THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
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Chapter 8.8, Problem 77P

a)

To determine

The maximum work potential (Φ1) of the helium at the initial state

a)

Expert Solution
Check Mark

Answer to Problem 77P

The maximum work potential (Φ1) of the helium at the initial state is 5.27kJ_.

Explanation of Solution

Write the expression for the initial mass (m1) of helium in the cylinder.

m1=P1VRT1 (I)

Here, initial pressure is P1, initial volume of the cylinder is V1, mass of helium gas is m and gas constant of helium gas is R.

Write the expression for the final mass (m2) of helium in the cylinder.

m2=me=m12 . (II)

Here, exit mass of helium from the cylinder is me.

Write the expression for the initial specific volume (v1) of helium.

v1=RT1P1 . (III)

Write the expression for the dead-state volume (v0) of helium.

v0=RT0P0 . (IV)

Here, dead-state temperature is T0 and dead-state pressure is P0.

Write the expression for the difference between initial entropy (s1) and dead-state entropy (s0) as;

s1s0=cplnT1T0RlnP1P0 . (V)

Here, specific heats of helium is cp.

Write the expression for the maximum work potential (Φ1) of the helium at the initial state.

Φ1=m1[P0(v1v0)T0(s1s0)] . (VI)

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the value of R for helium as 2.0769kPam3/kgK(2.0769kJ/kgK).

Substitute 200kPa for P1, 0.12m3 for V1, 20°C for T1 and 2.0769kPam3/kgK for R in equation (I).

m1=(200kPa)(0.12m3)(2.0769kPam3/kgK)(20°C)=24kPam3(2.0769kPam3/kgK)(20+273)K=24kPam3(2.0769kPam3/kgK)(293K)=0.03944kg

Substitute 0.03944kg for m1 in Equation (II).

m2=0.03944kg2=0.01972kg

Substitute 2.0769kPam3/kgK for R, 20°C for T1, and 200kPa for P1 in equation (III)

v1=(2.0769kPam3/kgK)(20°C)200kPa=(2.0769kPam3/kgK)(20+273)K200kPa=(2.0769kPam3/kgK)(293K)200kPa=3.043m3/kg

Substitute 2.0769kPam3/kgK for R, 20°C for T0, and 95kPa for P0 in equation (IV).

v0=(2.0769kPam3/kgK)(20°C)95kPa=(2.0769kPam3/kgK)(20+273)K95kPa=(2.0769kPam3/kgK)(293K)95kPa=6.406m3/kg

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the value of cp for helium as 5.1926kJ/kgK

Substitute 5.1926kJ/kgK for cp, 2.0769kPam3/kgK for R, 20°C for T1andT0 respectively, 95kPa for P0, and 200kPa for P1 in Equation (V).

s1s0=(5.1926kJ/kgK)(ln20°C20°C)(2.0769kPam3/kgK)(ln200kPa95kPa)=[(5.1926kJ/kgK)(0)](2.0769kPam3/kgK)(0.7444)=1.546kPam3/kgK

Substitute 0.03944kg for m1, 20°C for T0, 1.546kPam3/kgK for s1s0, 95kPa for P0, 3.043m3/kg for v1 and 6.406m3/kg for v2 in equation (VI).

Φ1=(0.03944kg)[95kPa(3.0436.406)m3/kg20°C(1.546kPam3/kgK)]=(0.03944kg)[95kPa(3.363m3/kg)(20+273)K(1.546kPam3/kgK)]=(0.03944kg)[319.485kPam3/kg(293K)(1.546kPam3/kgK)]=(0.03944kg)(319.485kPam3/kg+452.978kPam3/kg)

=(0.03944kg)(133.493kPam3/kg)=5.27kPam3(kJkPam3)=5.27kJ

Thus, the maximum work potential (Φ1) of the helium at the initial state is 5.27kJ_.

b)

To determine

The exergy destroyed during the process (Xdestroyed)

b)

Expert Solution
Check Mark

Answer to Problem 77P

The exergy destroyed during the process (Xdestroyed) is 0_.

Explanation of Solution

Write the mass balance equation for the system as,

minmout=Δmsystemme=m1m2 . (VII)

Here, mass of the helium at the inlet of cylinder is min, mass of the helium coming out from the cylinder is mout, net mass of the helium within the system is Δmsystem, mass of helium at the exit is me, initial and final mass of the helium is m1andm2 respectively.

Write the Energy balance for the uniform flow system as;

EinEout=ΔEsystemQinmehe+Wb,in=m2u2m1u1 . (VIII)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem.heat transfer during the process is Qin, enthalpy of helium at exit is he, boundary work done inside the cylinder is Wb,in, initial internal energy is u1 and final internal energy is u2.

Combine Equation (VII) and Equation (VIII) to get,

Qin=(m1m2)he+m2u2m1u1Wb,in=(m1m2)he+m2h2m1h1Wb,in=(m1m2)h1+m2h1m1h1Wb,in=(m1m2+m2m1)h10=0

Write the expression for the entropy balance for helium.

SinSout+Sgen=ΔSsystemmese+Sgen=m2s2m1s1Sgen=m2s2m1s1+mese (IX)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Substitute m1m2 for me in Equation (IX).

Sgen=m2s2m1s1+(m1m2)se

Sgen=m2s1m1s1+(m1m2)s1=(m2m1+m1m2)s1=0

Write the expression for the exergy destroyed during the process (Xdestroyed).

Xdestroyed=T0Sgen (X)

Conclusion:

Substitute 20°C for T0 and 0 for Sgen in equation (X)

Xdestroyed=(20°C)(0)=0

Thus, the exergy destroyed during the process (Xdestroyed) is 0_.

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