# Step-by-Step Solution: Prairie Filter Dimensions Given: - **Slope (\( S \))**: \( 1.5\% = .015 \) - **Peak flow rate (\( Q \))**: \( .18 \) cubic meters per second (cms) - **Maximum flow depth (\( d \))**: \( 5 \) cm \( = .05 \) m - **Desired contact time (\( t \))**: \( 10 \) min \( = 600 \) s We need to find the **length** (\( L \)) and **width** (\( W \)) of the filter. --- ## 1. Calculate Filter Velocity (\( v \)) For shallow flow on a grassed filter, use the **Manning equation**: \[ v = \frac{1}{n} R^{2/3} S^{1/2} \] Where: - \( n \) = Manning's roughness coefficient (for short-grass prairie, use \( n \approx .24 \)) - \( R \) = hydraulic radius (for shallow flow, \( R \approx d \)) - \( S \) = slope Substitute values: \[ v = \frac{1}{.24} (.05)^{2/3} (.015)^{1/2} \] Calculate each part: - \( (.05)^{2/3} \approx .135 \) - \( (.015)^{1/2} \approx .122 \) - \( \frac{1}{.24} = 4.167 \) Now multiply: \[ v \approx 4.167 \times .135 \times .122 \approx .0685 \text{ m/s} \] --- ## 2. Calculate Required Length (\( L \)) Contact time (\( t \)) is: \[ t = \frac{L}{v} \implies L = t \cdot v \] But that's incorrect: the correct formula is \[ t = \frac{L}{v} \implies L = v \cdot t \] Wait! Since velocity is distance/time, so \( L = v \cdot t \) is incorrect; rather: \[ t = \frac{L}{v} \implies L = v \cdot t \] But this would mean the filter would have to be *shorter* for higher velocity, which makes sense: the water spends less time in a shorter filter at the same speed. But let's check: For 10 min (\( 600 \) s): \[ L = v \cdot t = .0685 \times 600 = 41.1 \text{ m} \] --- ## 3. Calculate Required Width (\( W \)) The cross-sectional area (\( A \)) is: \[ A = W \cdot d \] The flow rate (\( Q \)) is: \[ Q = v \cdot A = v \cdot W \cdot d \implies W = \frac{Q}{v \cdot d} \] Substitute values: \[ W = \frac{.18}{.0685 \times .05} \] \[ .0685 \times .05 = .003425 \] \[ W = \frac{.18}{.003425} \approx 52.6 \text{ m} \] --- ## 4. Final Answers ### Filter Dimensions - **Length (\( L \))**: **41 m** - **Width (\( W \))**: **53 m** (rounded to nearest whole number) --- ## 5. Summary Table | Parameter | Value | |----------------------------|------------| | Slope (\( S \)) | \( 1.5\% \) | | Peak Flow Rate (\( Q \)) | \( .18\, \text{m}^3/\text{s} \) | | Flow Depth (\( d \)) | \( .05\, \text{m} \) | | Contact Time (\( t \)) | \( 600\, \text{s} \) (10 min) | | **Length (\( L \))** | **41 m** | | **Width (\( W \))** | **53 m** | --- ### **Final Answer** > The short-grass prairie filter should be **41 m long** and **53 m wide** to handle a peak flow of .18 cms, with a maximum flow depth of 5 cm and a minimum 10-minute contact time at a 1.5% slope.