Question

Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit and (b) the pressure difference between the inlet and the exit of the pump.

Answer 1

Answer:

(a) 66.9%

(b) 147.14 kPa

Explanation:

Given:

Elevation of the water tank z = 15 m

Water volume flow rate V = 70 L/s = 0.07 m³/s

Input electric power consumption by the pump, Welec, in = 15.4  kW

Assuming there are no frictional losses in the pipes and changes in kinetic energy, the efficiency of the pump-motor will be;

                                     η =   ΔEmech ÷ Welec,

Where;

η is the overall efficiency

ΔEmech is the workdone to move the water pumped from the lake to a storage tank 15m above

Welec is the Input electric power consumption by the pump

Solving for ΔEmech ;

                         ΔEmech = mgh

                         mass, m = density × volume  

Density of water = 1000 kg/m³

                          m = 1000 kg/m³ ×  0.07 m³/s

                          m = 70 kg/s

            ∴ ΔEmech = 70 × 9.8 × 15

                               = 10.3 kW

Substituting the values of ΔEmech and Welec to calculate the overall efficiency

             η  =  (10.3 kW ÷ 15.4  kW) × 100 %

                   = 0.6688 × 100 %

                   = 66.88 %

                   = 66.9 %

The overall efficiency of the pump-motor unit is = 66.9 %

(b) The pressure difference between the inlet and the exit of the pump is calculated to be;

                                  Pressure = ΔEmech ÷ V  

                                                  = 10.3 ÷ 0.07

                                                  = 147.14 kPa

Answer 2

This question involves the concept of potential energy, pressure difference, and electrical work.

(a) Efficiency of pump-motor unit is "66.9 %".

(b) The pressure difference between the inlet and the exit of the pump is "147.15 KPa".

(a) Pump Efficiency

The efficiency of the pump-motor unit can be given by the following formula:

$\eta = \frac{W_{P.E}}{W_{elect}}$

where,

• $\eta$ = efficiency = ?
• P.E = power due to potential energy = $\frac{mgh}{t}=\frac{\rho Vgh}{t}$
• $\rho$ = density of water = 1000 kg/m³
• g = 9.81 m/s²
• h = height = 15 m
• $\frac{V}{t}$ = volume flow rate = 70 L/s = 0.07 m³/s
• $W_{elect}$ = electrical power input =  15.4 KW = 15400 W

Therefore,

$\eta = \frac{\rho Vgh}{t\ W_{elect}}\\\\\eta=\frac{(1000\ kg/m^3)(0.07\ m^3/s)(9.81\ m/s^2)(15\ m)}{15400\ W}\\\\\eta =0.669 = 66.9\ \%$

(b) PRESSURE DIFFERENCE

The pressure difference between inlet and outlet of the pump can be found using the following equation:

$\Delta P = \rho gh\\\\\Delta P = (1000\ kg/m^3)(9.81\ m/s^2)(15\ m)$

$\Delta P = 147150\ Pa = 147.15\ KPa$

Learn more about the pressure difference here:

https://brainly.com/question/124508