pump question? 3

[banshee33]

hello , new to forum , not an engineer , just a wrench puller .i would like some one to explain something to me about a single stage ansi pump . if i have to pumps in parallel , only using one pump at a time . both pumps identical , suction size,discharge size, same impeller size . one pump has a electric motor with a 200 hp rating and the other has an electric motor with a 400 hp rating , both motors run the same rpm . what happens to the flow or head between the two ? only running one pump at a time .

 

[Pumpsonly]

Both pump will perform the same flow and head if they are identical as you said. The one with 400HP motor may give an insignificant better performance due to it running a few RPM more than the 200HP motor unit.

 

[banshee33]

i dont understand , if it was on a pump curve with hp required , and the 200 hp motor was sized right for the application , and a 400 hp motor was used , it would change were the pump is pumping on the curve , no? would the bep change? just hard to sink in to me

 

[Pumpsonly]

The centrifugal pump performance depends on the RPM, not the power of the driver.The difference it makes with a bigger HP motor is that the motor will run at less slip and closer to its synchronous speed than the small motor.

 

[banshee33]

So speed and impeller size dictate where the pump is on the curve, not hp . and the system will not change going from 200 to 400 hp . There's no laws on using hp ratings if 200 is required then 400 wont change anything in the system because speed and diameter determines that in pump laws . i dont get it lol sorry just mule headed!! if i have a carrige with 200 horses running a straight line and one with 400 running my velocity will change, how is velocity related to flow in a system.

 

[micalbrch]

The rpm of the motor determines the rpm of the pump and that determines the flow of the pump. The velocity (in the pipeline) is determined by flow and pipe diameter. If both are constant the velocity will not change. The 400 HP motor might allow the pump to work against a higher head/pressure. But if there is no high pressure 400 vs. 200 hp will not make a difference.

Assume you have truck and a small car. If you drive both with same constant speed, the much higher power of the truck does not make a difference. Not the best comparison but I cannot think of any better at the moment.

 

[hydtools]

The motor will deliver only the power required to drive the pump to meet the system demand as long as the motor is capable of delivery at least the system demand. If the system demand is 100hp, any motor capable of 100hp or more will work and deliver no more than the 100hp system demand. 200hp-rated motor will deliver only 100hp at system demand. 400hp-rated motor will deliver only 100hp at system demand. 1000hp-rated motor will deliver only 100hp at system demand.

Ted

 

[zdas04]

There is a tiny bit more to it than that. Load (hp) is a function of discharge head and flow rate. If the motor is running at full load and the discharge pressure requirement increases then the motor (and therefore the pump) must slow down until load matches capability. So in the 100 hp example above, you wouldn't see any difference between the 200 hp and 400 hp motors with an increase in discharge pressure.

If the actual load is closer to 200 hp, then a fairly small increase in either flow rate or discharge pressure will tend to slow the motor down (decreasing the hp available to do work, and decreasing the flow rate through the pump). With the 400 hp motor, there is room to increase current (and therefore hp) to prevent the motor from slowing down with increased load.

David

 

[hydtools]

zdas04, my example simply assumes constant system load, nothing more. You are adding 'what if' system load changes and resulting driver response. What you say is true, but complicates the response to the original post. Perhaps the exemplar 200hp pump/motor is running at 100hp system load.

Ted

 

[SNORGY]

Maybe the very simplistic but "close enough" way to put it is that the power required to be put into the system by the pump, corrected for the pump efficiency, will be the power required to be put into the pump by the motor, corrected for the motor efficiency. Whatever surplus power (in this case at least 200 hp) is available by the bigger motor simply isn't tapped into.

Regards,

SNORGY.

 

[bimr]

Note that operating the larger motor will result is an extended service lifetime, but the electrical efficiency is not optimal, since the electrical motor never operates at its optimal duty point. The power factor will normally be low due to the partial load.

 

[zdas04]

hydtools,
Your example simply assumed a set of conditions far from any interesting boundary. The OP never mentioned where on the pump curve they were operating. You can assume 100 hp and system load and discuss what (doesn't) happen as you swap pumps. I chose to assume a point closer to a limit because I found it more interesting.

