Need to find delta P & delta f 4

[tdw]

I haven't any real idea how to do this;
I'm working on a system that blows nitrogen (@25 deg. C) through copper 1.5" ID tubes at 40CFM and 10" water pressure.
The question is if I extend the 1.5" pipe by 30 feet and add 10 90 degree bends what will be the CFM and pressure outputs?
Is there a standard formula for this?
TDW

 

[m777182]

There is not a formula but a procedure:
1.calculate linear velocity in your tubes
v=(volumetric flow)/(total cross area of the tubes)
from the Darcy's equation
deltaP=(loss factor)*(density of N2)*(v**2)/2 ( where I use v**2 to express the square of v)
2.calculate
loss factor=2*deltaP/((density of N2)*(v**2))
3. calculate loss factor per unit lenght of tubes:
loss factor per unit lenght=loss factor/(lenght of tubes)
4.calculate new loss factor including additional lenght:
loss factor =(loss factor per unit lenght)*(new total lenght)
5. from a handbook choose the type of your bend and its corresponding particular loss factor,say loss_factor_bend
6. multiply by number N of bends
loss factor_bends=loss_factor_bend *N
7. sum loss factors of tubes and bends as new loss factor
8.look to Darcy equation and decide:
if you cannot afford higher deltaP then with the same deltaP and new loss factor calculate new velocity and new volumetric flow
if you would like to keep the same volumetric flow and velocity in tubes calculate new deltaP.
m777182

 

[kmpillai]

Dear friend,
Here is a method without referring the charts for friction factors. I am doing the complete calculation in SI unit system
Pipe dia = 1.5"(0.0381 m)
Length = 30 feet (9.144 m)
Flow rate = 40 cfm (0.0189 m3/s)
1. Calculate the velocity of the N2 in the pipe
Velocity(V) = Volume flow rate(Q)/flow area(A)
Q = 40 cfm = 0.0189 m3/min
A = pi/4 * d^2 = 0.00114 m2
V = 0.0189/0.00114 = 16.56 m/s

2. Calculate Reynolds number(Re)
Re = density*Velocity*Pipe ID/Dynamic viscosity at 25 deg C
Dynamic viscosity = 17.675e-6 Pa s at 25 deg C for N2
Re = 1.1736*16.56*0.0381/17.675e-6 = 41893.44
Since flow is turbulent and the friction factor is given by
f = 0.316/(Re)^0.25
= 0.316/(41893.44)^0.25
= 0.022
3. Pressure drop calculation
Head loss H = fLV^2/(2gd)
Where L is the pipe length + fittings equivalent length
Equivalent length for 90 deg bends is given by
Le/d = 32 for 90 deg elbows, standard radius
Le = 32* 0.0381 = 1.2192 m (4 ft)
For 10 90 deg bends = 1.2192*10 = 12.192 m (40 ft)
Total length L = 9.144 + 12.192 = 21.336 m (70 ft)

Head loss H = 0.022*21.336*(16.56^2)/(2*9.81*0.0381)
H = 172.89 m
delta P = density*g*H
= 1.1736*9.81*172.89 = 1990.44 N/m2 = 0.0199 bar
=0.2029 m of water
= 8" water column

Notes:
1.Equivalent length for other fittings also avialble
2. In the above problem 10 90 deg bends effect is more than the straight length (30 ft)portion. So try to minimise the number bends in the piping system.

Regards,
KMPillai