Answer:
[tex][/tex] Given:
• Flow rate, Q = 5 cfs
• Reservoir surface elevations:
* Upstream: 250 ft
* Downstream: 100 ft
• Distance between reservoirs: 2 miles = 10,560 ft
• Pipe material: Smooth steel
SOLUTION
1. Determine the head loss, hL, due to friction:
Using the Darcy-Weisbach equation:
hL = f [tex] \times[/tex] ([tex]\frac{L}{D}[/tex]) [tex] \times[/tex] ([tex]\frac{V²}{2g}[/tex])
where:
• f is the Darcy friction factor
• L is the pipe length
• D is the pipe diameter
• V is the flow velocity
• g is the acceleration due to gravity
2. Determine the flow velocity, V:
V = [tex]\frac{Q}{A}[/tex]
where A is the cross-sectional area of the pipe.
3. Determine the pipe diameter, D:
Substituting equations (2) and (3) into equation (1) and solving for D:
hL = f [tex] \times[/tex] ([tex]\frac{L}{D}[/tex]) [tex] \times[/tex] [tex]\frac{Q²}{2gA²}[/tex]
D = (f [tex] \times[/tex] L [tex] \times[/tex] Q²) / (2g [tex] \times[/tex] hL [tex] \times[/tex] A²)
We can assume a Darcy friction factor of f = 0.02 for smooth steel pipes.
4. Calculate the cross-sectional area, A:
A = [tex]\frac{πD²}{4}[/tex]
5. Calculate the head loss, hL:
hL = (250 - 100) ft = 150 ft
6. Substitute into the equation for D:
D = (0.02 [tex] \times[/tex] 10,560 ft [tex] \times[/tex] [tex]\frac{(5 cfs)²)}{2}[/tex] \times[/tex]32.2ft/s²}[/tex] [tex] \times[/tex] 150 ft [tex] \times[/tex] ([tex]\frac{(πD²}{4)²)}[/tex]
Solving for D gives:
D ≈ 0.52 ft = 6.24 inches
Therefore, the diameter of the smooth steel pipe that will carry 5 cfs of water between the two reservoirs is approximately 6.24 inches.
