Question 1.21: 100 kg of steam is enclosed in a piston-cylinder device, ini...

100 kg of steam is enclosed in a piston-cylinder device, initially at 300°C and 5 bar. It expands and cools to 200°C and 1 bar.

A) What is the change in internal energy of the steam in this process?
B) If the external pressure is constant at 1 bar, how much work was done by the steam on the surroundings?
C) Research and briefly describe at least two examples of machines, either historical or currently in use, that harness the energy in steam and convert it into work. Any form of work is acceptable; you needn’t confine your research to expansion work (which was examined in parts A and B).

InitialStateT1=300CP1=5barFinalStateT2=200CP2=1bar\begin{matrix}\boxed{\begin{matrix}\sf{Initial \,State}\\\\ \sf T_1=300^\circ \sf C\\\\ \sf P_1=5\,\sf bar\end{matrix}}& \longrightarrow & \boxed{\begin{matrix}\sf{Final \,State}\\\\ \sf T_2=200^\circ \sf C\\\\ \sf P_2=1\,\sf bar\end{matrix}}\end{matrix}

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Define the gas as the system.

A) From steam tables:

U^1=2803.2kJkgU^2=2658.2kJkg\begin{aligned}& \widehat{U}_{1}=2803.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\\& \widehat{U}_{2}=2658.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{aligned}

So

ΔU=M(U^2U^1)=(100 kg)(2658.22803.2kJkg)=14,500kJ\begin{aligned}\Delta U=M\left(\widehat{U}_{2}-\widehat{U}_{1}\right)=(100 \mathrm{~kg})\left(2658.2-2803.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=-\mathbf{1 4}, \mathbf{500}\,\mathbf{k J}\end{aligned}

B) From steam tables:

V^1=0.5226m3 kgV^2=2.1724m3 kg\begin{aligned}& \widehat{V}_{1}=0.5226 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}\\& \widehat{V}_{2}=2.1724 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}\end{aligned}

Thus the gas expands, and work is described by equation 1.22:

WEC=PdVW_{E C}=-\int P d V

Pressure opposing the motion is constant at 1 bar, so:

WEC=P(V2V1)=MP(V^2V^1)WEC=(100 kg)(1 bar )(2.17240.5226m3 kg)(105 Pabar)(1Nm2 Pa)(1 JNm)(1 kJ1000 J)=16,498kJ\begin{aligned}&\qquad\qquad\qquad\qquad W_{E C}=-P\left(V_{2}-V_{1}\right)=-M P\left(\hat{V}_{2}-\hat{V}_{1}\right) \\& W_{E C}=-(100 \mathrm{~kg})(1 \text { bar })\left(2.1724-0.5226 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}\right)\left(\frac{10^{5} \mathrm{~Pa}}{\mathrm{bar}}\right)\left(\frac{1 \frac{\mathrm{N}}{\mathrm{m}^{2}}}{\mathrm{~Pa}}\right)\left(\frac{1 \mathrm{~J}}{\mathrm{Nm}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right) \\& \qquad\qquad =-\mathbf{1 6}, \mathbf{4 9 8} \,\mathbf{k J}\end{aligned}

The negative sign indicates work is transferred from gas to the surroundings.

C) Steam power is still used in electrical generators that run on the Rankine cycle with water as the operating fluid. Historically, steam engines have been used to power systems like pumps, boats, trains, sawmills etc. The common feature in these machines is they require shaft work that can be supplied by steam turbines.

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