The universe that I live in is really short on systems with a "constant system load". If the OP has such a system and it is operating far from the hp limit of the 200 hp motor then I would be surprised that they had installed the second pump with a 400 hp motor. But people surprise me all the time.

David

 

[BigInch]

Hydraulic efficiency might be a smidge higher, perhaps not, but the most important thing you need to know is that the load factor of 50% on the 400 HP motor will not allow it to run very efficiently. Expect to have a 75% motor efficiency for that 400 HP motor; well below what they are capable of when loaded near their rated load. I would expect to see higher power consumption for the 400 HP motor running at 50% efficiency than you would have if running the 200 HP properly loaded ... and for that you will take maybe a 20% hit at the electric meter!

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[banshee33]

thanks for all the comments, i think i understand, i am not sure but i think the 200 hp motor runs most of the time so using the 400 hp motor when only close to 200 hp is required , the motor is not running efficient, and power loss is imminent.

 

[Pumpsonly]

David,

Re the last para of your first post of 14 Jan.
Increase in the load of a pump, will always increase the power draw from the motor,with reduction in speed due increase in slippage with a corresponding increase in the motor current even if the motor was already operating at it rated power.The motor will just being over loaded and burn out the insulation eventually if allow to run for extended period of time. The time take to cause damage is depending on how much is the over load, the ambient temp.and the insulation material rating.

 

[stanier]

banshee33,

Try

http://www.mcnallyinstitute.com

&

http://www.pump-zone.com

for more technical papers on pumps.

By the way banshee33 enough of the 'just a wrench puller". We are all part of teams bringing together systems and keeping them working. I have learnt a heck of a lot of stuff from guys on the tools. Even started there myself as a spotty 15 yo. My Dad was an electrical fitter and he taught me heaps. He had no academic training but started a factory during WWII injection moulding toys.

"Sharing knowledge is the way to immortality"
His Holiness the Dalai Lama.

http://waterhammer.hopout.com.au/

 

[bimr]

One big disadvantage of this operating scheme is the power cost, in particular the utility fees for demand charges for the big motor.

Depending on how they use electricity, electric utility customers are charged for different electric services. Along with a basic customer charge – which is a set fee paid monthly or seasonally – most customers pay for the energy they use (measured in kilowatt-hours, abbreviated kWh).

Larger users of electricity are also charged for something called demand (measured in kilowatts, abbreviated kW).

The demand charge is typically based on the largest motor in the facility.

Example: A customer runs a 400 horsepower (hp) pump for only five hours during July:
Demand Charge = 400 hp x .746 kW/hp x $8.03/kW = $2,396.15
Energy Charge = 400 hp x .746 kW/hp x 5 Hr x $0.034/kWh = $50.73

The demand charge will be a large part of the bill if the facility uses a lot of power over a short period of time, and a smaller part of the bill if the customer uses power at a more or less constant rate throughout the month.

The demand charge may be enough to buy a smaller motor.

The "Demand" charge is a premium price you pay to guarantee that all of the power your business needs will be there when you need it.

 

[BigInch]

Bimr, another good point. Is demand charge based on installed nameplate capacity, as your calculation implies, or on actual power used. I ask now, since the OP was talking about running a load of 200 HP on a 400 HP motor.

In your specific calculated scenario, I think that would force me to switch to IC engine.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[77JQX]

BigInch - at our facilities, demand charge is based on actual power used. I'm also curious about your statement on motor efficiency at 50% load being on 75% - I was under the impression that large induction motors had a flat efficiency curve down to around 25% load - that is, to not expect an efficiency penalty to running at 50% load. Where does the 75% efficiency value come from? Thanks.

 

[bimr]

Where I am located, the utility charges for both demand and consumption. A customer that sets a high demand requires more services from the utility--additional generating plant capacity, and more expense in lines, transformers and substation equipment.

The utility representative walks through the facility and sets the demand charge for the largest motor in the facility.