A) For an ideal gas:
\begin{gathered}\mathrm{d} \underline{\mathrm{H}}=\mathrm{C}_{\mathrm{p}}^* \mathrm{dT} \\\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\mathrm{C}_{\mathrm{p}}^{*} \mathrm{dT}\end{gathered}
Set up an energy balance around the compressor
\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{aligned}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}\\100 \frac{\mathrm{kg}}{\min }\left(\widehat{\mathrm{H}}_{\text {out }}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{W}}_{\mathrm{S}}\end{aligned}
Using Appendix D
\begin{aligned}&\underline{\mathrm{d}} \underline{\mathrm{H}}=\mathrm{C}_{\mathrm{p}}{ }^{*} \mathrm{dT}\\&\frac{\mathrm{C}_{\mathrm{P}}^{*}}{\mathrm{R}}=\mathrm{A}+\mathrm{BT}+\mathrm{CT}^{2}+\mathrm{DT}^{3}+\mathrm{ET}^{4}\\&\mathrm{d} \underline{\mathrm{H}}=\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+\mathrm{RDT}^{3}+\mathrm{RET}^{4}\right) \mathrm{dT}\\&\int_{\text {out }}^{\text {in }} \mathrm{d} \underline{\mathrm{H}}=\int_{\text {out }}^{\text {in }}\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+\mathrm{RDT}^{3}+R E T^{4}\right) \mathrm{dT}\\&\left(\underline{\mathrm{H}}_{\text {out }}-\underline{\mathrm{H}}_{\text {in }}\right)=\mathrm{RA}\left(\mathrm{T}_{\text {out }}-\mathrm{T}_{\text {in }}\right)+\frac{\mathrm{RB}}{2}\left(\mathrm{~T}_{\text {out }}{ }^{2}-\mathrm{T}_{\text {in }}{ }^{2}\right)+\frac{\mathrm{RC}}{3}\left(\mathrm{~T}_{\text {out }}{ }^{3}-\mathrm{T}_{\text {in }}{ }^{3}\right) \\& \qquad\qquad+\frac{\mathrm{RD}}{4}\left(\mathrm{~T}_{\text {out }}^{4}-\mathrm{T}_{\text {in }}{ }^{4}\right)+\frac{\mathrm{RE}}{5}\left(\mathrm{~T}_{\text {out }}{ }^{5}-\mathrm{T}_{\text {in }}{ }^{5}\right) \\& \mathrm{T}_{\text {out }}=300 \mathrm{~K} \\& \mathrm{~T}_{\text {in }}=250 \mathrm{~K}\end{aligned}
\begin{array}{|c|c|c|c|c|c|c|} \hline\mathbf{Name} & \mathbf{Formula} & \mathbf{A} & \bf B\times10^3 & \bf C\times10^5 & \bf D\times10^8 & \bf E\times10^{11}\\ \hline \mathrm{Nitrogen} & \rm{N}_2 & 3.539 & -0.261 & 0.007 & 0.157 & -0.099\\\hline \end{array}
\begin{aligned}&\left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)= \left(8.314 \frac{\mathrm{kJ}}{\mathrm{k~mol}~ \mathrm{K}}\right)\{(3.539)(300-250) \\& \qquad\qquad\quad +\frac{1}{2}\left(-2.61 \times 10^{-4}\right)\left(300^{2}-250^{2}\right)+\frac{1}{3}\left(7.00 \times 10^{-8}\right)\left(300^{3}-250^{3}\right) \\& \qquad\qquad\quad\left.+\frac{1}{4}\left(-1.57 \times 10^{-9}\right)\left(300^{4}-250^{4}\right)+\frac{1}{5}\left(9.9 \times 10^{-13}\right)\left(300^{5}-250^{5}\right)\right\}\\& \left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\text {in }}\right)=1454 \frac{\mathrm{kJ}}{\mathrm{kmol}}\\& \widehat{\mathrm{H}}_{\text {out }}-\widehat{\mathrm{H}}_{\mathrm{in}}=1454 \frac{\mathrm{kJ}}{\mathrm{kmol}}\left(\frac{\mathrm{kmol}}{28.02 \mathrm{~kg}}\right)=52.95 \frac{\mathrm{kJ}}{\mathrm{kg}}\\& 100 \frac{\mathrm{kg}}{\min }\left(\widehat{\mathrm{H}}_{\text {out }}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{W}}_{\mathrm{S}}\\& 100 \frac{\mathrm{kg}}{\min }\left(52.95 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=\dot{\mathrm{W}}_{\mathrm{S}}=5295 \frac{\mathbf{k J}}{\min }\end{aligned}
