Power Consumption of a Pump: Solve at 500kPa, 60m3/h, 80% Efficiency

In summary, the power consumed by the pump can be determined by using the formula P = q_v * p * h_p, where q_v is the volumetric flow rate, p is the pressure, and h_p is the energy head. This formula can be modified to include the 80% efficiency of the pump.
  • #1
jderulo
34
0

Homework Statement


Determine the power consumed by a pump when it is pumping a fluid of density 1,200kg/m^-3 60m3/h at 500kPa with an 80% overall efficiency.

Homework Equations


I believe it to be: [tex]P = q_m ρ g h_p[/tex]

The Attempt at a Solution


Stuck at the above part, I don't have enough info?
 
Physics news on Phys.org
  • #2
jderulo said:
I believe it to be: [tex]P = q_m ρ g h_p[/tex]
The formula looks useful. What is qm?

Instead of g and a height, you are given the pressure. How are those concepts related?
 
  • #3
mfb said:
The formula looks useful. What is qm?

Instead of g and a height, you are given the pressure. How are those concepts related?

qm is mass flow-rate. g is gravity, h isn't height but energy head..
 
  • #4
jderulo said:
g is gravity, h isn't height but energy head..
Similar enough.

The main idea is that you can write down a similar formula where you replace ρ, g and hp to include pressure instead of those two parameters.

I am surprised that it is the mass flow rate and not the volume flow rate.

Anyway, you need the delivered power output of the pump (that's what your formula can do).
Once you have that, you can see how you can include the 80% efficiency.
 
  • #5
mfb said:
Similar enough.

The main idea is that you can write down a similar formula where you replace ρ, g and hp to include pressure instead of those two parameters.

I am surprised that it is the mass flow rate and not the volume flow rate.

Anyway, you need the delivered power output of the pump (that's what your formula can do).
Once you have that, you can see how you can include the 80% efficiency.

My mistake it's volumetric flow. - but unsure how to do the first point.
 
Last edited:
  • #6
[tex]h_p = p / ρ g[/tex]

?

Am I right in thinking the final formula would be

[tex]P = q_v ρ g (p / ρ g)[/tex]
 
Last edited:
  • #7
jderulo said:
[tex]h_p = p / ρ g[/tex]

?

Am I right in thinking the final formula would be

[tex]P = q_v ρ g (p / ρ g)[/tex]
Right. You can simplify that formula, ρ and g will drop out and you are left with a simple product.
 

Related to Power Consumption of a Pump: Solve at 500kPa, 60m3/h, 80% Efficiency

What is the power consumption of a pump at 500kPa, 60m3/h, and 80% efficiency?

The power consumption of a pump is determined by the formula: Power (kW) = (Flow Rate (m3/h) x Pressure (kPa) x 1000) / (3600 x Efficiency). Plugging in the values for 500kPa, 60m3/h, and 80% efficiency, we get a power consumption of approximately 6.67 kW.

How do you calculate the efficiency of a pump?

Pump efficiency is calculated by dividing the output power (water flow rate x pressure) by the input power (electrical energy). This value is then multiplied by 100 to get a percentage. For example, if a pump has an output power of 100kW and an input power of 120kW, the efficiency would be calculated as (100/120) x 100 = 83.3%.

What is the significance of power consumption in pump selection?

Power consumption is an important factor to consider when selecting a pump because it can affect the operating costs and efficiency of the pump. A pump with a higher power consumption will require more energy and result in higher operating costs. Additionally, a pump with a lower efficiency will also have a higher power consumption, leading to decreased efficiency and potentially higher operating costs.

How does the flow rate and pressure affect the power consumption of a pump?

The flow rate and pressure are two key factors that determine the power consumption of a pump. Generally, an increase in either the flow rate or pressure will result in a higher power consumption. This is because the pump needs to work harder to maintain a higher flow rate and/or pressure. However, the efficiency of the pump also plays a role in determining the power consumption. A pump with a higher efficiency will require less power to achieve the same flow rate and pressure compared to a pump with a lower efficiency.

Is there a way to reduce the power consumption of a pump?

Yes, there are several ways to reduce the power consumption of a pump. One way is to select a pump with a higher efficiency, as this will require less power to achieve the desired flow rate and pressure. Another way is to properly size the pump for the specific application, as an oversized pump will consume more power than necessary. Additionally, regular maintenance and cleaning of the pump can help improve its efficiency and reduce power consumption.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
12
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Mechanical Engineering
Replies
8
Views
814
  • Engineering and Comp Sci Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
4K
Back
Top