Answer:
70.9 cents.
Explanation:
Let assume that water at inlet is a saturated liquid. The process within pump is modelled after the First Law of Thermodynamics:
[tex]\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0[/tex]
The power consumed by the pump is:
[tex]\dot W_{in} = \dot m \cdot (h_{out}-h_{in})[/tex]
The isentropic effciency of the pump is:
[tex]\eta_{s} = \frac{h_{out,s}-h_{in}}{h_{out}-h_{in}}[/tex]
Where,
[tex]h_{out} - h_{in} = \frac{h_{out,s}-h_{in}}{\eta_{s}}[/tex]
The power equation is modified into the following form:
[tex]\dot W_{in} = \dot m \cdot \frac{(h_{out,s}-h_{in})}{\eta_{s}}[/tex]
Specific enthalpies and entropies at inlet and outlet are obtained from steam tables and included below:
State 1 (Saturated Liquid)
[tex]h = 417.51\,\frac{kJ}{kg}[/tex]
[tex]s = 1.3028\,\frac{kJ}{kg\cdot K}[/tex]
State 2s (Subcooled Liquid)
[tex]h = 417.823\,\frac{kJ}{kg}[/tex]
[tex]s = 1.3028\,\frac{kJ}{kg\cdot K}[/tex]
The power consume by the pump is:
[tex]\dot W_{in} = (20\,\frac{kg}{s} )\cdot \left(\frac{417.823\,\frac{kJ}{kg}-417.51\,\frac{kJ}{kg} }{0.75} \right)[/tex]
[tex]\dot W_{in} = 8.347\,kW[/tex]
The hourly energy consumption is:
[tex]E_{hourly} = (8.347\,kW)\cdot (3600\,s)\cdot \left(\frac{1\,kWh}{3600\,kJ} \right)[/tex]
[tex]E_{hourly} = 8.347\,kWh[/tex]
The hourly cost is:
[tex]C_{hourly} = (0.085\,\frac{USD}{kWh} )\cdot (8.347\,kWh)[/tex]
[tex]C_{hourly} = 0.709\,USD[/tex]
