Question

Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting the changes in kinetic and potential energies and assuming the process to be reversible, determine the power input to the pump. Use steam tables. The power input to the pump is kW.

Answer 1

Answer:

$\dot W_{in} = 273.69\,kW$

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

$\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0$

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

$P = 20\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 251.42\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

Outlet (Subcooled Liquid)

$P = 6000\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 257.502\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

The power input to the pump is computed hereafter:

$\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 273.69\,kW$