THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 7.13, Problem 175RP

A piston–cylinder device contains steam that undergoes a reversible thermodynamic cycle. Initially the steam is at 400 kPa and 350°C with a volume of 0.5 m3. The steam is first expanded isothermally to 150 kPa, then compressed adiabatically to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the net work and heat transfer for the cycle after you calculate the work and heat interaction for each process.

Expert Solution & Answer
Check Mark
To determine

The net work done and net heat transfer by piston cylinder device.

Answer to Problem 175RP

The net work done by piston cylinder device is 20.8kJ.

The net heat transfer by piston cylinder device is 20.8kJ.

Explanation of Solution

Write the expression to calculate the mass of the steam in the cylinder.

m=ν1v1 (I)

Here, mass of the steam is m, initial volume is ν1, and initial molar volume is v1.

Write the expression for the volume at state 3.

ν3=mv3 (II)

Here, volume at state 3 is ν3, mass of the steam is m, and molar volume at state 3 is v3.

Write the expression to calculate the heat transfer in for the isothermal expansion process 1-2.

Qin,12=mT1(s2s1) (III)

Here, heat transfer in for process 1-2 is Qin,12, entropy at state1 is s1 and entropy at state 2 is s2.

Write the expression to calculate the work done out for the isothermal expansion process 1-2.

Wout,12=Qin,12m(u2u1) (IV)

Here, work done out for process 1-2 is Wout,12, internal energy state 1 is u1 and internal energy state 2 is u3.

Write the expression to calculate the work done in for the isentropic compression process 2-3.

Win,23=m(u3u2) (V)

Here, work done in for process 2-3 is Win,23 and internal energy state 3 is u3.

Write the expression to calculate the work done in for the constant pressure compression process 3-1.

Win,31=P3(ν3ν1) (VI)

Here, work done in for process 3-1 is Win,31, volume at state 1 is ν1 and volume at state 3 is ν3.

Write the expression to calculate the heat transfer out for the constant pressure compression process 3-1.

Qout,31=Win,31m(u1u3) (VII)

Here, heat transfer out for the process 3-1.

Write the expression to calculate the net work done by piston cylinder device.

Wnet,in=Win,31+Win,23Wout,12 (VIII)

Here, the net work done is Wnet,in

Write the expression to calculate the net heat transfer by piston cylinder device.

Qnet,in=Qin,12Qout,31 (IX)

Here, the net heat transfer is Qnet,in

Conclusion:

Refer to Table A-6, “Superheated water”.

Obtain the value of internal energy state 1 (u1) at the initial pressure of (P1) of 400kPa and at the initial temperature of (T1) of 350°C by using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (X)

Here, the variables denoted by x and y are temperature and internal energy.

Show temperature and initial internal energy values from the Table A-6.

Temperature (T1),in °CInternal energy (u1), in kJ/kg
3002805.1
350?
4002964.9

Substitute 300 for x1, 350 for x2, 400 for x3, 2805.1 for y1, and 2964.9 for y3 in Equation (X).

y2=(350300)(2964.92805.1)(400300)+2805.1=2885

The value of internal energy state 1 (u1) at the initial pressure of (P1) of 400kPa and at the initial temperature of (T1) of 350°C is 2885kJ/kg.

Refer to Table A-6, “Superheated water”.

Obtain the value of initial molar volume (v1) at the initial pressure of (P1) of 400kPa and at the initial temperature of (T1) of 350°C by using interpolation method.

Show temperature and molar volume values from the Table A-6.

Temperature (T1),in °CMolar volume (v1) , in m3/kg
3000.65489
350?
4000.77265

Substitute 300 for x1, 350 for x2, 400 for x3, 0.65489 for y1, and 0.77265 for y3 in Equation (X).

y2=(350300)(0.772650.65489)(400300)+0.65489=0.713770

The value of initial molar volume (v1) at the initial pressure of (P1) of 400kPa and at the initial temperature of (T1) of 350°C is 0.713770m3/kg.

Refer to Table A-6, “Superheated water”.

Obtain the value of entropy at state1 (s1) at the initial pressure of (P1) of 400kPa and at the initial temperature of (T1) of 350°C by using interpolation method.

Show temperature and entropy values from the Table A-6.

Temperature (T1),in °CEntropy (s1) , in kJ/kgK
3007.5677
350?
4007.9003

Substitute 300 for x1, 350 for x2, 400 for x3, 7.5677 for y1, and 7.9003 for y3 in Equation (X).

y2=(350300)(7.90037.5677)(400300)+7.5677=7.734

The value of entropy at state1 (s1) at the initial pressure of (P1) of 400kPa and at the initial temperature of (T1) of 350°C is 7.734kJ/kgK.

Similarly,

obtain the values of internal energy at state 2 (u2) , entropy at state 2 (s2) at the pressure of (P2) of 150kPa and at the initial temperature of (T1) of 350°C as 2888.0kJ/kg and 8.19683kJ/kgK respectively by interpolation method.

Obtain the values of internal energy at state 3 (u3) , molar volume at state 3 (v3) at the pressure of (P3) of 400kPa and at the entropy at state 2 (s2) of 8.19683kJ/kgK as 0.89148m3/kg respectively by interpolation method.

Substitute 0.5m3 for ν1 and 0.713770m3/kg for v1 in Equation (I).

m=0.5m30.713770m3/kg=0.700506kg

Substitute 0.700506kg for m and 0.89148m3/kg for v3 in Equation (II).

ν3=(0.700506kg)(0.89148m3/kg)=0.624487m3

Substitute 0.700506kg for m, 350° C for T1, 8.1983kJ/kgK for s2 and 7.734kJ/kgK for s1 in Equation (III).

Qin,12=(0.700506kg)(350° C)(8.1983kJ/kgK7.734kJ/kgK)=(0.700506kg)(350+273)K(8.1983kJ/kgK7.734kJ/kgK)=202.6kJ

Substitute 202.6kJ for Qin,12, 0.700506kg for m , 2888kJ/kg for u2 and 2885kJ/kg for u1 in Equation (IV).

Wout,12=202.6kJ(0.700506kg)(2888kJ/kg2885kJ/kg)=200.49kJ

Substitute 0.700506kg for m, 3132.9kJ/kg for u3 and 2888kJ/kg for u2 in Equation (V).

Win,23=(0.700506kg)(3132.9kJ/kg2888kJ/kg)=171.55kJ

The heat transfer during the process is zero, since isentropic compression process, entropy remains constant.

Substitute 400kPa for P3, 0.6243m3 for ν3 and 0.5m3 for ν1 in Equation (VI).

Win,31=(400kPa)(0.6243m30.5m3)=49.72kJ

Substitute 49.72kJ for Win,31, 0.700506kg for m, 2885kJ/kg for u1 and 3132.9kJ/kg for u3 in Equation (VII).

Qout,31=49.72kJ(0.700506kg)(2885kJ/kg3132.9kJ/kg)=223.38kJ

Substitute 49.72kJ for Win,31, 171.55kJ for Win,23 and 200.49kJ for Wout,12 in Equation (VIII).

Wnet,in=49.72kJ+171.55kJ200.49kJ=20.78kJ=20.8kJ

Thus, the net work done by piston cylinder device is 20.8kJ.

Substitute 202.6kJ for Qin,12, and 223.38kJ for Qout,31 in Equation (IX).

Qnet,in=202.6kJ223.38kJ=20.78kJ=20.8kJ

The negative sign indicates that the heat transfer occurs from system to surroundings.

Thus, the net heat transfer by piston cylinder device is 20.8kJ.

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Chapter 7 Solutions

THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<

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