Need Help Transposing the pressure drop formula

In summary, the Homework Equation states that the pressure drop in a system is limited to 0.3 bar when a pipe of equivalent length is delivered through it at a free air flow rate of 300 l/s. The equation was derived from the pressure drop in a system and the flow rate. The value of R was found to be 6.94 and Q was found to be 300 l/s.
  • #1
Jack Mc
7
0

Homework Statement



Hi, Iv got to find the minimum diameter of a pipe if the pressure drops in a system is to be limited to 0.3bar. when delivered through a pipe of equivalent length 160m

I have to use the formula

Pressure Drop = 800lQ^2/Rd^5.31

I believe I have to transpose this to find the min diameter.

Homework Equations



compressor delivers 300 l s–1 of free air into a pipe at a pressure of 6 bar gauge.

Using the pressure drop formula: pressure drop = 800lQ2 Rd 5.31

calculate the minimum diameter of pipe if the pressure drop in a system is to be limited to 0.3 bar when is delivered through a pipe of equivalent length 160 m .

The Attempt at a Solution



d^5.31=800lQ^2/R
 
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  • #2
Ok so far then take the 5.31th root of both sides.

If your calculator can't do that perhaps logarithms will help..

log(mn) = n * log(m)
 
  • #3
Actually I see you forgot to put the pressure drop on the right hand side. If ΔP is the pressure drop it should read..

d5.31=800lQ2/RΔP
 
  • #4
CWatters said:
Actually I see you forgot to put the pressure drop on the right hand side. If ΔP is the pressure drop it should read..

d5.31=800lQ2/RΔP

I see thank you. Ill give this a go.
 
  • #5
d^5.31=800x160x300^2/6.94x0.3
= d^5.31 = 55.331412.1
Is this the right lines?
 
  • #6
Looks right to me although I don't have my calculator to hand.
 
  • #7
I get..
d5.31 = 800x160x300^2/6.94x0.3 = 5.5 * 106
then
5.31*log(d) = Log(5.5 * 106)
so
d = 10(Log(5.5 * 106)/5.31)
= 68m
which is very large for a pipe!

Where did the equation come from?
Is the value of R correct?
Should the pressure drop be in bar or some other units?
 
  • #8
I got R by...

R=p2/p1

R=6+1.01/1.01

R=6.94
 
  • #9
Should the flow rate Q be in L/s or m3/s ?
 
  • #10
l = the equivalent length of pipe in metres
Q = the flow through the pipe in ℓs−1 free air
R = p2/p1 = the ratio of compression at the start of the pipe
d = the internal diameter of the pipe in mm
 
  • #11
Ah OK 68mm makes more sense than meters.
 

1. What is the formula for calculating pressure drop?

The formula for calculating pressure drop is ΔP = (ρ * v^2 * L)/D, where ΔP is the pressure drop, ρ is the fluid density, v is the fluid velocity, L is the length of the pipe, and D is the diameter of the pipe.

2. How do I transpose the pressure drop formula to solve for fluid velocity?

To solve for fluid velocity, you can rearrange the pressure drop formula to v = √(ΔP * D)/(ρ * L). This will allow you to calculate the fluid velocity when given the other variables.

3. Can the pressure drop formula be used for any type of fluid?

Yes, the pressure drop formula can be used for any type of fluid as long as you use the appropriate units for density and velocity. For example, if you are working with water, the density should be in kg/m^3 and the velocity in m/s.

4. Is the pressure drop formula only applicable for straight pipes?

The pressure drop formula can also be used for other types of pipes, such as curved or coiled pipes. However, in these cases, the length of the pipe (L) would need to be adjusted to account for the curvature.

5. What are some factors that can affect the accuracy of the pressure drop formula?

Some factors that can affect the accuracy of the pressure drop formula include changes in fluid properties (such as density and viscosity), changes in pipe roughness, and changes in flow conditions (such as changes in velocity or flow rate).

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