Question 5.8: Hot water at a temperature of 95 °C and a pressure of 100 kP...
Hot water at a temperature of 95 °C and a pressure of 100 kPa flows in a pipe at a flow rate of 1 kg/s, as shown in Figure 5.10. The pipe loses heat at a rate of 1 kW per meter along the pipe length. (a) Write the mass, energy, entropy, and exergy balance equations, (b) find the temperature of the water leaving the pipe after 100 m of pipe length, and (c) calculate thermal exergy loss, rate entropy generation rate, and exergy destruction rate. Take the pipe surface temperature as 60 °C and reference temperature (i.e. surroundings temperature) as 25 °C.
a) Write the mass, energy, entropy, and exergy balance equations.
\text { MBE }: \dot{m}_{1}=\dot{m}_{2}=\dot{m}\text { EBE : } \dot{m}_{1} h_{1}=\dot{m}_{2} h_{2}+\dot{Q}_{\text {out }}
\text { EnBE : } \dot{m}_{1} S_{1}+\dot{S}_{\text {gen }}=\dot{m}_{2} S_{2}+\dot{Q}_{\text {out }} / T_{\text {su }}
\text { ExBE : } \dot{m}_{1} e x_{1}=\dot{m}_{2} e x_{2}+\dot{E} x_{\dot{Q}_{\text {out }}}+\dot{E} x_{d}
b) Find the temperature of the water leaving the pipe after 100 m of pipe length.
\dot{Q}_{o u t}=1 \frac{k W}{m} \times 100 m\dot{Q}_{ out }=100 kW
at P_{1}=100 kPa \text { and } T_{1}=95^{\circ} C , h_{1}=398.1 \frac{ kJ }{ kg }
\dot{m}_{1} h_{1}=\dot{m}_{2} h_{2}+\dot{Q}_{\text {out }}
1 \frac{k g}{s} \times 398.1 \frac{k J}{k g}=1 \frac{k g}{s} \times h_{2}+1 \frac{k W}{m} \times 100 m
From the above written energy balance equation, one obtains h_{2}=298.1 \frac{ kJ }{ kg }
By assuming no pressure loss occurs in the pipe, P_1 = P_2 = 100 kPa and h_{2}=298.1 \frac{ kJ }{ kg }
then one can use the property tables (such as Appendix B-1d) or EES to find the temperature:
T_{2}=71.19^{\circ} Cc) Calculate the exergy associated with the heat loss rate, entropy generation rate, and exergy destruction rate.
The exergy content of the heat loss is calculated as follows:
\dot{E} x_{\dot{Q}_{\text {out }}}=\left(1-\frac{T_{0}}{T_{\text {su }}}\right) \dot{Q}_{\text {out }}Note that the surface temperature of the pipe is taken as 60 °C, respectively to calculate the thermal exergy rate (or exergy rate of the heat losses) as follows:
\dot{E} x_{\dot{Q}_{\text {out }}}=\left(1-\frac{298}{60+273}\right) \times 100 kW\dot{E} x_{\dot{Q}_{\text {out }}}=10.51 k W
In order to find the entropy generated by the system, one can use the entropy balance equation along with the water inlet and exit properties from the property tables (such as Appendix B-1d) or EES. Take s_{1}=1.25 \frac{ kJ }{ kgK } and s_{2}=0.9696 \frac{ kJ }{ kgK }
\dot{m}_{1} S_{1}+\dot{S}_{\text {gen }}=\dot{m}_{2} S_{2}+\dot{Q}_{\text {out }} / T_{s u}
\dot{S}_{g e n}=\dot{m}_{2} S_{2}-\dot{m}_{1} S_{1}+\dot{Q}_{o u t} / T_{s u}
\dot{S}_{\text {gen }}=1 \frac{k g}{s} \times(0.9696-1.25)\left(\frac{k J}{k g K}\right)+\frac{100 k W}{(60+273) K}
\dot{S}_{g e n}=0.020 \frac{k W}{K}
One can then simply plug in the values in the exergy destruction relation as follows:
\dot{E} x_{d}=\dot{S}_{\text {gen }} T_{0}\dot{E} x_{d}=0.020 \frac{k W}{K} \times 298.15 K
\dot{E} \dot{x}_{d}=5.96 kW