Electrical Power to Pump Gasoline Between Gauges: 04-BS-7, December 2019, Q4

Question 4: Electrical Power to Pump Gasoline Between Gauges (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Suction / discharge pipe diameter	150 mm / 100 mm
Gauge elevation (discharge above C/L, suction below C/L)	1.5 m / 0.5 m
Discharge / suction gauge pressure	150 kPa / −30 kPa (vacuum)
Flow rate, Q	0.035 m³/s
SG gasoline / pump efficiency	0.75 / 75%

Energy equation between the suction and discharge gauge points, elevations measured from the pump centreline.

Find. The electrical power required to drive the pump.

Approach. Apply the general energy equation between the suction (1) and discharge (2) gauge points to find the pump head added, convert to hydraulic power via $\rho gQH$, then divide by the pump efficiency for electrical power.

Velocities at each gauge. $$V_1=\frac{Q}{A_1}=\frac{0.035}{\tfrac{\pi}{4}(0.150)^2}=\boxed{1.981\ \text{m/s}}\qquad V_2=\frac{Q}{A_2}=\frac{0.035}{\tfrac{\pi}{4}(0.100)^2}=\boxed{4.456\ \text{m/s}}$$
Pump head from the energy equation. With $\rho_{gas}=750$ kg/m³, and elevations $z_1=-0.5$ m, $z_2=+1.5$ m relative to the pump centreline, $$H_{pump}=\frac{p_2-p_1}{\rho_{gas}g}+(z_2-z_1)+\frac{V_2^2-V_1^2}{2g}$$ $$H_{pump}=\frac{150{,}000-(-30{,}000)}{(750)(9.81)}+\big(1.5-(-0.5)\big)+\frac{4.456^2-1.981^2}{2(9.81)}$$ $$H_{pump}=24.47+2.00+0.804=\boxed{27.28\ \text{m}}$$
Hydraulic and electrical power. $$P_{hyd}=\rho_{gas}gQH_{pump}=(750)(9.81)(0.035)(27.28)=7024\ \text{W}$$ $$P_{elec}=\frac{P_{hyd}}{\eta}=\frac{7024}{0.75}=\boxed{9.37\ \text{kW}}$$

Quantity	Value
V₁ / V₂	1.981 / 4.456 m/s
Pump head, H	27.28 m
Hydraulic power	7.02 kW
Electrical power required	9.37 kW

Concept & Theory

Topic. Pump head and power from the general energy equation.

Key relations. Energy equation with a pump term $\tfrac{p_1}{\rho g}+z_1+\tfrac{V_1^2}{2g}+H_{pump}=\tfrac{p_2}{\rho g}+z_2+\tfrac{V_2^2}{2g}$; power $P=\rho gQH$; efficiency divides hydraulic power UP to electrical input power.

Why this works. The pump must supply energy to overcome three separate effects between the two gauge points: the pressure rise (by far the largest term here, since suction is even in vacuum), the elevation gain, and the kinetic-energy gain from the smaller discharge pipe — treating all three consistently in one energy balance avoids missing any of them.

Common pitfall. Forgetting the suction gauge's NEGATIVE sign (a vacuum) — the pressure the pump must supply is $p_2-p_1=150-(-30)=180$ kPa, not $150-30=120$ kPa.

Source. White, Fluid Mechanics, Ch. 3 (Energy equation with pumps and turbines).

Question 4 of 13: Electrical Power to Pump Gasoline Between Gauges

Question 4: Electrical Power to Pump Gasoline Between Gauges (5 marks)