6-1. Assume that refuse has a C/N ratio of 24:1. If raw sludge with a C/N of 16:1 is to be added to refuse to reach a required C/N of 20:1, how much of each-refuse and sludge-is needed? (Note: Use a wet weight basis.)

Structural Analysis
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Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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### Problem 6-1

**Given Problem:**
Assume that refuse has a C/N (Carbon to Nitrogen) ratio of 24:1. If raw sludge with a C/N ratio of 16:1 is to be added to the refuse to reach a required C/N of 20:1, how much of each—refuse and sludge—is needed?

**Note:** Use a wet weight basis.

**Solution:**
To determine the amount of refuse and sludge needed to achieve a C/N ratio of 20:1, perform the following calculations:

1. **Let R and S** represent the weights of refuse and sludge respectively (since the problem specifies using a wet weight basis).

2. Use the given C/N ratios to develop a system of equations.

   We know:
   - C/N of refuse (R) = 24:1
   - C/N of sludge (S) = 16:1
   - Desired C/N ratio when mixed = 20:1

3. Because we are dealing with ratios and weight, the combined carbon to nitrogen ratio of the mixed refuse and sludge should equal the target C/N ratio:
   
   \[
   \frac{24R + 16S}{R + S} = 20
   \]

4. Multiply both sides by \(R + S\) to eliminate the denominator:
   
   \[
   24R + 16S = 20(R + S)
   \]

5. Expand and simplify:

   \[
   24R + 16S = 20R + 20S
   \]

   \[
   24R - 20R = 20S - 16S
   \]

   \[
   4R = 4S
   \]

   \[
   R = S
   \]

Thus, equal weights of refuse and sludge are needed to achieve the required C/N ratio of 20:1 on a wet weight basis. So if you mix equal parts by weight of refuse and sludge, the desired C/N ratio will be achieved.

This solution shows how to use equations to balance the different C/N ratios of two materials to achieve a desired mixture.
Transcribed Image Text:### Problem 6-1 **Given Problem:** Assume that refuse has a C/N (Carbon to Nitrogen) ratio of 24:1. If raw sludge with a C/N ratio of 16:1 is to be added to the refuse to reach a required C/N of 20:1, how much of each—refuse and sludge—is needed? **Note:** Use a wet weight basis. **Solution:** To determine the amount of refuse and sludge needed to achieve a C/N ratio of 20:1, perform the following calculations: 1. **Let R and S** represent the weights of refuse and sludge respectively (since the problem specifies using a wet weight basis). 2. Use the given C/N ratios to develop a system of equations. We know: - C/N of refuse (R) = 24:1 - C/N of sludge (S) = 16:1 - Desired C/N ratio when mixed = 20:1 3. Because we are dealing with ratios and weight, the combined carbon to nitrogen ratio of the mixed refuse and sludge should equal the target C/N ratio: \[ \frac{24R + 16S}{R + S} = 20 \] 4. Multiply both sides by \(R + S\) to eliminate the denominator: \[ 24R + 16S = 20(R + S) \] 5. Expand and simplify: \[ 24R + 16S = 20R + 20S \] \[ 24R - 20R = 20S - 16S \] \[ 4R = 4S \] \[ R = S \] Thus, equal weights of refuse and sludge are needed to achieve the required C/N ratio of 20:1 on a wet weight basis. So if you mix equal parts by weight of refuse and sludge, the desired C/N ratio will be achieved. This solution shows how to use equations to balance the different C/N ratios of two materials to achieve a desired mixture.
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