A pump moves water horizontally at a rate of \(0.02 \mathrm{m}^{3} / \mathrm{s}\) Upstream of the pump where the pipe dianeter is \(90 \mathrm{mm}\), the pressure is \(120 \mathrm{kPa}\). Downstream of the pump where the pipe diameter is \(30 \mathrm{mm}\), the pressure is \(400 \mathrm{kPa}\). If the loss in energy across the purnp due to fluid friction effects is \(170 \mathrm{N} \cdot \mathrm{m} / \mathrm{kg}\), determine the hydraulic efficiency of the pump.

We first need to calculate the velocities of the water at the pump inlet and outlet. The volumetric flow rate (Q) is given as 0.02 m³/s. The areas of the inlet and outlet cross-sections can be calculated from the given diameters using the formula \(A = \pi d² / 4\), where d is the diameter. So, the inlet velocity (u1) can be calculated as \(u1 = Q / A1\) and the outlet velocity (u2) as \(u2 = Q / A2\).

Next, we calculate the power input and output to the pump. The energy loss across the pump due to fluid friction effects is given as 170 N m / kg. This energy must be supplied by the pump, so the power input to the pump (Pin) is the sum of this frictional loss and the change in mechanical energy of the fluid. The mechanical energy change includes changes in kinetic energy and pressure energy, which can be calculated by \(\Delta K = 0.5 ρ (u2² - u1²)\) and \(\Delta P = ρ g h + (P2 -P1)/\rho\), respectively. Then \(Pin = \Delta K +\Delta P\). Meanwhile, the power output from the pump (Pout) can be calculated by multiplying the volumetric flow rate by the pressure difference across the pump: \(Pout = Q (P2 - P1)\).

Finally, the hydraulic efficiency (\η) of the pump can be calculated by dividing the useful power output by the total power input: \η = Pout / Pin.