The maximum work potential ( Φ 1 ) of the helium at the initial state

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Answer 1

Question

Chapter 8.8, Problem 75P

a)

To determine

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

a)

Expert Solution

Answer to Problem 75P

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

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;

s1−s0=cplnT1T0−RlnP1P0 . (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(v1−v0)−T0(s1−s0)] . (VI)

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the value of R for helium as 2.0769 kPa⋅m3/kg⋅K(2.0769 kJ/kg⋅K).

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

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

Substitute 0.03944 kg for m1 in Equation (II).

m2=0.03944 kg2=0.01972 kg

Substitute 2.0769 kPa⋅m3/kg⋅K for R, 20°C for T1, and 200 kPa for P1 in equation (III)

v1=(2.0769 kPa⋅m3/kg⋅K)(20°C)200 kPa=(2.0769 kPa⋅m3/kg⋅K)(20+273) K200 kPa=(2.0769 kPa⋅m3/kg⋅K)(293 K)200 kPa=3.043 m3/kg

Substitute 2.0769 kPa⋅m3/kg⋅K for R, 20°C for T0, and 95 kPa for P0 in equation (IV).

v0=(2.0769 kPa⋅m3/kg⋅K)(20°C)95 kPa=(2.0769 kPa⋅m3/kg⋅K)(20+273) K95 kPa=(2.0769 kPa⋅m3/kg⋅K)(293 K)95 kPa=6.406 m3/kg

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

Substitute 5.1926 kJ/kg⋅K for cp, 2.0769 kPa⋅m3/kg⋅K for R, 20°C for T1 and T0 respectively, 95 kPa for P0, and 200 kPa for P1 in Equation (V).

s1−s0=(5.1926 kJ/kg⋅K)(ln20°C20°C)−(2.0769 kPa⋅m3/kg⋅K)(ln200 kPa95 kPa)=[(5.1926 kJ/kg⋅K)(0)]−(2.0769 kPa⋅m3/kg⋅K)(0.7444)=−1.546 kPa⋅m3/kg⋅K

Substitute 0.03944 kg for m1, 20°C for T0, −1.546 kPa⋅m3/kg⋅K for s1−s0, 95 kPa for P0, 3.043 m3/kg for v1 and 6.406 m3/kg for v2 in equation (VI).

Φ1=(0.03944 kg)[95 kPa(3.043−6.406) m3/kg−20°C(−1.546 kPa⋅m3/kg⋅K)]=(0.03944 kg)[95 kPa(−3.363 m3/kg)−(20+273) K(−1.546 kPa⋅m3/kg⋅K)]=(0.03944 kg)[−319.485 kPa⋅m3/kg−(293 K)(−1.546 kPa⋅m3/kg⋅K)]=(0.03944 kg)(−319.485 kPa⋅m3/kg+452.978 kPa⋅m3/kg)

=(0.03944 kg)(133.493 kPa⋅m3/kg)=5.27 kPa⋅m3(kJkPa⋅m3)=5.27 kJ

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

b)

To determine

The exergy destroyed during the process (Xdestroyed)

b)

Expert Solution

Answer to Problem 75P

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

Explanation of Solution

Write the mass balance equation for the system as,

min−mout=Δmsystemme=m1−m2 . (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 m1 and m2 respectively.

Write the Energy balance for the uniform flow system as;

Ein−Eout=ΔEsystemQin−mehe+Wb,in=m2u2−m1u1 . (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=(m1−m2)he+m2u2−m1u1−Wb,in=(m1−m2)he+m2h2−m1h1−Wb,in=(m1−m2)h1+m2h1−m1h1−Wb,in=(m1−m2+m2−m1)h1−0=0

Write the expression for the entropy balance for helium.

Sin−Sout+Sgen=ΔSsystem−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+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 m1−m2 for me in Equation (IX).

Sgen=m2s2−m1s1+(m1−m2)se

Sgen=m2s1−m1s1+(m1−m2)s1=(m2−m1+m1−m2)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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