Cooling tower fan loads
We recently had an "Energy Expert" come through the plant to do a survey. He told us that instead of running one 75hp cooling tower fan at 90% we should run three at 30% and our total load would go from 56kw to 2kw. We have some excellent control tech and a top of the line BAS. He says the system can be programmed to perform this load spreading automatically and that we would get these big savings. I have a hard time believing this. Is this a standard practice? thanks. km
While it is true that a motor with a VSD runs at dramatically reduced kW consumption when running at the lower end of the frequency scale, I doubt the expert's numbers. I do not believe for a minute that one 75 HP tower fan motor is going to draw 56 kW. Nor do I believe that three 75 HP motors running at a minimum speed will draw 2kW. With a VSD, a 100 HP motor is equivalent to 12.5 HP when it is running at 1/2 speed (30 Hz.). So your 75 HP motor running at 30 Hz would be the equivalent of 9.375 HP or about 28 HP for all three.
The other thing one has to consider when slowing a tower fan is at what point is it ineffective to slow the motor any further. When you say 30% are you talking about 30% of FLA or 30% of 60 Hz.?
I believe the energy expert is on the right track. I just don't believe the numbers as stated here.
a regular ol' 75 hp, 3 phase motor draws 55 kW...but do you actually draw 55kW from the motor? most motors are fairly over-sized...some more than others. i agree with KY. it should be pretty easy to measure the power the motors draw at different loads and make a determination. also, double check with the tower/motor manufacturer on 'how low can you go'. keeping motor bearings greased at lower speeds is more difficult. many motor manufacturers have minimum speed requirements to maintain the lubrication.
"I got both hands on the wheel
And my big foot on the gas"....Chickenfoot
Originally Posted by KnewYork
Take the easy way out and go to one of your drives, dial it back to 30% (18 hz) or 30 hz (50%) and check your kw consumption using the drive panel or your own meters. I don't have time to run the numbers so I can't tell you what you actually will be running at the reduced speeds, but I think you'll find that you can only spread it so thin.
My first question to you would be why are you not running them all at the same time, anyway? When you leave a fan off and have common tower cells, you mix heated water in the cold deck and defeat your purpose by driving the on-line fan speed up to compensate. Do you have isolated cells?
I agree what everyone else is saying. I think your "expert" needs to recheck his numbers. Also, in addition to what everyone else is saying, CFM is not proportional to speed on a fan. So, 50% RPM doesn't equal 50% CFM. I want to say it is squared meaning when you reduce the speed by half, CFM reduces by 4 times, but im not sure on that.
We haven't been told what kind of fan this is. The fan laws apply to centrifugal fans. I would assume that a tower of this size would have propeller fans.
Originally Posted by R123
Here's the fan law that would have some bearing on the OP's question:
The first fan law relates the airflow rate to the fan rotational speed: Volume flow rate, Q, is directly proportional to the fan rotational speed, N.
(Q1/Q2) = (D1/D2)3 (N1/N2)
Since the diameter of the fan is not changing then the flow is going to be directly proportional to the speed.
Dear KY: I did not know the fan laws distinguished between different types of fans, But you are correct, they are propeller fans.
Originally Posted by KnewYork
I appreciate the comments. Thank you. kjm
The fan laws do have approximation to other types of fans, but as stated the laws are for geometrically similar centrifugal fans.
Originally Posted by kjm4422
Here's a website that shows the laws and the formulas.
I will weigh in with my $.02 here.
As previously stated much depends on the type of fan. A prop fan is positive displacement so the standard centrifugal laws dont apply (that most people are familiar with do not apply, a seperate set of laws applies, see below.) However with that said, it is always better to use more cells than less when using a cooling tower. This will provide a larger surface area for your heat exchanger (since a cooling tower is a specialized heat exchanger) and require you to flow less air to meet the same setpoint. However, you have to be careful about minimum flows over the tower. If you go less than a certain flow rate (50%) I believe is the rule of thumb, you will start to have carry over due to not enough water in the fill.
As far as the laws for positive displacement speed changes, I believe the power consumption tracks linearly with speed. Ie... if you are consuming 50hp to drive the fan at 1800 rpm and reduce to 900 rpm you will need 25 hp, as opposed to a centrifugal device where you would need 6.25 hp. Definately double check this, but I believe thats the rule for positive displacement.
Fyi.. typically prop fans are used on cooling towers that are induced draft, because you need a smaller motor when using an induced draft fan as opposed to a forced draft fan. this is because you are not pressurizing the fill, you are reducing the pressure across it.
So while your energy expert is on the right track, I dont think his numbers are accurate.
Everyone is an expert nowadays.
Are you referring to water flow or air flow? We aren't discussing water flow, but if we were wouldn't it stand to reason that the fan speed would be reduced as the water flow was reduced (as in a Hartman Loop). I have seen many tower fans turning at less than 30 Hz...in fact, many towers here will run without the fan running at all for periods of time in the winter.
Originally Posted by Robs98ss
I don't understand what this has to do with the OPs question. We aren't discussing positive displacement fans. And as far as fact checking...I think the person who posts it should do that.
Originally Posted by Robs98ss
I'm not trying to bust your b**ls. My opinion is that we should try to stay on topic. The OP asked about fan speeds.