Give step-by-step solution with explanation and final answer:" N ne
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Give step-by-step solution with explanation and final answer:
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Asked by: Test User 100044
Created at: 2025-05-05 13:17:24
Asked by: Test User 100044
Created at: 2025-05-05 13:17:24
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Let's break down the problem step by step.
### **Given Data**
- **Pressure at top of container** (\(P_{top}\)): \(2 \, \text{atm}\)
- **Height of fluid column** (\(h_0\)): \(10\, \text{m}\)
- **Vertical distance to outlet** (\(d\)): \(5\, \text{m}\)
- **Radius of tube** (\(r\)): \(2\, \text{cm} = 0.02\, \text{m}\)
- **Fluid**: Water (\(\rho = 1000\, \text{kg/m}^3\))
- **Acceleration due to gravity** (\(g\)): \(9.81\, \text{m/s}^2\)
- **Atmospheric Pressure** (\(P_{atm}\)): \(1\, \text{atm} = 101325\, \text{Pa}\)
---
### **Step 1: Apply Bernoulli’s Equation**
Apply Bernoulli's equation between the water surface (point 1) and the outlet (point 2):
\[
P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2
\]
Assume the water surface velocity (\(v_1\)) is negligible (\(v_1 \approx 0\)).
Let:
- \(P_1 = 2\, \text{atm} = 2 \times 101325 = 202650\, \text{Pa}\)
- \(P_2 = 1\, \text{atm} = 101325\, \text{Pa}\) (pressure at outlet open to atmosphere)
- \(h_1 = h_0 = 10\, \text{m}\)
- \(h_2 = d = 5\, \text{m}\)
- \(v_2 = v\) (velocity at outlet)
Plug in values:
\[
P_1 + \rho g h_1 = P_2 + \frac{1}{2}\rho v^2 + \rho g h_2
\]
\[
202650 + 1000 \times 9.81 \times 10 = 101325 + \frac{1}{2} \times 1000 \times v^2 + 1000 \times 9.81 \times 5
\]
Calculate each term:
- \(1000 \times 9.81 \times 10 = 98100\)
- \(1000 \times 9.81 \times 5 = 49050\)
So:
\[
202650 + 98100 = 101325 + 49050 + 500 v^2
\]
\[
300750 = 150375 + 500 v^2
\]
\[
300750 - 150375 = 500 v^2
\]
\[
150375 = 500 v^2
\]
\[
v^2 = \frac{150375}{500} = 300.75
\]
\[
v = \sqrt{300.75} \approx 17.34\, \text{m/s}
\]
---
### **Step 2: Calculate Flow Rate (\(Q\))**
\[
Q = \text{Area} \times \text{Velocity}
\]
- Area of tube, \(A = \pi r^2 = \pi (0.02)^2 = \pi \times 0.0004 = 0.001257\, \text{m}^2\)
- Flow rate: \(Q = 0.001257 \times 17.34 = 0.0218\, \text{m}^3/\text{s}\)
---
### **Final Answers**
- **Velocity at outlet (\(v\)):** \(\boxed{17.34\, \text{m/s}}\)
- **Flow rate (\(Q\)):** \(\boxed{0.0218\, \text{m}^3/\text{s}}\)
---
**Summary of Solution Steps:**
1. **Apply Bernoulli's Equation** to find the velocity at the outlet.
2. **Calculate Flow Rate** using the tube's area and the outlet velocity.
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