Answer:
The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.
Explanation:
Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as
v = (ΔPR²/4μL) [1 - (r²/R²)]
where v = fluid velocity at any point in the radial direction
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
But the maximum velocity of the fluid occurs at the middle of the pipe when r=0
Hence, maximum veloxity is
v(max) = (ΔPR²/4μL)
So, velocity at any point in the radial direction is
v = v(max) [1 - (r²/R²)]
At the point r = (R/2)
r² = (R²/4)
(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)
So,
1 - (r²/R²) = 1 - (1/4) = (3/4)
Hence, v at r = (R/2) is given as
v = v(max) × (3/4)
Hope this Helps!!!
Answer:
Small points of colored light arranged in a grid, each is formed from three colored lights: red,green,and blue. (RGB), nothing is absorbed,nothing is reflected, just see pure colored lights.
Answer:
h_f = 15 ft, so option A is correct
Explanation:
The formula for head loss is given by;
h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))
Where;
h_f is head loss due to friction in ft
L is length of pipe in ft
Q is flow rate of water in gpm
C is hazen Williams constant
D is diameter of pipe in inches
We are given;
L = 1,800 ft
Q = 600 gpm
C = 120
D = 8 inches
So, plugging in these values into the equation, we have;
h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))
h_f = 14.896 ft.
So, h_f is approximately 15 ft
Answer:
C. 14.55
Explanation:
12 x 10 = 120
120 divded by 10 is 12
so now we do the left side
7 x 3 = 21 divded by 10 is 2
so now we have 14
and the remaning area is 0.55
so 14.55
