VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:(1 point) The flow system shown in the figure is activated at time ¢ = 0. Let Q;(t) denote the amount of solute present Fe i eo J in the ith tank at time ¢. Assume that all the flow rates are a constant 10 L/min. It follows that the volume of solution in each tank remains constant; assume this volume to be 1000 L. The concentration of solute in the inflow to Tank 1 (from a source other than Tank 2) is 0.3 kg/L, and the concentration of solute in the inflow to Tank 2 (from a source other than Tank 1) is 0 kg/L. Assume each tank is mixed perfectly. a. Set up a system of first-order differential equations that models this situation. Qf | (] a | = — 4 ——4 + = Qf | J] |e lo |
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
(1 point)
The flow system shown in the figure is activated at time ¢ = 0. Let Q;(t) denote the amount of solute present Fe i eo J
in the ith tank at time ¢. Assume that all the flow rates are a constant 10 L/min. It follows that the volume of
solution in each tank remains constant; assume this volume to be 1000 L. The concentration of solute in the
inflow to Tank 1 (from a source other than Tank 2) is 0.3 kg/L, and the concentration of solute in the inflow to
Tank 2 (from a source other than Tank 1) is 0 kg/L. Assume each tank is mixed perfectly.
a. Set up a system of first-order differential equations that models this situation.
Qf | (] a |
= — 4 ——4 + =
Qf | J] |e lo |
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Asked by: Pagal
Created at: 2025-07-27 10:39:35
Asked by: Pagal
Created at: 2025-07-27 10:39:35
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Let's break down the problem step by step:
---
### **Step 1: Analyze the System**
- There are two tanks, each with a constant volume of 1000 L.
- Flow rate between tanks and into/out of tanks is **10 L/min**.
- **Inflow to Tank 1:** Concentration \( = 0.3 \) kg/L.
- **Inflow to Tank 2:** Concentration \( = 0 \) kg/L.
- Let \( Q_1(t) \) and \( Q_2(t) \) be the amount of solute (in kg) in tanks 1 and 2 at time \( t \).
#### For each tank, the rate of change of solute (\( Q_i' \)) is:
\[
\text{Rate in} - \text{Rate out}
\]
**Tank 1:**
- **In:** \( 10 \) L/min from outside (concentration \( 0.3 \) kg/L) \( \rightarrow 10 \times 0.3 = 3 \) kg/min.
- **In:** \( 10 \) L/min from Tank 2 (concentration \( Q_2/1000 \) kg/L) \( \rightarrow 10 \times (Q_2/1000) \) kg/min.
- **Out:** \( 10 \) L/min to Tank 2 (concentration \( Q_1/1000 \) kg/L) \( \rightarrow 10 \times (Q_1/1000) \) kg/min.
So:
\[
Q_1' = 3 + 10 \frac{Q_2}{1000} - 10 \frac{Q_1}{1000}
\]
**Tank 2:**
- **In:** \( 10 \) L/min from Tank 1 (concentration \( Q_1/1000 \) kg/L) \( \rightarrow 10 \times (Q_1/1000) \) kg/min.
- **In:** \( 10 \) L/min from outside (concentration \( 0 \) kg/L) \( \rightarrow 0 \) kg/min.
- **Out:** \( 10 \) L/min to Tank 1 (concentration \( Q_2/1000 \) kg/L) \( \rightarrow 10 \times (Q_2/1000) \) kg/min.
So:
\[
Q_2' = 10 \frac{Q_1}{1000} - 10 \frac{Q_2}{1000}
\]
**Explanation:**
This step analyzes the inflow and outflow for each tank to write down the differential equations.
---
### **Step 2: Write in Matrix Form**
Let's simplify the coefficients:
\[
10/1000 = 0.01
\]
So the equations become:
\[
Q_1' = 3 - 0.01 Q_1 + 0.01 Q_2
\]
\[
Q_2' = 0.01 Q_1 - 0.01 Q_2
\]
Write in matrix form:
\[
\begin{bmatrix}
Q_1' \\
Q_2'
\end{bmatrix}
=
\begin{bmatrix}
-0.01 & 0.01 \\
0.01 & -0.01
\end{bmatrix}
\begin{bmatrix}
Q_1 \\
Q_2
\end{bmatrix}
+
\begin{bmatrix}
3 \\
0
\end{bmatrix}
\]
**Explanation:**
This step rewrites the differential equations in the standard first-order system matrix form.
---
### **Final Answer**
\[
\boxed{
\begin{bmatrix}
Q_1' \\
Q_2'
\end{bmatrix}
=
\begin{bmatrix}
-0.01 & 0.01 \\
0.01 & -0.01
\end{bmatrix}
\begin{bmatrix}
Q_1 \\
Q_2
\end{bmatrix}
+
\begin{bmatrix}
3 \\
0
\end{bmatrix}
}
\]
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