Working Out The Cost

[canuckamuck]

I need some help with a cost analysis for my friend.

He wants to hook use a 2hp pump to bring water from the river to irrigate a 30 x 30 meter commercial vegetable garden. The amount of lift would be about 20 meters.

What he needs to figure out is the hourly cost of running the pump to determine if it is worthwhile.

Like I said the pump is 2 Hp, and it also has a 1.5 KW rating. I did not jot down any other info, but I can get it if necessary.

I know this should be posted in the farming section, but I thought there might be some non farmers that know how to work this out. Thanks

[Tywais]

1 hp = 746 watts thus 2 hp = 1492 watts or round up 1.5kW/hour running. Lot's of variables in that the watts drawn depends on the pump loading. Using max load of 1.5kW and unit cost of electricity is 4 baht then 6 baht/hour running costs.

[torrenova]

So if it has a max power consumption of 1.5kw then the cost is price of electric x 1.5 x number of hours used. No ?

[soundman]

So if it has a max power consumption of 1.5kw then the cost is price of electric x 1.5 x number of hours used. No ?

Theoretically correct if the motor is working at full capacity (drawing 1.5kw in this example) for the duration of the measured time.

Pumps are hard examples to figure out theoretical consumption on paper.

If you switch the pump on with no water in the pipe the motor will draw minimal electricity. If you have a long skinny pipe outlet as opposed to a short fat one the current draw will be higher. If you are drawing up to rated suction height & pumping to maximum specified head height the pump will be running at maximum rated capacity. If the outlet hose is blocked the pump will be running over rated capacity & likely burn out.

Cheers,

Soundman.

[canuckamuck]

Thanks everyone that was way too easy, and very quick. Looks like my friend can grow veggies now.

A virtual mug of suds for all of you.

[chownah]

1 hp = 746 watts thus 2 hp = 1492 watts or round up 1.5kW/hour running.

Wouldn't that be 1.k kW-hours/hour?....or just plain 1.5kW?

Chownah

P.S. Sorry, for some reason using power units for energy or visa versa just pushes one of my buttons and I can't help myself.

Chownah

[naka]

So if it has a max power consumption of 1.5kw then the cost is price of electric x 1.5 x number of hours used. No ?

Theoretically correct if the motor is working at full capacity (drawing 1.5kw in this example) for the duration of the measured time.

Pumps are hard examples to figure out theoretical consumption on paper.

If you switch the pump on with no water in the pipe the motor will draw minimal electricity. If you have a long skinny pipe outlet as opposed to a short fat one the current draw will be higher. If you are drawing up to rated suction height & pumping to maximum specified head height the pump will be running at maximum rated capacity. If the outlet hose is blocked the pump will be running over rated capacity & likely burn out.

Cheers,

Soundman.

If you have a long skinny pipe outlet as opposed to a short fat one the current draw will be higher

Not so ... With a centrifugal type pump (almost all water pumps are) then partially choking the outlet

will cause the pump to semi-cavitate and so the current will drop, not rise.

With centrifugal pumps the current draw is mainly dictated by "work done" e.g. flow and lift.

If the outlet hose is blocked the pump will be running over rated capacity & likely burn out.

The pump in this case may burn out, but not because it is running at over rated capacity as it will not be, more

likely because of lack of cooling to the bearings or motor.

Naka.

[chownah]

So if it has a max power consumption of 1.5kw then the cost is price of electric x 1.5 x number of hours used. No ?

Theoretically correct if the motor is working at full capacity (drawing 1.5kw in this example) for the duration of the measured time.

Pumps are hard examples to figure out theoretical consumption on paper.

If you switch the pump on with no water in the pipe the motor will draw minimal electricity. If you have a long skinny pipe outlet as opposed to a short fat one the current draw will be higher. If you are drawing up to rated suction height & pumping to maximum specified head height the pump will be running at maximum rated capacity. If the outlet hose is blocked the pump will be running over rated capacity & likely burn out.

Cheers,

Soundman.

If you have a long skinny pipe outlet as opposed to a short fat one the current draw will be higher

Not so ... With a centrifugal type pump (almost all water pumps are) then partially choking the outlet

will cause the pump to semi-cavitate and so the current will drop, not rise.

With centrifugal pumps the current draw is mainly dictated by "work done" e.g. flow and lift.

If the outlet hose is blocked the pump will be running over rated capacity & likely burn out.

The pump in this case may burn out, but not because it is running at over rated capacity as it will not be, more

likely because of lack of cooling to the bearings or motor.

Naka.

I believe you are mistaken. It is blocking the inlet (not the outlet) to the pump that will cause cavitation and I'm not at all sure if during cavitation the pump will draw less current or not. Blocking the outlet pipe or the outlet hose will cause a greater pressure that the pump is working against making the pump work harder....a pump working harder will draw more current which will raise the motors internal temperature...failure if it occurs will be from melting of the insulation in the motor windings...I think.

Chownah

[gregb]

I need some help with a cost analysis for my friend.

What he needs to figure out is the hourly cost of running the pump to determine if it is worthwhile.

As someone else said, the hourly cost is easy. You can pretty much assume the motor will work as hard as it can unless it breaks, has some kind of governor, or the power is turned off. So that works out to 6 baht per hour if you are paying 4 baht for electricity.

That's not the interesting question though. The interesting question is how many liters of water can be pumped during this hour. For that, you need to specify a whole host of parameters, the most important of which are the efficiency of the pump, the length and diameter of all PVC pipe in the irrigation system, and any head loss on the delivery system (sprinklers, drip, open valve, etc.)

My recommendation is simply to install the pump and gather this information experimentally. It's a very difficult calculation to make with a lot of different variables. You can however, easily establish the maximum theoretical possible value by looking at:

work = mass x gravity x height

You know the work, 1.5 kWh (5.4 MJ). You've specified 20m as the height, and g is a constant 9.8 m/s^2. All you have to do is calculate the mass. So, theoretically, if you had no losses anywhere in the system, you could get a maximum of 27.5 cu. meters of water per hour for this.

In the real world, I'd be surprised if you can get 50% over the entire system even with a very expensive pump, so I would put a ceiling at around 12 - 13 cu. meters on 6 baht per hour, and in my experience you'll probably only get about 6 cubes per hour unless you take great care engineering your system.

So tell your friend to figure about 1 baht per cube of water in electrical costs for calculations, and he'll have to experiment to find the real number. Physics places the lower cost limit at about 22 satang per cube, but you can't actually get this low. (All numbers assume electrical cost of 4 baht/kwh)