How can I calculate the volume inside a pump case?

[PacoCardenas]

I have to determine the heat increase on a pump that its discharge valve is closed, so it will be operating at shutoff pressure for 10 seconds.

I have the design information of this pump, the curve of the pump, with its outline drawing and data sheet.

is it posible to calculate the case water volume with this information?

I do not have the exact dimensions of the case and impeller.

Thanks in advance.

 

[MikeHalloran]

If you have the actual pump, blank off the ports and fill it with water.

If you don't, try winging a solid model from whatever information you can get.

Mike Halloran
Pembroke Pines, FL, USA

 

[PacoCardenas]

The pump is installed on plant, so it is difficult to access to it, do you have idea if with the pumped fluid information, speed, curve and that stuff, could i get that information?

 

[MikeHalloran]

The least awful source I can conjecture is installation drawings, assuming you can't get the actual volume from the pump manufacturer.


Mike Halloran
Pembroke Pines, FL, USA

 

[LittleInch]

At 10 seconds, the heat rise will be so low its not worth bothering about. The combined water and metal mass is a big heat sink.

My guess would be outer volume divided by 3, but I can't see the value of getting this info.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[DubMac]

Given inconsistencies with castings, any calculated volume figure would not provide the degree of accuracy to determine temp rise anyway.

Lucky for you, there is no need to figure the case volume to determine Temperature rise. Use the handy formula:


Temp rise (DegF) = [(Total Head, ft) / 778 ] X [ (1- efficiency )-1 ]

 

[ccfowler]

My view of this question may be somewhat different while sharing much with LittleInch.

If this were a very high energy pump, such as a multi-thousand horsepower boiler feed pump, that would suffer from a very rapid and damaging temperature rise during a shut-off condition, nobody in their right mind would be contemplating seeking an answer to the question of pump body volume. Instead, the concern would be assuring a protective minimum cooling flow.

Any pump running in a shut-off condition will experience some temperature rise because it will be functioning as a mechanical water heater (or a mechanical heater of whatever the pumped medium may be). Most likely, even if the temperature rise would be a few degrees per minute, the temperature rise during ten seconds will likely not be enough to be a matter of concern.

If there is some sound basis for concern about the temperature rise, monitor the temperature with a suitable sensor and run the pump for a suitable period of time to measure its temperature rise characteristics during a shut-off condition. Most likely, you will be surprised at how slowly the temperature rises.

I know of cases where a pump has been deliberately run in a shut-off condition for periods of 15 to 30 minutes as part of a somewhat odd batch process situation. Part of the reason for doing this was to have a spurt of warmer water followed by the normal cooler water flow. The simplicity of eliminating a water heater and addition hardware and controls was most attractive. Obviously, low energy pumps were involved, and pump life and maintenance problems were entirely acceptable for the situation.

Valuable advice from a professor many years ago: First, design for graceful failure. Everything we build will eventually fail, so we must strive to avoid injuries or secondary damage when that failure occurs. Only then can practicality and economics be properly considered.

 

[Artisi]

I agree with ccfowler, even if this unit was multithousand kW uniy, a 10 sec run at CV temperature rise would not be a consideration, mechanical/ hydraulic considerations would be more of a concern.
For interest, what are the pump details
.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[LittleInch]

Dubmac - The equation appear to give a negative answer ((1-eff)-1) will always be negative if eff is expressed as a decimal fraction.

Also there does not seem to be a unit of time. Some idea of where this came from would be interesting. I always thought temp rise would be a curve which flattened out when input of heat equalled release of heat to environment??

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[DubMac]

LittleInch, of course my equation will yield a negative result; that's because it is wrong. I disobeyed the most important rule school ever taught me: "Check your work! At least twice; more times if it is important."

The portion of the equation with efficiency should read: (1/e) - 1, not (1-e) - 1.

Very perceptive of you to look for the time factor. It actually is in there and well disguised within the variable "e". Thinking of a pump curve, the efficiency can be translated to a capacity for a given pump; and that'll put the M in your GPM..


You are also correct that the temp rise curve DOES flatten out once you get into acceptable operating efficiencies. The "input of heat" from hydraulic losses isn't released to the environment as heat, but is converted back into pressure energy as the discharge valve is opened. Only the heat from frictional losses is continually released either into the fluid or environment. It of course becomes a negligible factor as flow is increased.

If someone must have an equation with time factor that's easier to find, then use: Temp Rise (DegF) = [ [BrakeHP-HydraulicHP] X 2545 ] / [Capacity, lbs/hr].

Many moons ago I worked for Worthington and was blessed to have access to their incredible library of pump knowledge, and design engineers. These equations were standard fare, and as to their derivation, I accepted and copied down everything I was told and never questioned my authorities at Worthington (and with very good reason). I would imagine some sort of document search could reveal the source; would guess Mr. Karasik had his hand in it. Worthington constantly put out technical literature in newsletters and "Did you know..." flyers that companies nowadays would keep confidential. What a shame.

 

[electricpete]

Some idea of where this came from would be interesting. I always thought temp rise would be a curve which flattened out when input of heat equalled release of heat to environment??

It is the principle used in the Yates meter.

Derivation:
* calculate dt which is defined as steady state temperature rise from input to output of pump

PowerOut = Q * Head * rho (where rho = density)
SHP = PowerIn = PowerOut/E = Q * Head * rho / E
Losses = SHP * (1-E) = Q * Head * rho * (1-E) / E
Also losses = Q Cv dt
equating two expressions for losses:
Q Cv dt = Q * Head * rho * (1-E) / E
dt = (rho/Cv) * Head (1-E) / E = (rho/Cv) * Head (1/E -1)

The rest is units.

* It doesn't relate well to the current situation. In this derivation dt is the steady state temperature rise from suction to discharge while pumping fluid. You're looking for a temperature rise of some representative temperature of the pump fluid above initial temperature over time during shutoff condition. It's Apples and oranges (other than the fact that both dt's arise from pump losses). If you knew the total losses at shutoff L (watts) and a representative (conservatively-low) lumped heat capacity C (Joule/degree), you might neglect heat transferred out and estimate a conservative bounding-high value of dt((time)) = (L /C) * time .... I gather maybe that's where you were already headed in attempting to determine volume (to estimate C)


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(2B)+(2B)' ?