Temperature change after compression

[cannondale]

I'm looking for some help.
To calculate the temperature change of a fluid (water for example), when its pressurised to 345bar from atmosphere or 3bar maybe, and then what the temperature change would be when brought back to atmosphere again.
Pressure rise gradient for a closed valve condition for a reciprocating positive displacement pump. I'm assuming it'll be the same the gradient of the calculated pulsation frequency of the pump, but with some damping effect based on the bulk modulus, and volume/ length of pipe between the pump and the valve?
This is all to evaluate a pressure relief valves response to a change in pressure.

Any help much appreciated, as this is outside my field of knowledge. I'm learning but I'm kind of stuck right now.

 

[weldstan]

Gas compression produces a rise in temp per the Ideal Gas Law, PV = nRT. It does not apply to a liquid.

 

[cannondale]

Hmmmm I understand that there's a substantial difference between gases and liquids. However, why doe the fluid get very hot after running tests on a pump in a test bay with a circulation of fluid from a tank? I know the fluid gets hot, and I'm presuming there's a temp change where the compression/decompression takes place.

 

[zdas04]

Gas compression is closer to the Adiabatic Gas Law than the Ideal Gas Law.

The only mechanism for a liquid to increase in temperature during pressure increases is mechanical heat transferred in from the pump or fluid friction. Both are generally negligible. Heat of compression is not a thing in liquids (even when you add enough pressure to reduce the volume measurably any heat of compression would be lost to the atmosphere).

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[BigInch]

Heating of a liquid can be estimated by the inefficiency of the pump. If a pump is 67% efficient, then 100%-67%, or 33% of that work can be roughly assumed as going into heating of the liquid.

 

[LittleInch]

Cannondale.

I'll echo the previous posters. What you are experiencing is the heating of water due to friction, either within the pump or within the piping or the pressure regulating valve. In a relatively small volume test set up with water constantly flowing round a loop and little opportunity for it to cool down it can get really quite hot. However this is friction, not due to compression.

Remember any pump has a hydraulic efficiency of no better than 75% and probably less. Hence look at the electrical power of your pump, take 25 to 30% of it and that is the heat going into the water. This could easily be a few kW constantly. Then think about if you put a heating element of that power in your tank and it should then become a bit clearer to you.


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

 

[zdas04]

BigInch,
That only applies to the pump-end efficiency (I know you know that, but the "pump" in many people's minds is properly called the "pump skid" or "the pump and driver"). I have seen people try to dump the heat lost in the driver into the process fluid and get a wrong number.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[BigInch]

This online calculation uses total volume in the system and pump efficiency (or rather inefficiency) to calculate the heating of the fluid. (assuming a closed circuit)

http://www.freecalc.com/ptemfram.htm

 

[cannondale]

I'm considering the issue when using a PD pump. I've always used 95% eff fluid end, and 85% eff for the mechanical/driver. so approx. 5% as a heat/noise estimation. Acknowledged the friction through the choke (relief valve) would be where id expect the majority of the heat to come from. When I'm considering the smaller pumps to consume around the 300kw mark, and the larger ones to be 2000kw I know the heat build up can be substantial in just a 5m3 tank recirculating at 50m3/hr.
I was hoping for an example of how I might calculate the temp rise, but think I'm understanding its a lot more complex than I'd maybe given credit for . Might just stick to the humble 5% rule. There's still the 5% energy into a given flow rate?

 

[cannondale]

Hi BigInch,

That's exactly what I was looking for, but wanted to try and understand the math behind it.

 

[LittleInch]

Just remember the golden rule that energy can't be lost, only change state.

Hence start with electrical power in, loose some in the motor heat / efficiency (5-10%, some in the pump (varies depending on your pump type and size, but 10% looks typical) then what work the fluid does. If the flow velocity isn't big or distance not long, but you have a big pressure drop over a valve, then tat's where most of you potential energy "stored" in the high pressure fluid is released as heat, mainly into the fluid.

Those look mighty big pumps for a non cooled test loop....

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

 

[cannondale]

LoL, yes they are pretty big numbers. They are for a pump test set up. Specifically a 375kw test in 30degC, approx. 70-80% humidity, and no facility for cooling the test fluid. Only maybe, more tanks and/or more water in the loop. Its not often we have to do it, but needs must an all that. This is diesel engine driven btw.
The calc online comes up with +37degC after 1hr at normal rates.

 

[LittleInch]

Care to share your numbers - flow, head, pressure loss, water volume etc.

375 kW and only a 37 deg rise after 1 hour?? Volume must be pretty significant for that???

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

 

[cannondale]

http://www.freecalc.com/ptemfram.htm

Fluid is water
Flow rate 450lpm
Pump Head 2750m
Pump Eff 95%
Tank volume 5000ltr
Recic rate 1hr

result +36.61 C

That's a typical example of a smaller pump test. However I've convinced them to use a 32m3 tank now, so approx. +5.7C in an hour which is much more manageable.

 

[LittleInch]

That works for me as well. A high head low flow pump. I only get about 215kW shaft power but 32 cubic m tank sounds a much better idea.

Hope you now understand the theory a bit better.



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

 

[cannondale]

I was kind of hoping for a mathematic solution rather than an online calc, but I got the answer I was looking for none the less, I just cant use it in a report very easily.

 

[BigInch]

Pump work * (1- pump efficiency), convert to calories, calculate temperature increase of fluid.

 

[LittleInch]

Ok, here's how I did it.

Some assumptions:
Your pump is set up with inlet at a nominal pressure (0-2 barg), your pump is flowing at the head you gave. This pressure is then dissapeted through a control valve (choke valve?) back ont pot he tank from which it came essentially at 0 barg.

This is then an energy equation. For all intents and purpose's the energy imparted into the fluid by the pump (shaft power) is then dissipated by friction (heat) in a small loop, mainly by friction inside the control valve resulting in increased temperature of the water. Assume no heat loss to the outside.

Shaft power = Q x H x density x g / (3.6 x 10^6 x eff (0.x) ) Units are m3/hr, m, kg/m3, 9.81 = KW

Heating of water - Specific heat of water is 4.186 kJ/kg/K.

So for you case, to raise 32,000 litres = 32,000kg of water 1 deg C needs 133,952 kJ.

Shaft power for your pump = 27 x 2750 x 1000 x 9.81 / (3,600,000 x 0.95) = 212.98 kW = kJ/sec.

So number of seconds to raise your water 1 deg C = 133952/213 = 628 seconds = 10.48 minutes. Hence rise in one hour = 60/10.48 = 5.72 C

Life is not as precise as this and some energy will dissipate not directly into the water, your pump may not run exactly as you want or be as efficient, but you should only get a temperature rise between 5 and 7 degrees C with that amount of water.

If you need any back up for the formulae or numbers just google it.

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

 

[cannondale]

thank you, v much appreciated.

 

[Latexman]

This isn't exactly what is desired, but it's close. See attachment. From my 5th (1980) Edition of the Durco Pump Engineering Manual, plus some artwork (scribbles) from one of the latexkids many years ago. It's available for sale on-line for $45. Mine was $10 back in the day. Just Google "Durco Pump Engineering Manual.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.