How many liters per minute does a 1.5 HP pump pumps?

How many liters per minute does a 1.5 HP pump pumps?

I am doing a water level system using flow sensors. I want to know what is the max liters of water a 1.5 HP pump pumps in a minute. Pump is 230V, 50Hz, 1 phase.

You can either calculate the hydraulical work performed by the pump, which is a theoretical limit for the flow. Or refer to data sheets of existing pumps (after you selected a specific pump type).

For the hydraulical work, you'll consider hydrostatical pressure (by static head), pipe pressure drop and internal pump pressure drop. Multiply with expectable pump efficiency.

Mechanical engineering text books have ususually example calculations.

I calculated the L/min using wet method mentioned elsewhere on the web. My 1.5 HP pump pumps approx 14 L/min and I have purchased 40L/min flow sensors.

My system will have a 3.2" TFT and 2 flow sensors. The reservoir and high level tank water levels will be displayed graphically. I have used Mega 2560 + 3.2" TFT + TFT Shield + 2 X flow sensors.

Pumps come in all shapes and sizes and the flow depends on the 'head' it has to pump to but I am suspicious that your 1.5HP pump only manages 14L/min. Mine here shifts 25m3 per hour or about 415L/min. Double check your figures, I'm not saying you are wrong but the big difference concerns me.

Brian.

It is pumping to 3rd floor and it is around 48 feet high. Maybe that is why it is 14 L/min. I am doing it for my house and I have measured it practically by measuring the output of the pump.

A simple centrifugal pump (Grundfoss JP6) achieves about 75 l/min over 20 m height, at not more than 25 % effciency.

Hello!

Basically there are 2 different issues.
1. The height (usually you use pump to put water or any liquid higher). This is easy to calculate,
the power taken by the liquide transfer is mgh/t (mass x gravitation x height / time).

2. The load loss (I don't know the word in english). It means that it depends on which tube you use.
if your tube is big enough, then the pressure difference will be negligible. Careful: the load loss is proportional
to 1/d^5 (inversely proportional to the 5th power of the tube diameter).
And it's of course proportional to the length of the tube.

For example, if you want to move 1 l/s of water with a tube (smooth inside) of 1 cm internal radius (therefore
20 mm internal, smooth tube), then you need 0.05 bar (that's a value I have memorised because of it's simplicity).

Now a 10 mm radius tube tells you that the surface is 3.14 cm^2, therefore you need 3.18 meter of tube
to hold one liter. Then for the pressure, as you need 0.05 bar per meter, you will need 0.159 bar for your
3.18 meter. As for the force, 0.159 bar over a surface of 3.14 cm^2 yields 0.49 kgf or 4.9 N. And these
4.9N working over 3.18m yield about 15W.

Now let's redo the calculation for a diameter of 14mm. You may remember that root(2) is 0.7. This tube
of 14mm will therefore have half of the surface, and therefore you have to double the length to have one liter.
As you needed 0.5 bar for 20mm tube, and as the pressure vs diameter is a 1/d^5 relation, you will therefore
need 5.65 times the pressure (5.65 is (root(2))^5). Therefore you will need 2.82 bar per meter.
But this time, you double the length, and you need therefore 6.36 meter therefore 18 bars.

As you have only 1.57 cm^2, the force is now 28kgf or 280N. And over 6.36m, you will need 1800W, just
to push the water into the tube.

Well, it's late and I have the impression that the difference is huge. There might be a miss somewhere in
my calculation and I should verify. But for sure, even a small variation of the diameter makes a huge
difference.

Dora.

In a first order, flow resistance is inverse proportional to fourth power of tube diameter.

But the size water supply pipes will be ususally chosen for moderate pressure drop at rated flow.

Hello!

FvM said:

In a first order, flow resistance is inverse proportional to fourth power of tube diameter.

Click to expand...

We are maybe not talking about the same thing. I say that for a given flow
(in liters per second), the resistance is inverse proportional to the fifth power
of the tube diameter. Here is the proof:

It's generally admitted that fluide flow in a tube can be described with the
following relation:

dP/dL = l * r * V^2 / 2D (Equation A)

where:
dP is the difference of pressure between both ends of the tube
dL is the length of a piece of the tube for the current experiment
-> But let's leave this under the forem dP/dL which will be linear
pressure loss of our tube.
l is a general factor of pressure loss (a constant in standard units)
r is the volumic mass
V is the speed of the fluid
D is the diameter of the tube

As r and l are both constants, we can call the product K, and then we get:

dP/dL = K V^2/2D (Equation B)

Now we are talking about a constant flow (let's call it Q for quantity, let's remind
that it's a quantity per second and that it's a constant, 1l/s in my example above),
and we want to know the difference of pressure between both ends as a function of
the diameter.

As Q is constant we can express V from Q and D.

As V is the speed of the fluid, it is the length per second. It is therefore Q/S,
S being the surface.
S = pi D^2/4

Therefore V = Q/S = 4Q / pi D^2

When squared, this yields:

V^2 = 16 Q^2 / pi^2 D^4

And therefore, replacing this in Eqiation B:

dP/dL = K * 16 Q^2 / (pi^2 D^4) * (2D)

And we get finally:

dP/dL = 8 K Q^2 / pi^2 D^5

That's about it.

Dora.

Thank you. I will try to calculate mathematically with the formula you provided and will update.

We are maybe not talking about the same thing. I say that for a given flow
(in liters per second), the resistance is inverse proportional to the fifth power
of the tube diameter. Here is the proof:

It's generally admitted that fluide flow in a tube can be described with the
following relation:

dP/dL = l * r * V^2 / 2D (Equation A)

Click to expand...

Everything depends of course on the validity of "Equation A". I presume it's a simplified description of pressure drop in turbulent flow range. Comparing it with exact calculations for water, I agree that fifth power is a good estimation. Fourth power is in contrast an exact value for laminar flow. But I see that water flow is turbulent for all meaningfull flow rates in supply pipes. In so far I agree to your conclusions.

Hello!

Fourth power is in contrast an exact value for laminar flow.

Click to expand...

Let's keep our feet on earth. A laminar flow at 1l/s does not exist. Or it would be in
a _REALLY_ huge pipe. I remember something about 50 cm internal diameter. I may be
wrong for the number but anyway the diameter is huge. Anyway for water distribution
pipes, say within a few cm diameter, the flow is clearly not laminar.

Now as for Equation A, it's been established by Darcy-??? (I don't remember the second guy).
It's a well established relationship in fluids physics and maybe one of the most known and used
and there is no reason to doubt of its accuracy.

There is also another (independent) formula from a guy named Calmant, Callement, Calmont,
something like that, and which says that

dP/dL = Q^k/1000D^5.

I remember the equation only approximately. I remember that k is very close to 2 (in which case the
formula would be close to the Darcy formula, and anyway in our case Q is constant, so the variation
implies only D), but the denominator is clearly D^5.

A quick simulation with a fluid mechanics program gives me the following results.
1. My favorite pipe (see above): 1m, 1cm radius (internal) and 1l/s -> 0.05 bar pressure loss.
2. Half diameter of above, everything else unchanged. -> 1.43 bar.

1.43 bar is 28.6 times more than 0.05, therefore quite closer to a fifth power (32) than 4th power (16).

Dora

Let's keep our feet on earth. A laminar flow at 1l/s does not exist.

Click to expand...

From a more general fluid mechanics viewpoint, laminar is as common as turbulent flow, e.g. in many gas flow problems. But as I already admitted, not for water pipes. I didn't consider this in post #8.