A) 1.5
B) 2.0
C) 2.5
D) 3.0
Correct Answer: A
Solution :
\[{{T}_{3}}=300K,\]\[{{T}_{4}}=180\,K\] \[{{r}_{p}}=\frac{{{p}_{2}}}{{{p}_{1}}}\,\left( \frac{{{T}_{3}}}{{{T}_{4}}} \right){{\,}^{\frac{\gamma }{\gamma \,-\,1}}}\,={{\left( \frac{300}{180} \right)}^{3.5}}=5.977\] \[COP=\frac{1}{{{r}_{p}}^{\gamma \,-\,1/\gamma }-1}=\frac{1}{(5.977){{\,}^{0.4\,\,/1.4}}-1}=1.5\] \[COP=\frac{1}{\frac{{{T}_{3}}}{{{T}_{4}}}-1}=\frac{1}{\frac{300}{180}-1}=1.5\]You need to login to perform this action.
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