Max HP from 5 HP motor

[Sahib]

obviously
since suction lift is usually in feet

NPSHA = Atmospheric pressure(converted to head) + static head + surface pressure head - vapor pressure of your product - loss in the piping, valves and fittings

• Static head = 5 feet
• Atmospheric pressure = pressure x 2.31/sg. = 14.7 x 2.31/1 = 34 feet absolute
• Gage pressure = 0
• Vapor pressure of 68°F. water converted to head = pressure x 2.31/sg = 0.27 x 2.31/1 = 0.62 feet
• Looking at the friction charts:
  • 100 gpm flowing through 2 inch pipe shows a loss of 17.4 feet for each 100 feet of pipe or 17.4/10 = 1.74 feet of head loss in the piping
  • The K factor for one 2 inch elbow is 0.4 x 1.42 = 0.6 feet
• Adding these numbers together, 1.74 + 0.6 = a total of 2.34 feet friction loss in the pipe and fitting.

NPSHA (net positive suction head available) = 34 + 5 + 0 - 0.62 - 2.34 = 36.04 feet

Of course, it assumes the pump suction is below water level.

 

[Ingenieur]

Of course, it assumes the pump suction is below water level.

not necessarily

 

[Sahib]

not necessarily

But you have taken static head as positive number.

 

[Ingenieur]

But you have taken static head as positive number.

I misunderstood
you meant in the posted example
I thought you meant in all cases
sorry

 

[Besoeker]

Well, my good golly gosh, how far off topic can this go............

 

[Jraef]

Well, my good golly gosh, how far off topic can this go............

Said the pot to the kettle... :lol:

 

[Ingenieur]

Well, my good golly gosh, how far off topic can this go............

hmy:

practice what thy preach

 

[Phil Corso]

Here is a relatively simple rule-of-thumb to calculate temp-rise when operating a motor at higher than its nameplate rating:

The change in temp-rise above ambient is the rated value times the square of the service factor!

Example... if motor has a rated temp-rise of 40^o above ambient, and is be operated at a 1.15 service-factor, then, the expected temp-rise will be about 1.15² x 40^oC... say 53^oC!

Phil Corso

 

[GoldDigger]

Derived from I squared heating being higher by square and temp rise being proportional to rejected heat?

Sent from my XT1585 using Tapatalk

 

[junkhound]

Here is a relatively simple rule-of-thumb to calculate temp-rise when operating a motor at higher than its nameplate rating:

The change in temp-rise above ambient is the rated value times the square of the service factor!

Example... if motor has a rated temp-rise of 40^o above ambient, and is be operated at a 1.15 service-factor, then, the expected temp-rise will be about 1.15² x 40^oC... say 53^oC!

Phil Corso

Ah, but if one puts 10X the amount of cooling air thru the motor and uses it at a SF of 2X, is the temp rise only 40*4/10 = 16C ? Just kicking the can......:angel:

Or, going by a typical forced air chart for static fins on electronic heat sinks, 10X air = about 1/3 dT, and it is asymptotic at about 1/3, more airflow than 10X helps very little on basic finned heat sinks from low flow design point, probably a function of boundary layer.

Or, how maybe much does the added turbulence by the rotor aid in internal cooling.

FWIW external frame temp of example motor has a temp rise of just 12 C, with airflow of > 1000 CFM, no idea what the normal airflow in a 184T frame motor is. Don't have an internal thermocouple installed.

 

[GoldDigger]

Ah, but if one puts 10X the amount of cooling air thru the motor and uses it at a SF of 2X, is the temp rise only 40*4/10 = 16C ? Just kicking the can......:angel:

Or, going by a typical forced air chart for static fins on electronic heat sinks, 10X air = about 1/3 dT, and it is asymptotic at about 1/3, more airflow than 10X helps very little on basic finned heat sinks from low flow design point, probably a function of boundary layer.

Or, how maybe much does the added turbulence by the rotor aid in internal cooling.

FWIW external frame temp of example motor has a temp rise of just 12 C, with airflow of > 1000 CFM, no idea what the normal airflow in a 184T frame motor is. Don't have an internal thermocouple installed.

Just kicking the can back, that 16C would be correct if the only resistance to heat flow was at the interface between the motor parts and the ambient air.
There will also be a substantial temperature drop between center of winding and the parts of the case where the air flows.
How the numbers actually pencil out for that case, I do not know. But there are definitely limits to what you can do be adding air cooling at room temperature.

 

[Phil Corso]

Derived from I squared heating being higher by square and temp rise being proportional to rejected heat?

Right on!

Phil

 

[Phil Corso]

How the numbers actually pencil out for that case, I do not know. But there are definitely limits to what you can do be adding air cooling at room temperature.

The major components are carcass, rotor, stator, bearings, bearing housings, and shaft... and they certainly won't increase in temp at the same rate!

Phil