Centrifugal pump problem 4

[ChipFuller]

I have a situation that I’m trouble shooting. We have process where are trying to maintain some small process vessels at 100 degrees F. The process is a closed loop water system running through a shell & tube heat exchanger which is heated by 15 psi steam on modulating valve. The centrifugal pump is 3 HP pump running at 50 psi head differential and 10 gpm.

Here’s the problem. The temperature keeps rising past the 100 degree control point. I think the problem is the pump is adding too much heat to the system. After looking at the pump curves, the pump is operating at close to dead head conditions. Here’s my questions.

What is the formula for calculating the heat transfer rate?
The local pump representative suggested a Gould multistage pump with a ¾ horsepower pump to reduce the heat input. Does this sound like a logical suggestion? How much will this reduce the heat input?
Is there any other suggestions?

Thanks

 

[RWF7437]

Of my calculations for water are correct, your pump/motor combination is
only : e=10 gpm x (50 psi x 2.31 ft/psi)/ (3960 x 5 hp) = ~ 6% efficient. If 94% of the power is being wasted as heat and can be transfered to the fluid you are pumping ( is it water ?) it is hardly surprising that it is being heated up. You have the equivalent of a 4.65 hp heater. What would the output of such a heater be in BTU/ hour ?

This very low efficiency suggests that the pump is ill suited to the task or that there is a partially closed valve somewhere or that something else is amiss. Pumps are usually chosen to operate at or near their peak efficiency; not near their shutoff head.

good luck

 

[d23]

Chip,

A 3 HP pump will not add much heat to your system. If you assume all of the pump inefficiency is converted to heat and is only transferred to the liquid then the formula would be:

Fahrenheit rise = ((BHP – WHP)*42.41)/(lbs/min * Specific heat)

Where:

BHP = Brake HP
WHP = Water HP = (Q GPM * 8.33 * Fluid Sp GR * H Ft) / 33000
Lbs / min = Q GPM * 8.33 * Average Specific Gravity
Specific Heat for Fresh Water at 4 degree C = 1.00
Specific Gravity for Water at 4 degree C = 1.00
42.42 = Constant conversion of HP to BTU

This is a worst-case formula. It puts all in heat in your fluid.

Hope this helps
D23

 

[d23]

Chip,

Should have told you if its water you are adding less than 2 degree F.

D23

 

[RWF7437]

Correcting my error.

Your pump / motor combination is only about 9%-10% efficient. Still sounds very low. Although it may not be the cause of your heat problem it is cause for concern.

 

[RWF7437]

d23,

Don't your calculations show that the pump may be adding about 1 to 1.5 °F PER MINUTE to this "closed loop" system ? If the starting temperature of the fluid, which let's assume is water, is 30 °F it would only take about 47 to 70 minutes to raise the temp by 100 °F. ???

Add to this the heat being added by the steam and we're pretty quickly over the desired temperature, aren't we ?

 

[quark]

Chip,

There is a temperature rise across the pump as already suggested, by about 1.37F. 3/4HP pump as suggested will take care of the hydraulics in a better way. However, I think your main problem may be the steam control valve.

With your permission, I will convert the data to SI and do some calculations.

Water flowrate = 10USGPM = 0.63L/s
Initial temperature = 30⁰C (assumption)
Final temperature = 100F = 37.78⁰C
Heat input required = 0.63kg/s x 4.18kJ/kg C x (37.78-30)C = 20.49 kJ/s

Steam pressure = 15psia (I assume) = 1.034bara
Latent heat of steam = (2676.43-421.458) = 2254.97kJ/kg
Therefore, steam flowrate should be 20.49/2254.97 = 0.009kg/s = 32kgs/hr. Check the size and control characteristics. Also, check if your trap is leaking.

 

[d23]

RWF7437

You are correct a closed loop will continually add heat.

Operating this pump near shut-in will also affect pump run life however; without knowing the actual system I wouldn’t want to guess at that.

D23

 

[ChipFuller]

Thanks everyone for there replies.

One thing I tried probably but not long enough is that I shut off the steam supply so I knew for sure that heat was not being inputted into the system. I ran it for about 15 minutes and the temperature rose about 2 degrees from 92 to 94.

I've seen the formulas showing that d23 provided. What I don't is that temperature rise calculted in degress per minute or degrees per hour.

One thing about this system is that as I add heat from the pump I'm also loosing heat from system due heat loss from piping. If I knew how much Btu/hr the pump was adding then I could decide if a smaller pump would solve the problem.

Thanks

 

[BigInch]

In any case, you have the wrong pump now. Its probably worth changing out.

http://virtualpipeline.spaces.msn.com

 

[ChipFuller]

I guess what I'm trying to figure out is if a 3/4 HP pump will prevent this from happening. This is why I'm trying to figure out the Btu/hr for the current pump plus how much heat 3/4 HP will put into the system.

 

[dcasto]

A 3HP pump at near dead head does not draw 3HP. The pump curve may be at 30% to 40% eff.

 

[TBP]

If the correct pump overheats your system, the process heating load would have to be the square root of nothing. 50 PSI pressure differential on systems like this is huge. This sounds like a job for a little low-head circulating pump, like those used for hot water heating loops in buildings. Think of the system like a ferris wheel. The ride operator puts people on so that the weight coming down is about the same as the weight going up. This means a relatively small motor can do the job. It's the same with closed-loop circulating systems.

Something else you may wish to check - I'm assuming that the pump is shut off during downtime. Is the steam positively shut off? If you're relying on the temp control valve on the steam to do the job, here's what typically happens. The sensor for the temp control valve will be on the HX outlet. When the pump is off, there's no flow. Over time, the system cools, the sensor detects this, and tells the steam valve to open, just like it's supposed to. By the time the water around the probe hits set-point, the water actually in the HX is at a FAR higher temp. When the pump is started, the first thing you get is a slug of VERY hot water moving through your system. This situation won't happen in a few minutes. It's normally something seen overnight, or over a weekend. The two solutions to this situation are keep the pump operating so there's always a flow over the probe, or have a positive shut off on the steam supply that is separate from the temp control valve.

 

[quark]

I've seen the formulas showing that d23 provided. What I don't is that temperature rise calculted in degress per minute or degrees per hour.

That is easy to do if you have a recirculating system. The calculation of temperature rise is done based upon the flow in gpm. That means the temperature of so many gallons of water rises by so many degrees in a minute. As this is a recirculating system, neglecting heat losses from the water, you can calculate the temperature rise in an hour by multiplying temperature rise value per minute by 60. If this is an open discharge system and the liquid is not circulated back to the pump suction then it is only temperature rise. No time basis is required to be mentioned here.

 

[pumpking]

I dont understand your system sufficiently, but if your local pump guy is saying you can go from 3hp down to 3/4hp,then the pump is clearly oversized. We had a situation pumping ferric chloride in a chemical milling machine (constant recirc through spray nozzles) and that kept increasing the temperature in a similar way. The pump internals are critically important in eliminating the error - but bottom line, the pumps were too big for the job !!

 

[ChipFuller]

Thanks everyone for your response. I was able to do some testing on the system by turning the steam off and running the system. I let the water in the system cool off overnight so when I started the pump the water temperature was 70 degrees F after 1 hour the heat had risen 20 degrees.

The more I think about this it makes sense because this similar to a hydraulic system except in this case the fluid is water rather than hydraulic oil. In both cases the systems are closed loop. With hydraulic system you have to cool of the oil because of energy from the pump so this setup would be the same way.