## Step-by Solution for Question7 ** Restated:**What diameter castiron pipe is to carry water a rate of 10 cfs between two reservoirs if the reservoirs are 2 mi apart and the elevation difference between the water surfaces in the reservoirs is 20 ft? --- ### **Step 1: List Given Data** - Flow rate, \( Q = 10 \) cfs (cubic feet per second) - Pipe length, \( L = 2 \) miles \( = 2 \times 528 = 10,560 \) ft - Elevation head difference, \( h_f = 20 \) ft - Pipe material: cast-iron (assume Hazen-Williams coefficient \( C \approx 130 \) for old cast iron; adjust as needed) --- ### **Step 2: Use the Hazen-Williams Formula** The Hazen-Williams formula (U.S. customary units) for head loss is: \[ h_f = 10.67 \frac{L}{C^{1.85} D^{4.87}} Q^{1.85} \] where: - \( h_f \) = head loss (ft) - \( L \) = length (ft) - \( C \) = Hazen-Williams coefficient - \( D \) = pipe diameter (ft) - \( Q \) = flow rate (cfs) --- ### **Step 3: Substitute Known Values** \[ 20 = 10.67 \frac{10,560}{130^{1.85} \cdot D^{4.87}} \cdot 10^{1.85} \] --- ### **Step 4: Calculate Intermediate Values** - \( 130^{1.85} \approx 5129 \) - \( 10^{1.85} \approx 70.8 \) Plug these in: \[ 20 = 10.67 \frac{10,560}{5129 \cdot D^{4.87}} \cdot 70.8 \] Calculate the numerator: \[ 10.67 \times 10,560 \times 70.8 \approx 8,004,570 \] So: \[ 20 = \frac{8,004,570}{5129 \cdot D^{4.87}} \] --- ### **Step 5: Solve for \( D^{4.87} \)** \[ 20 \cdot 5129 \cdot D^{4.87} = 8,004,570 \] \[ 102,580 \cdot D^{4.87} = 8,004,570 \] \[ D^{4.87} = \frac{8,004,570}{102,580} \] \[ D^{4.87} \approx 78.09 \] --- ### **Step 6: Solve for \( D \) (Diameter in ft)** \[ D = (78.09)^{1/4.87} \] \[ \ln D = \frac{1}{4.87} \ln 78.09 \] \[ \ln 78.09 \approx 4.358 \] \[ \ln D = \frac{4.358}{4.87} \approx .895 \] \[ D = e^{.895} \approx 2.45 \text{ ft} \] --- ### **Step 7: Convert to Inches** \[ D = 2.45 \text{ ft} \times 12 = 29.4 \text{ in} \] --- ## **Final Answer** ### **The diameter of the cast-iron pipe needed is approximately:** \[ \boxed{2.45 \ \text{ft} \ (29.4 \ \text{in})} \] --- ### **Summary Table** | Quantity | Value | |-------------------|--------------| | Pipe Diameter (ft)| 2.45 ft | | Pipe Diameter (in)| 29.4 in | --- **Assumptions:** - Hazen-Williams \( C = 130 \) (typical for old cast iron) - Minor losses neglected - Water at typical temperatures If you need a different formula (like Darcy-Weisbach), let me know!