• Support PF! Buy your school textbooks, materials and every day products Here!

Pump Power Consumption

  • Thread starter jderulo
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
34,270
10,318
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
34
0
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
34,270
10,318
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
34
0
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
34
0
[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
34,270
10,318
[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 Threads on Pump Power Consumption

  • Last Post
Replies
17
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
8
Views
1K
Replies
12
Views
2K
  • Last Post
Replies
22
Views
1K
Replies
2
Views
792
Replies
10
Views
2K
Replies
12
Views
3K
Replies
2
Views
711
Replies
5
Views
2K
Top