Since this is an ideal gas, we know that enthalpy is only influenced by Temperature. Therefore, the Energy balance yields a Q that is equal but opposite to our Shaft Work
Set up an energy balance around the heat exchanger
\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{m}}_{\text {in }}\left(\widehat{\mathrm{H}}_{\text {in }}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\operatorname{gh}_{\text {in }}\right)-\dot{\mathrm{m}}_{\text {out }}\left(\widehat{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+\mathrm{gh}_{\text {out }}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\100 \frac{\mathrm{kg}}{\min }\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{Q}}\\\left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)=-1454 \frac{\mathrm{kJ}}{\mathrm{kmol}}\\\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=-1454 \frac{\mathrm{kJ}}{\mathrm{kmol}}\left(\frac{\mathrm{kmol}}{28.02 \mathrm{~kg}}\right)=-52.95 \frac{\mathrm{kJ}}{\mathrm{kg}}\\100 \frac{\mathrm{kg}}{\min }\left(-52.95 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=\dot{\mathrm{Q}}=\bf-5295 \bf\frac{k J}{min}\end{gathered}
B) Set up an energy balance around the compressor
\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\text {out }}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}\\100 \frac{\mathrm{kg}}{\min }\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{W}}_{\mathrm{S}}\end{gathered}
Using Figure 2-3, Find \widehat{\mathrm{H}}_{\mathrm{in}}
Nitrogen at 250 \mathrm{~K} and 1 \mathrm{bar} \rightarrow 408 \frac{\mathrm{kJ}}{\mathrm{kg}}
Find \widehat{\mathrm{H}}_{\mathrm{out}}
Nitrogen at 300 \mathrm{~K} and 10~ \mathrm{bar} \rightarrow 460 \frac{\mathrm{kJ}}{\mathrm{kg}}
100 \frac{\mathrm{kg}}{\min }\left(460 \frac{\mathrm{kJ}}{\mathrm{kg}}-408 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=\dot{\mathrm{W}}_{\mathrm{S}}=\bf 5200 \frac{\mathbf{kJ}}{\mathbf{min}}
Set up an energy balance around the heat exchanger
\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\dot{\mathrm{m}}_{\text {in }}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\text {in }}\right)-\dot{\mathrm{m}}_{\text {out }}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\text {out }}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\100 \frac{\mathrm{kg}}{\min }\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{Q}}\end{gathered}
\widehat{\mathrm{H}}_{\mathrm{in}} for the heat exchanger is the same as \widehat{\mathrm{H}}_{\text {out }} from the compressor \rightarrow 450 \frac{\mathrm{kJ}}{\mathrm{kg}}
Using Figure 2-3, Find \widehat{\mathrm{H}}_{\text {out }}
Nitrogen at 250 \mathrm{~K} and 10 \mathrm{bar} \rightarrow 402 \frac{\mathrm{kJ}}{\mathrm{kg}}
100 \frac{\mathrm{kg}}{\min }\left(402 \frac{\mathrm{kJ}}{\mathrm{kg}}-460 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=\dot{\mathrm{Q}}=\bf-5800 \bf\frac{k J}{min}