Question
Water at $15^{\circ} \mathrm{C}$ is pumped at a rate of $20 \mathrm{~L} / \mathrm{s}$ using a $5 \mathrm{~kW}$ pump. If the efficiency of the pump is $80 \%,$ what is the head added to the water as it passes through the pump?
Step 1
First, we need to find the actual power output of the pump, which is 80% of the 5 kW input power. To do this, we multiply the input power by the efficiency: Actual power output = Input power × Efficiency Actual power output = 5 kW × 0.8 Actual power output = 4 kW Show more…
Show all steps
Your feedback will help us improve your experience
Ankur S and 54 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The pressure rise across a water pump is 35 psi when the volume flow rate is 500 gpm. If the pump efficiency is 80 percent, determine the power input to the pump.
A water pump develops a total head of 200 ft. The motor efficiency is 87.5%. If the power rate is 1.5 cents per kw-hr, what is the power cost in cents per minute for pumping 1000 gal/hr
A $10 \mathrm{H} . \mathrm{P}$ pump, working at $80 \%$ efficiency is used for drawing water from a well. The water level in the well is $6 \mathrm{~m}$ below the location of the pump. The water outlet from pump is horizontal and the water flow rate through the pump is 50 litre per second. (i) What is the energy spent by the pump in 10 hour? (ii) What is the speed with which water will come out of the pump?
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD