# Calculate pressure loss due to bends in a pipe?

1. Dec 28, 2011

### daftdave11

how we doing i found the following information regarding finding the pressure drop in a pipe due to 90° bends. i was wondering could this be adapted for say a bend of 60° or 45°. what would be needed to change to the below information to allow this to happen?? thanking ou in advance. this would be a great help to me if i could find the solution to this.. thanks

K_B = (n-1)(.25*\pi f_T \frac{r}{d} + .5 K) + K
Where:
K_B = Resistance coefficient for overall pipe bend.
n = # of 90° bends (for a single 180° bend, n=2).
\pi = well......pi
f_T = Friction factor in turbulent zone.
r= radius of bend (in same units as d).
d= inside diameter of pipe (same unit as r).
K= Loss coefficient for a 90° bend based on table below.

90° Bend Loss Coefficients:
r/d = 1, K = 20f_T
r/d = 1.5, K = 14f_T
r/d = 2, K = 12f_T
r/d = 3, K = 12f_T
r/d = 4, K = 14f_T
r/d = 6, K = 17f_T
r/d = 8, K = 24f_T
r/d = 10, K = 30f_T
r/d = 12, K = 34f_T
r/d = 14, K = 38f_T
r/d = 16, K = 42f_T
r/d = 20, K = 50f_T

2. Dec 29, 2011

### daftdave11

anyone??

3. Dec 29, 2011

### Q_Goest

You have the values given by the Crane paper for the total resistance of a 90 degree bend. You can multiply those resistance values by the following equation to get the resistance for all other bends between 0 and 180 degrees:
y = 1.838E-7*A3 - 8.756E-5*A2+1.748E-2*A
Where y = angle factor
A = angle of pipe bend from 0 to 180 degrees

This equation comes from a graph out of a book on fluid flow from a company I used to work for. I took the graph and did a curve fit to it. You should find that at 90 degrees the angle factor comes to 1.00 and at 180 degrees, the angle factor comes to 1.38.