 +273) K]=52.03 kJ/kg

Substitute 52.03 kJ/kg for (h2−h1)ideal, 0 for Zh2, 0.18 for Zh1, 0.2081 kPa⋅m3/kg⋅K for R, and 151 K for Tcr in Equation (VII).

h2−h1={52.03 kJ/kg−[(0−0.18)(0.2081 kPa⋅m3/kg⋅K)(151 K)]}=52.03 kJ/kg+5.6561 kJ/kg=57.6861 kJ/kg≃57.69 kJ/kg

Substitute 57.69 kJ/kg for h2−h1, 0.2081 kPa⋅m3/kg⋅K for R, 0.99 for Z2, 273 K for T2, 0.95 for Z1, and 173 K for T1 in Equation (X).

u2−u1={57.69 kJ/kg−(0.2081 kPa⋅m3/kg⋅K)[(0.99×273 K)−(0.95×173 K)]}=57.69 kJ/kg−22.0419 kJ/kg=35.648 kJ/kg

Substitute 35.1 kg for m and 35.648 kJ/kg for u2−u1 in Equation (XI).

Qin=35.1 kg(35.648 kJ/kg)=1251.2465 kJ≃1251 kJ

Thus, the heat transfer is 1251 kJ.

Answer 21

(c)

To determine

The heat transfer.

Answer 22

(c)

Expert Solution

Answer to Problem 99RP

The heat transfer is 1251 kJ.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write formula for enthalpy departure factor (Zh).

Zh=(hideal−h)T,PRTcr (V)

Here, the enthalpy at ideal gas state is hideal, the enthalpy and normal state is h, the gas constant of propane is R, and the critical temperature is Tcr; The subscripts T,P indicates the correspondence of given temperature and pressure.