Operating Point Calculation for a Centrifugal Pump in a Process System

[Jayohtwo]

Homework Statement

A centrifugal pump has the following pressure – capacity characteristics:

It is planned to use this for a process having the following system characteristic:

(a) Determine the operating point for this pump with this system.

(b) The actual flowrate required by the process is 50 m3 h–1. If the overall efficiency of the pump at this flowrate is 70%, determine the power consumed when the liquid being pumped has a density of 1200 kg m–3.

Homework Equations

For Part a I think I need to plot the data given onto a graph and find where the lines intersect.

Part b: P_o = q_v ρ g H_p

where H_p is the system characteristic

H_p = (H_t2 - H_t1) + h_f + h_m

The Attempt at a Solution

(a)

After graphing the data given in excel the lines intersect at a flowrate of 100 m³h^-1 and a Pressure of 465 KPa

(b)
I haven't figured out how to calculate the system characteristic as I'm not sure where to find the initial total head and final total head. I must have misread or overlooked something in my notes on how to calculate it. I'll be reading through my notes again. Plaese let me know if I've got anything wrong that would be much appreciated thanks.

 

[CWatters]

Part a) looks correct.

For part b)... What is the pressure required to push 50 m³/h through the system? This will be the pressure the pump has to deliver eg the pressure or head across the pump.

 

[Jayohtwo]

Part a) looks correct.

For part b)... What is the pressure required to push 50 m³/h through the system? This will be the pressure the pump has to deliver eg the pressure or head across the pump.

Hi CWatters, thanks for the reply. The pressure required according to the tables would be 400KPa. So I've got density, flow rate and pressure now and looking through some notes and other resources I think I was missing some equations.

b)
y = ρg =1200 x 9.81 = 11772

where y is specific weight.

h = (P₂ - P₁)/y = 400x10³/11772 = 33.98m

Po = qv ρ g Hp = 50 x 1200 x 9.81 x 33.98 = 20000628W

Actual Power output = 70% of P_o = 0.7 x 20000628 = 14000439.6 W

I hope this is right, I'm a bit out of practice with these questions on pumps.

 

[mechengstu123]

Hi i used this equation P=Q x dP/3600 and i came out with 3.89kW @ 70% eff. Sounds about right for size of pipework and distance pumped.

 

[Niall11]

Hi I am struggling on this question myself and seem to be having the same problem in that I don't have an equation for finding Hp using the data given, jayohtwo can I ask where you found h = (P2 - P1)/y ?
Thanks

 

[CWatters]

I was thinking along same lines as mechengstu123 but got different numbers..

Power = Pressure * Flow Rate

50m³h^–1 = 13.9 L/S

From the table the pressure at 50 m³ h^–1 = 400kPa

So the Hydraulic power required is..

13.9 * 400 = 5,560W

The efficiency is only 70% so the motor power required is..

5560/0.70 = 7,942W

 

[Rogue]

Apologies for resurrecting an old thread.
I am also stuck on part b of this question.

I had used one of the formulas above, using 50 as Hp.
However, as I move on, it's obvious that this value is incorrect.
Please can someone offer a bit of guidance or a nudge in the right direction?

 

[CWatters]

Why use "50" for the Hydraulic Pressure? The problem statement (b) says 50 is the required flow rate not the pressure.

Edit: Sorry I see Hp = System Characteristic not Hydraulic Pressure.

 

[Rogue]

I don't know to be honest.
But looking at your previous suggestion, it doesn't seem to take fluid density into account, and I'm really unsure where to take this one.

 

[CWatters]

I have to admite that fluid dynamics isn't my field but unless someone else can see a problem I stand by my calculations in my post #6.

I guess I could have been better with the units and accuracy but I think the method is correct. I should have written..

Power (W) = Pressure (Pascals) * Flow Rate (cubic meters per second)
= 400,000 * 0.0139
= 5,555.5W

which for some reason I rounded up to 5560W in #6. Anyway..

Taking into account pump efficiency that gives..

5,555.5/0.7 = 7,936W

References..

http://www.wikicalculator.com/formu...flow-rate-in-systems-with-fluid-flow)-526.htm
https://hydraulicsonline.com/resources/hydraulic-calculations/(has same equation but uses different units).

 

[CWatters]

Ok here is another reference that does take into account density...

https://www.engineeringtoolbox.com/pump-fan-power-d_632.html
...but in the end it arrives at effectively the same equation. eg

P = Q (p2 - p1) …………………….(8)

where Q is the volumetric flow rate and (p2-p1) is the pressure increase provided by the pump.

 

[CWatters]

Is there more to the question than stated in the OP?

 

[cjl]

I agree with CWatters - the power required by a pump is, fundamentally, just the pressure rise multiplied by the volumetric flow rate. The fluid density appears to be a bit of a red herring. At 50 cubic meters per hour, and 400kPa pressure, I get exactly the same numbers as CWatters does - just under 5.6kW fluid power and 7.93kW after accounting for efficiency.