vimming pool whose volume is 10,000 gal contains water that is 0.01% chlorine. Starting at t=0, city water containing 0.001% chlorine is pumped into the pool at a rate of 4 gal/min. The pool water flows out at the same rate. What is the centage of chlorine in the pool after 1 hour? When will the pool water be 0.002% chlorine?
vimming pool whose volume is 10,000 gal contains water that is 0.01% chlorine. Starting at t=0, city water containing 0.001% chlorine is pumped into the pool at a rate of 4 gal/min. The pool water flows out at the same rate. What is the centage of chlorine in the pool after 1 hour? When will the pool water be 0.002% chlorine?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement**
A swimming pool whose volume is 10,000 gallons contains water that is 0.01% chlorine. Starting at \( t = 0 \), city water containing 0.001% chlorine is pumped into the pool at a rate of 4 gallons per minute. The pool water flows out at the same rate.
**Questions:**
1. What is the percentage of chlorine in the pool after 1 hour?
2. When will the pool water be 0.002% chlorine?
**Graph/Diagram Explanation:**
There are no graphs or diagrams provided in this problem.
**Step-by-Step Solution:**
Let's denote:
- \[ V \] = Volume of the pool (10,000 gal)
- \[ C_{\text{initial}} \] = Initial concentration of chlorine in the pool (0.01% or 0.0001)
- \[ C_{\text{city}} \] = Concentration of chlorine in the city water (0.001% or 0.00001)
- \[ F \] = Flow rate of water into and out of the pool (4 gal/min)
We aim to determine:
1. The percentage of chlorine in the pool after 1 hour.
2. The time when the pool water will be 0.002% chlorine.
**1. Finding the percentage of chlorine in the pool after 1 hour:**
We start with the differential equation for the mixing problem:
\[ \frac{dC}{dt} = \text{(rate in) - (rate out)} \]
Since the rate of water flowing in and out is the same:
\[ \text{Rate in} = F \times C_{\text{city}} \]
\[ \text{Rate out} = F \times \frac{C(t)}{V} \]
Here, \( C(t) \) is the concentration of chlorine at time \( t \). Thus, the differential equation becomes:
\[ \frac{dC}{dt} = 4 \times 0.00001 - 4 \times \frac{C(t)}{10000} \]
\[ \frac{dC}{dt} = 0.00004 - 0.0004C(t) \]
To solve this, we use separation of variables:
\[ \frac{dC}{0.00004 - 0.0004C(t)} = dt \]
Integr](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb54c94dc-8fa5-45f1-a08b-382951c33f95%2F1f270373-00a3-490a-b5bb-d4a7cdc27ee0%2F5it3owo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
A swimming pool whose volume is 10,000 gallons contains water that is 0.01% chlorine. Starting at \( t = 0 \), city water containing 0.001% chlorine is pumped into the pool at a rate of 4 gallons per minute. The pool water flows out at the same rate.
**Questions:**
1. What is the percentage of chlorine in the pool after 1 hour?
2. When will the pool water be 0.002% chlorine?
**Graph/Diagram Explanation:**
There are no graphs or diagrams provided in this problem.
**Step-by-Step Solution:**
Let's denote:
- \[ V \] = Volume of the pool (10,000 gal)
- \[ C_{\text{initial}} \] = Initial concentration of chlorine in the pool (0.01% or 0.0001)
- \[ C_{\text{city}} \] = Concentration of chlorine in the city water (0.001% or 0.00001)
- \[ F \] = Flow rate of water into and out of the pool (4 gal/min)
We aim to determine:
1. The percentage of chlorine in the pool after 1 hour.
2. The time when the pool water will be 0.002% chlorine.
**1. Finding the percentage of chlorine in the pool after 1 hour:**
We start with the differential equation for the mixing problem:
\[ \frac{dC}{dt} = \text{(rate in) - (rate out)} \]
Since the rate of water flowing in and out is the same:
\[ \text{Rate in} = F \times C_{\text{city}} \]
\[ \text{Rate out} = F \times \frac{C(t)}{V} \]
Here, \( C(t) \) is the concentration of chlorine at time \( t \). Thus, the differential equation becomes:
\[ \frac{dC}{dt} = 4 \times 0.00001 - 4 \times \frac{C(t)}{10000} \]
\[ \frac{dC}{dt} = 0.00004 - 0.0004C(t) \]
To solve this, we use separation of variables:
\[ \frac{dC}{0.00004 - 0.0004C(t)} = dt \]
Integr
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