Question 5.14: An electric motor–pump unit, as shown in Figure 5.16, is use...
An electric motor–pump unit, as shown in Figure 5.16, is used to raise the pressure of water from ambient conditions to a pressure of 900 kPa and a temperature of 30 °C. The water entering the pump is at a temperature of 25 °C, at atmospheric pressure, and the mass flow rate is 0.5 kg/s. (a) Write the mass, energy, entropy, and exergy balance equations for the pump, (b) calculate the pump work rate, and (c) find the entropy generation rate and exergy destruction rate.
a) Write the mass, energy, entropy, and exergy balance equations.
\operatorname{MBE}: \dot{m}_{\text {in }}=\dot{m}_{\text {out }}=\dot{m}EBE : \dot{m}_{i n} h_{i n}+\dot{W}_{i n}=\dot{m}_{o u t} h_{o u t}
EnBE : \dot{m}_{\text {in }} s_{\text {in }}+\dot{S}_{\text {gen }}=\dot{m}_{\text {out }} s_{\text {out }}
\operatorname{ExBE}: \dot{m}_{i n} e x_{i n}+\dot{W}_{i n}=\dot{m}_{o u t} e x_{o u t}+\dot{E} x_{d}
b) Calculate the pump work rate.
We use the EBE to calculate the specific power consumed by the compressor as follows:
\dot{m}_{i n} h_{i n}+\dot{W}_{i n}=\dot{m}_{o u t} h_{o u t}Here, the properties of the stream entering and leaving the pump are found from water properties tables (such as Appendix B-1d) or by using Engineering Equation Solver (EES), which contains the database for the properties of most of the fluids through a large range of temperatures and pressures. The properties of each state point are shown in Table 5.4.
The work consumption rate of the pump can be calculated by rearranging the EBE as follows:
\dot{W}_{\text {in }}=\dot{m}_{2} h_{2}-\dot{m}_{1} h_{1}Substituting the given mass flow rate and the specific enthalpy values from Table 5.4 into the rearranged EBE yields the following work consumption rate:
\dot{W}_{i n}=\left(0.5 \times 126.5 \frac{k J}{k g}\right)-\left(0.5 \times 104.8 \frac{k J}{k g}\right)\dot{W}_{\text {in }}=10.82 k W
Table 5.4 The properties of the inlet and the exit streams.
| Stream | P (kPa) | T ( °C) | h (kJ/kg) | s (kJ/kgK) | ex (kJ/kg) |
| Reference | 101.3 | 25 | 104.8 | 0.3669 | 0 |
| State point 1 | 101.325 | 25 | 104.8 | 0.3669 | 0 |
| State point 2 | 900 | 30 | 126.5 | 0.4363 | 0.9742 |
c) Determine the entropy generation and exergy destruction rates.
The entropy generation rate can be calculated by using the entropy balance equation, as follows:
\dot{m}_{1} s_{1}+\dot{S}_{g e n}=\dot{m}_{2} s_{2} \rightarrow \dot{S}_{g e n}=\dot{m}_{2} s_{2}-\dot{m}_{1} s_{1}The properties to their corresponding state points can be found in Table 5.4 and are used to calculate the entropy generation rate as follows:
\dot{S}_{\text {gen }}=0.5 \frac{ kg }{ s }\left(0.4363-0.3669 \frac{ kJ }{ kg K}\right)\dot{S}_{g e n}=0.0347 k W / K
The exergy destruction rate can be calculated through the exergy balance of the pump as follows:
\dot{m}_{i n} e x_{i n}+\dot{W}_{i n}=\dot{m}_{o u t} e x_{o u t}+\dot{E} x_{d}0.5 \frac{k g}{s} \times 0 \frac{k J}{k g}+10.82 k W=0.5 \frac{k g}{s} \times\left(0.9742 \frac{k J}{k g}\right)+\dot{E} x_{d}
\dot{E} x_{d}=10.33 kW