Poiseuille equation for water flow rate

In summary, the water pressure at the council tubby is around 1.5 bar, and using Poiseuille equation, one can estimate the flow rate at a typical kitchen tap. My assumption was that the pipe diameter was the same from the council tubby to where the water comes out of the tape, but when I substituted those values into Poiseuille's equation, I got an estimate of 0.0375 L/min. My order of magnitude calculation was off, and it would be better to use Bernoulli equation instead.
  • #1
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Homework Statement
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Relevant Equations
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For this problem,

The mains water pressure at the council tubby (just before it enters a house) is of the order of 1.5 bar. Using Poiseuille equation, estimate the flow rate in a typical home at the kitchen tap. You will need to make reasonable estimates on several parameters, clearly state these assumptions (note that this estimate does not need to agree with your measurement in Q1, but it shouldn’t be orders-of-magnitude different).

My assumption is that the pipe diameter is the same from the council tubby to where the water comes out of the tape.

##P_2 - P_1 ≈ 1 bar ≈ 100000 Pa##
##η ≈ 10^{-3} Pa \cdot s##
##L ≈ 10 m##
##π ≈ 3##
##r ≈ 1 cm ≈ 0.01 m##

However, when I substitute those values into Poiseuille's equation I get ##Q ≈ 0.0375 ≈ 0.04 \frac{m^3}{s}##. When I convert it to ##\frac{L}{min}## I get ##2400 \frac{L}{min}## However, Q should be in the order of ##10 \frac{L}{min}##. My order of magnitude calculation is orders of magnitude off.

I am not sure how to make the calculation more reasonable. If I increase the length to say 100m, it still dose not give a value in between ##10 \frac{L}{min}##.

I am wondering whether my assumption is wrong, that is, the pipe diameter varies from the council tubby to the where the water comes out. I think ##r## should be initially larger than what I said it to be, and there decreases to the value I measured when it reach's the tap cyclinder. However, I am not sure whether it possible to account for that in Poiseuille equation. Maybe I should use Bernoulli equation.

Any guidance would me much appreciated!

Many thanks!
 
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  • #2
Please quote the equation. Doesn't it have a term for the pipe radius? you don't show your value for that.
 
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  • #3
For a normal family home in USA, the diameter of the pipe connecting it to the city main via water meter is 1-inch or 3/4-inch.

domestic-water-piping-table3.jpg
 
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  • #4
haruspex said:
Please quote the equation. Doesn't it have a term for the pipe radius? you don't show your value for that.
Thank you for your reply @haruspex! The equation is ##Q = (P_2 - P_1)\frac{πr^4}{8ηL}##

Many thanks!
 
  • #5
Lnewqban said:
For a normal family home in USA, the diameter of the pipe connecting it to the city main via water meter is 1-inch or 3/4-inch.

View attachment 324058
Thank you for your reply @Lnewqban !

Sorry as @haruspex pointed out, I forgot to say ##r ≈ 1 cm ≈ 0.01 m##

Many thanks!
 
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  • #6
At 10 l/min, the Reynolds number is > 10000, and the flow is turbulent, you can’t use the poiseuille equation.
 
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  • #7
Actually, it's even worse because the flow leaves the tap at atmospheric pressure. The change in pressure from the mains to the tap is the full 1.5 bar (which is low). In my home its regulated down to about 50 psi from the mains (which are much higher).

Pipes snake through a home ( decreasing in diameter as they go in a trunk/branch system), so the length can add up, but 10 m doesn't seem unreasonable as a length. However, having a nearly 1 inch line at the sink tap is pretty out there. In my home lines for my sink are 3/8ths NPS ( I.D. ##\approx## 0.5 in. ). But even doing 10 meters of ##r = 0.25 \rm{in}## that equation predicts ##\approx 575\rm{ \frac{L}{min}}##.

As @Chestermiller points out its validity requires laminar flow ##Re = \frac{ 4 \rho Q}{\pi D \mu} < 2000 ## and it not satisfied for either scenario. So, if you are forced to use this its a poorly designed question.
 
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  • #8
Chestermiller said:
At 10 l/min, the Reynolds number is > 10000, and the flow is turbulent, you can’t use the poiseuille equation.
erobz said:
Actually, it's even worse because the flow leaves the tap at atmospheric pressure. The change in pressure from the mains to the tap is the full 1.5 bar (which is low). In my home its regulated down to about 50 psi from the mains (which are much higher).

Pipes snake through a home ( decreasing in diameter as they go in a trunk/branch system), so the length can add up, but 10 m doesn't seem unreasonable as a length. However, having a nearly 1 inch line at the sink tap is pretty out there. In my home lines for my sink are 3/8ths NPS ( I.D. ##\approx## 0.5 in. ). But even doing 10 meters of ##r = 0.25 \rm{in}## that equation predicts ##\approx 575\rm{ \frac{L}{min}}##.

