Head of fluids & Engine Efficieny.

[MikeNZ]

Homework Statement

A pump delivers water at a rate of 60L/s with a head of 65m. It runs for 8 hours per day. The efficiency of the pump is 75%. The efficiency of the electric motor driving the pump is 80%.

i) What is the value of the head in kPa
ii) Calculate the input power kW that is required to enable pump to supply the pressure as determined in part (i) above.

Homework Equations

pressure = density x gravity x height (take gravity as 9.81 meters per second)

The Attempt at a Solution

I know to find efficiency of the system you multiply the efficiency of both the pump and the motor - but the 'head' has stumped me - I know about head LOSS, but when it says 'a head of 65m' I'm lost. I've looked all through my handouts and can only find information about head loss. Not sure what this 65m head is referring to.

Thanks.

 

[KataKoniK]

Hello,

The term "head" in this context refers to the pressure that the pump is able to generate. It is a measure of the potential energy of the water as it is being pumped. In this case, the pump is able to generate a head of 65 meters, meaning that the water is being pumped to a height of 65 meters above its original level.

To convert this head to kPa, we can use the equation "pressure = density x gravity x height", as you mentioned in your post. The density of water is approximately 1000 kg/m^3, and gravity is 9.81 m/s^2. Plugging these values in, we get:

Pressure = 1000 kg/m^3 x 9.81 m/s^2 x 65 m = 637,650 Pa

To convert this to kPa, we divide by 1000, giving us a head of approximately 637.7 kPa.

To calculate the input power required for the pump, we can use the equation "power = flow rate x head / efficiency", where the flow rate is given as 60 L/s, head is 65 m, and efficiency is 75% (0.75). Plugging in these values, we get:

Power = 60 L/s x 65 m / (0.75 x 0.8) = 5,200 watts = 5.2 kW

This is the input power that is required for the pump to supply the pressure determined in part (i).

I hope this helps! Let me know if you have any other questions.

 

[Mene]

I can provide a response to this content by clarifying the concept of "head" in fluid mechanics. In this context, "head" refers to the height or vertical distance that the pump is able to lift the water. It is a measure of the potential energy of the water. In this case, the pump is able to lift the water to a height of 65m.

To convert this to kPa, we can use the equation provided in the homework: pressure = density x gravity x height. Since we are dealing with water, we can use a density of 1000 kg/m^3 and gravity of 9.81 m/s^2. Thus, the pressure in kPa would be (1000 kg/m^3) x (9.81 m/s^2) x (65m) = 637.65 kPa.

To calculate the input power required for the pump, we can use the equation P = Q x ΔP, where P is power, Q is flow rate, and ΔP is pressure difference. Since we are given the flow rate of 60L/s and the pressure in kPa, we need to convert the flow rate to m^3/s and the pressure to Pa. This gives us 0.06 m^3/s and 637650 Pa. Plugging these values into the equation, we get a required input power of 38.259 kW.

In conclusion, the value of the head in kPa is 637.65 kPa and the required input power for the pump is 38.259 kW. It is important to note that these calculations assume ideal conditions and do not take into account any losses in the system. it is important to consider all factors and assumptions when analyzing and solving problems in fluid mechanics.