2. The tank in right figure is cylindrical and has a vertical axis. Its horizontal cross-sectional area is 100 ft². The hole in the bottom has a cross-sectional area of 1 ft². The interface between the gasoline and the water remains perfectly horizontal at all times. That interface is now 10 ft above the bottom. How soon will gasoline start to flow out the bottom? Assume frictionless flow. 1848-1 10 ft 10 ft Gasoline Water
2. The tank in right figure is cylindrical and has a vertical axis. Its horizontal cross-sectional area is 100 ft². The hole in the bottom has a cross-sectional area of 1 ft². The interface between the gasoline and the water remains perfectly horizontal at all times. That interface is now 10 ft above the bottom. How soon will gasoline start to flow out the bottom? Assume frictionless flow. 1848-1 10 ft 10 ft Gasoline Water
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
can you please solve question 2
Transcribed Image Text:1. The system in right figure consists of a water reservoir with a
layer of compressed air above the water and a large pipe and nozzle.
The pressure of the air is 50 psig, and the effects of friction can be
neglected. What is the velocity of the water flowing out through the
nozzle?
50 psig
2. The tank in right figure is cylindrical and has a vertical axis. Its
horizontal cross-sectional area is 100 ft². The hole in the bottom has
a cross-sectional area of 1 ft². The interface between the gasoline and the
water remains perfectly horizontal at all times. That interface is now 10 ft
above the bottom. How soon will gasoline start to flow out the bottom?
Assume frictionless flow.
30 ft
10 ft
10 ft
1
Gasoline
Water
iiili
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i am having a hard time seeing the unit conversion for step 2
i am trying and this is what i have so far
if there is any unit conversion that u didnt write out for step 3 plz let me know so i can plug in for better viewer
![The image displays a series of fluid mechanics calculations, primarily focusing on the Bernoulli equation and the calculation of velocity.
Equation Breakdown:
1. **Bernoulli Equation:**
\[
0 + 0 + z_1 = \frac{P_1}{\rho} + \frac{v_2^2}{2g} + z_2
\]
2. **Pressures and Unit Conversion:**
\[
1091.8 \text{ psi or } \frac{\text{lb}}{\text{in}^2}
\]
3. **Calculation Setup:**
\[
20 = \frac{1091.8 \frac{\text{lb}}{\text{in}^2}}{62.43 \frac{\text{lb}}{\text{ft}^3}} + \frac{V_{\text{bottom}}^2}{2 \left( g \rightarrow \frac{32.2 \text{ ft}}{\text{s}^2} \right)}
\]
4. **Unit Conversion Consistency:**
\[
\frac{\text{lb}}{\text{in}^2} \times \frac{\text{ft}^3}{\text{lb}} \times \frac{144 \text{ in}^2}{\text{ft}^2} = \text{ft}
\]
5. **Velocity Calculation:**
\[
V_{\text{bottom}} = \sqrt{2 \left( \frac{32.2 \text{ ft}}{\text{s}^2} \times 17.488 \right)}
\]
\[
V_{\text{bottom}} = 33.56 \frac{\text{ft}}{\text{sec}}
\]
**Explanation:**
- **Bernoulli Equation:** Describes the behavior of a fluid under varying conditions of flow and height and is used to find the velocity at the bottom.
- **Units:** Demonstrates conversion of pressure units and volume consistency to maintain coherent results.
- **Velocity Calculation:** Involves the square root of doubled gravitational acceleration and an unknown constant (17.488) to determine velocity, resulting in \(33.56 \text{ ft/sec}\).](https://content.bartleby.com/qna-images/question/63f5865a-7713-44c4-be73-1c5678c4385b/3fcd8f5a-4e73-4a85-a5ea-4562d42f69e5/j0vcdb_processed.jpeg)
Transcribed Image Text:The image displays a series of fluid mechanics calculations, primarily focusing on the Bernoulli equation and the calculation of velocity.
Equation Breakdown:
1. **Bernoulli Equation:**
\[
0 + 0 + z_1 = \frac{P_1}{\rho} + \frac{v_2^2}{2g} + z_2
\]
2. **Pressures and Unit Conversion:**
\[
1091.8 \text{ psi or } \frac{\text{lb}}{\text{in}^2}
\]
3. **Calculation Setup:**
\[
20 = \frac{1091.8 \frac{\text{lb}}{\text{in}^2}}{62.43 \frac{\text{lb}}{\text{ft}^3}} + \frac{V_{\text{bottom}}^2}{2 \left( g \rightarrow \frac{32.2 \text{ ft}}{\text{s}^2} \right)}
\]
4. **Unit Conversion Consistency:**
\[
\frac{\text{lb}}{\text{in}^2} \times \frac{\text{ft}^3}{\text{lb}} \times \frac{144 \text{ in}^2}{\text{ft}^2} = \text{ft}
\]
5. **Velocity Calculation:**
\[
V_{\text{bottom}} = \sqrt{2 \left( \frac{32.2 \text{ ft}}{\text{s}^2} \times 17.488 \right)}
\]
\[
V_{\text{bottom}} = 33.56 \frac{\text{ft}}{\text{sec}}
\]
**Explanation:**
- **Bernoulli Equation:** Describes the behavior of a fluid under varying conditions of flow and height and is used to find the velocity at the bottom.
- **Units:** Demonstrates conversion of pressure units and volume consistency to maintain coherent results.
- **Velocity Calculation:** Involves the square root of doubled gravitational acceleration and an unknown constant (17.488) to determine velocity, resulting in \(33.56 \text{ ft/sec}\).
Solution
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