As @Chestermiller points out its validity requires laminar flow ##Re = \frac{ 4 \rho Q}{\pi D \mu} < 2000 ## and it not satisfied for either scenario. So, if you are forced to use this its a poorly designed question.
Thank you for your replies @Chestermiller and @erobz !

I will refer to this thread when I email the profs on Monday. I will post what the outcome is.

Many thanks!
 
  • #9
I found the actual flow rate Q to be ##Q = \frac{1}{9.97} = 0.101 L/s (3 s.f.) = 6.03 L/min (3 s.f.) = 0.000101 m^3/s (3 s.f.)##
 
  • #12
Callumnc1 said:
Thank you for your reply @Lnewqban!

Was there a specific part that you wanted me to see?

Many thanks!
Mainly as a reference about typical pipe's diameters and water supply flows in a family home.
 
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  • #13
Lnewqban said:
Mainly as a reference about typical pipe's diameters and water supply flows in a family home.
Thank you for your reply @Lnewqban!

Yeah, it turns out the order of magnitude calculation is likely to be off due the bends in the pipe which decrease the fluid momentum that the Poiseuille equation dose not account for. I'm also right about the different diameter part.

Many thanks!
 
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  • #14
Callumnc1 said:
Thank you for your reply @Lnewqban!

Yeah, it turns out the order of magnitude calculation is likely to be off due the bends in the pipe which decrease the fluid momentum that the Poiseuille equation dose not account for. I'm also right about the different diameter part.

Many thanks!
Bends/fittings in the pipe are typically referred to as "minor head loss" terms, they can add up, but they aren't typically dominant. How about elevation head?
 
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  • #15
erobz said:
Bends in the pipe are typically referred to as "minor head loss", they can add up but they aren't typically dominant. How about elevation head?
Thank you for your reply @erobz! That is true you mention about elevation head. Where I was performing the experiment was 2 floors high and the house is about 200 m above sea level.

Many thanks!
 
  • #16
Callumnc1 said:
Thank you for your reply @erobz! That is true you mention about elevation head. Where I was performing the experiment was 2 floors high and the house is about 200 m above sea level.

Many thanks!
The mains typically enter in the basement. So, if you are doing this on the second floor you could have used 12 m of elevation head just getting to the height of the faucet. Also, your 10 m of pipe are probably also underestimated, and its not likely 1 " pipe, probably more like 1/2" pipe (or smaller).

What do you get if you take 12 m of elevation head out of ##\Delta P##, and change the length to 20 m of pipe at a radius of 7 mm?
 
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  • #17
erobz said:
The mains typically enter in the basement. So, if you are doing this on the second floor you could have used 12 m of elevation head just getting to the height of the faucet. Also, your 10 m of pipe are probably also underestimated, and its not likely 1 " pipe, probably more like 1/2" pipe (or smaller).

What do you get if you take 12 m of elevation head out of ##\Delta P##, and change the length to 20 m of pipe at a radius of 7 mm?
Thank you for your reply @erobz! I will do that problem once I have a bit more time from uni.

Many thanks!
 

FAQ: Poiseuille equation for water flow rate

What is the Poiseuille equation for water flow rate?

The Poiseuille equation, also known as the Hagen-Poiseuille law, describes the volumetric flow rate of an incompressible and Newtonian fluid through a long cylindrical pipe. The equation is given by Q = (πΔP r^4) / (8ηL), where Q is the volumetric flow rate, ΔP is the pressure difference between the two ends of the pipe, r is the radius of the pipe, η is the dynamic viscosity of the fluid, and L is the length of the pipe.

How does the radius of the pipe affect the flow rate according to the Poiseuille equation?

According to the Poiseuille equation, the flow rate (Q) is directly proportional to the fourth power of the radius (r) of the pipe. This means that even a small increase in the radius of the pipe results in a significant increase in the flow rate. Specifically, if the radius is doubled, the flow rate increases by a factor of 16 (2^4).

What assumptions are made in the Poiseuille equation?

The Poiseuille equation assumes that the fluid is incompressible and Newtonian, meaning its viscosity is constant and does not change with the rate of flow. It also assumes laminar flow, which occurs at low Reynolds numbers where the fluid flows in parallel layers with no disruption between them. Additionally, the equation assumes the pipe is rigid, cylindrical, and has a constant cross-sectional area.

How does the viscosity of the fluid impact the flow rate in the Poiseuille equation?

In the Poiseuille equation, the flow rate (Q) is inversely proportional to the dynamic viscosity (η) of the fluid. This means that as the viscosity increases, the flow rate decreases. Viscosity represents the internal friction within the fluid, so higher viscosity fluids resist flow more than lower viscosity fluids.

Can the Poiseuille equation be applied to turbulent flow?

No, the Poiseuille equation cannot be applied to turbulent flow. The equation is derived under the assumption of laminar flow, which occurs at low Reynolds numbers. Turbulent flow, which occurs at high Reynolds numbers, involves chaotic and irregular fluid motion that cannot be accurately described by the Poiseuille equation. For turbulent flow, other models and equations, such as the Darcy-Weisbach equation, are used.

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