Let's solve the problem **step by step**. --- ## **Step 1: Write the Known Values** - **Discharge, \( Q \):** \( 15\, \text{ft}^3/\text{s} \) - **Maximum head above crest, \( H \):** \( 18\, \text{in} = 1.5\, \text{ft} \) - **Height of crest above channel bottom, \( P \):** \( 3\, \text{ft} \) - **Weir type:** Contracted rectangular weir --- ## **Step 2: Weir Flow Equation** For a *contracted rectangular weir*, the discharge equation is: \[ Q = C_d \cdot \left( L - .2H \right) H^{3/2} \] Where: - \( Q \) = Discharge (\(\text{ft}^3/\text{s}\)) - \( C_d \) = Discharge coefficient (typically 3.33 for ft, .40 for m) - \( L \) = Length of weir (ft) - \( H \) = Head over crest (ft) --- ## **Step 3: Insert the Known Values** Given: - \( Q = 15 \) - \( H = 1.5 \) - \( C_d = 3.33 \) So, \[ 15 = 3.33 \cdot (L - .2 \times 1.5)\cdot (1.5)^{3/2} \] --- ## **Step 4: Calculate \( H^{3/2} \)** \[ H^{3/2} = (1.5)^{3/2} = (1.5)^{1} \times (1.5)^{.5} = 1.5 \times \sqrt{1.5} = 1.5 \times 1.225 = 1.837 \] --- ## **Step 5: Substitute and Solve for \( L \)** \[ 15 = 3.33 \cdot (L - .3) \cdot 1.837 \] Calculate the product \(3.33 \times 1.837\): \[ 3.33 \times 1.837 = 6.117 \] So, \[ 15 = 6.117 \cdot (L - .3) \] \[ \frac{15}{6.117} = L - .3 \] \[ 2.453 = L - .3 \] \[ L = 2.453 + .3 = 2.753\, \text{ft} \] --- ## **Step 6: Final Answer** \[ \boxed{L = 2.75\, \text{ft}} \] --- ### **Summary Table** | Parameter | Value | |--------------------------|-------------| | Discharge (\(Q\)) | 15 ft³/s | | Head over crest (\(H\)) | 1.5 ft | | Crest height (\(P\)) | 3 ft | | Weir length (\(L\)) | **2.75 ft** | --- **Required length of the contracted weir is \( \boxed{2.75~\text{ft}} \).**