Answer:
a) -8 lb / ft^3
b) -70.4 lb / ft^3
c) 54.4 lb / ft^3
Explanation:
Given:
- Diameter of pipe D = 12 in
- Shear stress t = 2.0 lb/ft^2
- y = 62.4 lb / ft^3
Find pressure gradient dP / dx when:
a) x is in horizontal flow direction
b) Vertical flow up
c) vertical flow down
Solution:
- dP / dx as function of shear stress and radial distance r:
(dP - y*L*sin(Q))/ L = 2*t / r
dP / L - y*sin(Q) = 2*t / r
Where dP / L = - dP/dx,
dP / dx = -2*t / r - y*sin(Q)
Where r = D /2 ,
dP / dx = -4*t / D - y*sin(Q)
a) Horizontal Pipe Q = 0
Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)
dP / dx = -8 + 0
dP/dx = -8 lb / ft^3
b) Vertical pipe flow up Q = pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)
dP / dx = 8 - 62.4
dP/dx = -70.4 lb / ft^3
c) Vertical flow down Q = -pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)
dP / dx = -8 + 62.4
dP/dx = 54.4 lb / ft^3
Answer:
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