Volume Flow Rate in a Pipe for Viscous Fluid

[jde23]

Homework Statement

Hi, my first post here! I have a Thermodynamics resit over the summer and I'm trying to get my head around viscous flow, I'm stuck at a question and need help! Thanks :)

Glycerine discharges to the atmosphere through a circular pipe 100mm in diameter. The gauge pressure 50m from the exit is 458 kPa. Data for glycerine: ρ=1260 kg/m^3, μ=0.9 kg/ms.

Determine the volume flow rate Q for the following cases:
(i) a horizontal pipe (verify the flow is laminar!) (Ans: 25 litres/s)
(ii) a pipe inclined upwards by 20° (Ans: 13.5 litres/s)
(iii) a pipe inclined downwards by 20° (Ans: 36.5 litres/s)

Homework Equations

I thought this equation was the right one to use but my answer is totally wrong when I plug the numbers in, it's not a units thing either...

Q = (∏d^4 / 128μ) * (ΔPloss / length)

The Attempt at a Solution

I assumed the gauge pressure at the pipe exit is equal to atm pressure, so 100kPa - is this correct? The rest is just plugging numbers in. Am I missing something obvious here...?

 

[rock.freak667]

Have you tried using Bernoulli's equation?

 

[LawrenceC]

Your formula works. I get 25 for part i. Are you mixing gauge pressure and absolute pressure?

 

[jde23]

Hi LawrenceC, yes I think I am! Embarrassing moment, whoops...

Thanks!

 

[rezeew]

Hello! I can help you with your question on volume flow rate in a pipe for viscous fluid. First, let's start with the equation you have used: Q = (∏d^4 / 128μ) * (ΔPloss / length). This is the correct equation to use for calculating the volume flow rate for a viscous fluid in a pipe.

Now, let's look at the data given in the question. We have the diameter of the pipe (100mm), the gauge pressure (458 kPa), and the properties of glycerine (ρ=1260 kg/m^3, μ=0.9 kg/ms). The first thing to note is that the gauge pressure given is not the pressure at the pipe exit, but rather 50m away from the exit. This means that there is a pressure loss (ΔPloss) between the exit and the point where the pressure is measured.

To calculate the pressure loss, we can use the Bernoulli's equation, which states that the total energy of a fluid in a pipe remains constant. In this case, we can assume that the velocity of the fluid is negligible, so the equation simplifies to: P1 + ρgh1 = P2 + ρgh2, where P1 is the pressure at the exit, P2 is the pressure 50m away, ρ is the density of glycerine, g is the acceleration due to gravity, and h1 and h2 are the heights of the two points.

Using this equation, we can calculate the pressure at the exit (P1) by rearranging the equation: P1 = P2 + ρgh2 - ρgh1. Plugging in the values from the question, we get P1 = 508 kPa. This is the pressure at the exit of the pipe, which is equal to the atmospheric pressure.

Now, let's calculate the pressure loss (ΔPloss). This can be done by subtracting the gauge pressure (458 kPa) from the pressure at the exit (508 kPa), so ΔPloss = 508 kPa - 458 kPa = 50 kPa.

Finally, we can plug all these values into the equation Q = (∏d^4 / 128μ) * (ΔPloss / length) to calculate the volume flow rate. For the first case (horizontal pipe), we