A pump of a water distribution system at 25°C is powered by a 15 kW electric motor whose efficiency is 90 percent. The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are both 5 cm and the elevation difference across the pump is negligible. If the absolute pressures at the inlet and outlet of the pump are measured as 100 kPa and 300 kPa, respectively, determine the friction loss in the system.

Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:

Mass balance

[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (1)

[tex]\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}}[/tex] (2)

[tex]\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}}[/tex] (3)

Energy balance

[tex]\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0[/tex] (4)

Where:

[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.
[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.
[tex]\dot V_{in}[/tex] - Inlet volume flow, in cubic meters per second.
[tex]\dot V_{out}[/tex] - Outlet volume flow, in cubic meters per second.
[tex]\nu_{in}[/tex] - Inlet specific volume, in cubic meters per kilogram.
[tex]\nu_{out}[/tex] - Outlet specific volume, in cubic meters per kilogram.
[tex]\eta[/tex] - Pump efficiency, no unit.
[tex]\dot W_{el}[/tex] - Electric motor power, in kilowatts.
[tex]h_{in}[/tex] - Inlet specific enthalpy, in kilojoules per kilogram.
[tex]h_{out}[/tex] - Outlet specific enthalpy, in kilojoules per kilogram.
[tex]\dot W[/tex] - Work losses due to friction, in kilowatts.

Data from steam tables

From steam tables we get the following water properties at inlet and outlet:

Inlet

[tex]p = 100\,kPa[/tex], [tex]T = 25\,^{\circ}C[/tex], [tex]\nu = 0.001003\,\frac{kJ}{kg}[/tex], [tex]h = 104.927\,\frac{kJ}{kg}[/tex], Subcooled liquid

Outlet

[tex]p = 300\,kPa[/tex], [tex]T = 25\,^{\circ}C[/tex], [tex]\nu = 0.001003\,\frac{kJ}{kg}[/tex], [tex]h = 105.128\,\frac{kJ}{kg}[/tex], Subcooled liquid

Calculation of the friction loss in the system

If we know that [tex]\dot V_{in} = 0.05\,\frac{m^{3}}{s}[/tex], [tex]\nu_{in} = 0.001003\,\frac{m^{3}}{kg}[/tex], [tex]h_{in} = 104.927\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 105.128\,\frac{kJ}{kg}[/tex], [tex]\eta = 0.90[/tex] and [tex]\dot W_{el} = 15\,kW[/tex], then the friction loss in the system is:

[tex]\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}[/tex]

[tex]\dot W_{f} = \left(\frac{0.05\,\frac{m^{3}}{s} }{0.001003\,\frac{m^{3}}{kg} } \right)\cdot \left(104.927\,\frac{kJ}{kg}-105.128\,\frac{kJ}{kg}\right) + (0.90)\cdot (15\,kW)[/tex]

[tex]\dot W_{f} = 3.480\,kW[/tex]

The friction loss in the system is 3.480 kilowatts. [tex]\blacksquare[/tex]

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