Selection Of Pump

Sounds like homework.

Procurement would not even bother looking at temperatures!

You might want to find the power that is needed to get the water where you need it and apply your efficiencies and voila there will be the power.

With a bit of effort you can do this!

Watch out, its a trick question. The density of water at the given temperature is the most important missing parameter!

Have fun!

"Temperature of water inside tank is around 34.9 Degree" - get real! How accurate do you think the efficiency figures are? ±5%? Maybe, if you're very lucky.

Read something about total dynamic head.

You may approximately use 1000kg/m³ density. if you want the exact find a steam tables on the net, find v_f @ 34.9C Pressure at 14.7psi density is just the inverse of v_f.

eff. =WHP/BHP

Speak to your chosen pump supplier(s) and ask them. Be prepared to also provide information on inlet and outlet pipe sizing, layout and tank connection as well as any other flow restrictions. Please note that 34.9 C is close enough to 35 C for any practical engineering calculations that you should be able to find or calculate the density from steam tables (they often have this sort of information for feedwater). Lastly, unless this is homework pumping on the order of 700 litres of water per second seems to be a fairly important undertaking, enough to warrant more than the opinion of posters on an anonymous forum.

issues on anti-trust is critical

Homework alert!

Your first assignment is to convert 2500000 Kg/hr into volume.

Because I don't believe you have confidence in yourself to read a pump curve and since you have to buy your pump, contact a pump supplier to ensure correct sizing.

It could be either you or your instructor's fault.

Blame him for not teaching you the very best way that you would learn and/or yourself for not listening or not learning the basics ever since.

Change your mind, you could either be on track or off track on whatever you pursue.

The easiest way might not be beneficial for your future.

Thanks all for the replies. There are some serious errors found in the given values in my OQ. If time permits, I will let you know the correct values.

Don't bust a gut.

Contact a pump supplier, directly.

Coming here just wastes our time and yours.

Your pump supplier (if they are any good) will insist on having the needed information and doing the calculations themselves.

That way, you'll get the pump you need, not the one you THINK you need.

If this is homework, we don't do homework here!

Eeny meeny miny moe...

CR4 ADMIN: Deleted Post

Vulgar/Rude/Improper Behavior: This post was deleted because it did not adhere to the behavioral policies of the site. Please review Section 14 of the CR4 Site FAQ and the CR4 Rules of Conduct.

You provide an object lesson for potential posters.

1. Don't post stupid questions.

2. Don't insult the people you ask for help.

3. Rudeness never wins friends, but may influence your enemies.

4. Never argue with a fool. On lookers may not be able to tell the difference.

Good Suggestions.

yES sIR

Nah, just kidding around. It's a humor seeing Anon#1 and 2 argue.

I suggest a dart board.

With child proof darts of course.

If your having a tantrum they should bounce off of the dart board instead of sticking. This would be good choice for random selections.

The shaft power is the flow rate multiplied by the rise in pressure divided by the pump efficiency.

The motor power is the shaft power divided by the motor efficiency.

The only tools you need are your pocket calculator and a telephone to talk to potential vendors.

Assuming I did my algebra properly, you hope to move more water than the combined average water flow of the Dettifoss and the Jockfallet waterfalls with one pump and to pump this up by 35 meters. I want to know how big is this mythical open tank?

Dear Friend, The calculation is as follows and your

GIVEN DATA:

FLOW : 25,00,000 Kgs./Hr. = 694.44 Kgs,/Sec. Head : 35 Metres Efficiency : Pump -70% Motor - 90% Water Temp.: 34.9 Deg.C Hence from Tables Density = 994.0357 Kgs/M^3 (Or) 0.994039

CALCULATIONS:

Assumed Data – Suction Velocity – 1.0 M/sec. Delivery Velocity – 2.0 M/Sec.

Power Required is (W X H) / 75 x (Pump Eff.) x (Motor Eff.) Since 75 Kg. M is one H.P. absolute shaft power.

Hence

(694.444) x (35)/ (75) (0,7) x (0.9) = 661,37 H.P = 496.028 KW

Rounded to 500 KWe

3. PIPE Size

a) Suction Pipe

Flow in (M^3/Sec) =

(Weight in Kg./Sec.)/1000)) /(Density) = Area x Velocity

(694.44 Kg.Sec)/(0.994) = {0.698635} M ^3/Sec. = Area x 1,0 M/sec

Therefore Area = 0.6986 sq. m = (( 22/7)/4) x D^2

Therefore Suction Pipe Dia., = 0.6369 Metre = 636.9 MM Rounded to 650 MM

b) Delivery Pipe: Velocity = 2.0 Metres.

Dia. Of Delivery pipe calculated similarly as 478.8 MM rounded to 480 MM

DHAYANANDHAN.S