Question 7.3: Let there be a heat pump that exchanges the heat Qh with a h......
Let there be a heat pump that exchanges the heat {\mathit{Q}}_{h} with a hot temper-ature source {\mathit{T}}_{h}, and {\mathit{Q}}_{c} with a cold temperature source {\mathit{T}}_{c}, as shown in Fig. E.7.5
Find the following:
(a) Define F and P.
(b) Carry out an exergy balance.
(c) Determine the exergy efficiency and unit exergy consumption.
(a) The product of the heat pump is the exergy transferred by heat to the heat source
P=\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}while the fuel is the power consumed plus the exergy extracted from the cold source, which is
F=\left(1-{\frac{T_{0}}{T_{c}}}\right)Q_{c}+W=\left(1-{\frac{T_{0}}{T_{c}}}\right)Q_{c}+(Q_{h}-Q_{c})Evidently, if the cold source were the environment, then {{F}}=W.
(b) The equation of exergy balance is
\left(1-{\frac{T_{0}}{T_{c}}}\right)Q_{c}+W=\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}+D\rightarrow F=P+D(c) The exergy efficiency is
\varphi={\frac{P}{F}}={\frac{\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}}{\left(1-{\frac{T_{0}}{T_{c}}}\right)Q_{c}+W}}=1-{\frac{D}{\left(1-{\frac{T_{0}}{T_{c}}}\right)Q_{c}+W}}When the cold source is the environment, then T_{c}=T_{0} and, therefore, the efficiency is
\varphi={\frac{\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}}{W}}=1-{\frac{D}{W}}so that the relation of the heat pump exergy efficiency with the COP is
\varphi=\left(1-{\frac{T_{0}}{T_{h}}}\right)C O PThe unit exergy consumption is the inverse of efficiency and reflects the exergy needed per exergy unit of the product, which is
k={\frac{F}{P}}={\frac{\left(1-{\frac{T_{0}}{T_{c}}}\right)Q_{c}+W}{\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}}}=1+{\frac{D}{\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}}}and if the cold source is the environment
k={\frac{F}{P}}={\frac{W}{\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}}}=1+{\frac{D}{\left(1-{\frac{T_{0}}{T_{h}}}\right)Q_{h}}}