### 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 = 20:13. 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.

Given:- The C/N ratio of the refuse = 24:1 The C/N ratio of the sludge added to refuse = 16:1 The…

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.)

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Collection Systems

The collection systems form a part of sanitary engineering, which is a sub-branch of civil engineering. It is also known as public health engineering and wastewater engineering. It is a methodology followed to remove and properly dispose of human waste, along with creating systems to provide safe and clean usable water. These measures ensure a hygienic and disease-free community. Sanitary engineering, apart from building sewer disposal units, is also involved in plumbing, fire protection, hydraulics, life safety, environmental protection, information technology, microbiology, pathology, and so on.

Question

### 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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