- #1
daftdave11
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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
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