VIPSolutionsAct like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:‘The tank below has water flowing in from the left inlet, and water flowing out from the bottom and right outlets. Assume that the water density is constant, and assume that all inflow and outflow velocities are steady and constant over their respective areas. At what rate is the water level rising or falling? 2m ameter Water Ime 12m ] | [ho mm diameter 15s
Question:
Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
‘The tank below has water flowing in from the left inlet, and water flowing out from the bottom
and right outlets. Assume that the water density is constant, and assume that all inflow and
outflow velocities are steady and constant over their respective areas.
At what rate is the water level rising or falling?
2m ameter
Water
Ime 12m
] | [ho mm diameter
15s
This is AI generated Response.
Asked by: manish kumar
Created at: 2025-10-05 23:28:45
Asked by: manish kumar
Created at: 2025-10-05 23:28:45
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Let's break down the problem **step by step**:
## **Step 1: Find the Volumetric Flow Rates**
The volumetric flow rate, \( Q \), is given by:
\[
Q = V \cdot A
\]
where:
- \( V \) = velocity (m/s)
- \( A \) = cross-sectional area (m\(^2\))
### **Inlet (Left Inlet)**
- Diameter = \( 100\, \text{mm} = .1\, \text{m} \)
- Velocity = \( 3\, \text{m/s} \)
\[
A_{in} = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.1)^2 = .00785\, \text{m}^2
\]
\[
Q_{in} = 3 \times .00785 = .02356\, \text{m}^3/\text{s}
\]
### **Outlet 1 (Bottom Outlet)**
- Diameter = \( 150\, \text{mm} = .15\, \text{m} \)
- Velocity = \( 1.5\, \text{m/s} \)
\[
A_{out1} = \frac{\pi}{4} (.15)^2 = .01767\, \text{m}^2
\]
\[
Q_{out1} = 1.5 \times .01767 = .0265\, \text{m}^3/\text{s}
\]
### **Outlet 2 (Right Outlet)**
- Diameter = \( 75\, \text{mm} = .075\, \text{m} \)
- Velocity = \( 1.2\, \text{m/s} \)
\[
A_{out2} = \frac{\pi}{4} (.075)^2 = .00442\, \text{m}^2
\]
\[
Q_{out2} = 1.2 \times .00442 = .00531\, \text{m}^3/\text{s}
\]
---
## **Step 2: Calculate Net Flow Rate**
\[
\text{Net flow rate} = Q_{\text{in}} - (Q_{\text{out1}} + Q_{\text{out2}})
\]
\[
= .02356 - (.0265 + .00531)
\]
\[
= .02356 - .03181
\]
\[
= -.00825\, \text{m}^3/\text{s}
\]
- **Negative value** means the water level is **falling**.
---
## **Step 3: Rate of Water Level Change**
Tank area:
- Diameter = \( 2\, \text{m} \)
- Area \( A_{\text{tank}} = \frac{\pi}{4} (2)^2 = \pi \text{ m}^2 \approx 3.142 \text{ m}^2 \)
\[
\frac{dh}{dt} = \frac{\text{Net flow rate}}{A_{\text{tank}}}
\]
\[
= \frac{-.00825}{3.142}
\]
\[
= -.00263\, \text{m/s}
\]
---
## **Final Answer**
### **The water level is falling at a rate of \(\boxed{.00263\, \text{m/s}}\) (to 3 significant figures).**
If you need this in mm/s:
\[
.00263\, \text{m/s} \times 100 = 2.63\, \text{mm/s}
\]
---
### **Summary Table**
| Flow | Value (m³/s) | Direction |
|-----------|--------------|-------------------|
| Inlet | .02356 | Into tank (+) |
| Outlet 1 | .0265 | Out of tank (–) |
| Outlet 2 | .00531 | Out of tank (–) |
| **Net** | **–.00825** | Water level falls |
---
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