Flow from a tank. For the tank shown in figure below, compute the velocity of flow from the nozzle and the volume flow rate for a range of depth from 3.0 m to 0.5 m in steps of 0.5 m. The diameter of the jet at the nozzle is 50 mm. Depth h (m) V, (m/s) Q (m/s) HINT Use Bernoulli's equation between points 1 and 2. 3.0 2.5 P,ly + Z, + V,2/2g = P,) y + Z, + V,2/2g 2.0 Assume the datum to pass horizontally through point 2. Therefore, 1.5 h 1.0 PY+ Z, + V,/2g = P/ y+ Z, + V,?/2g 0.5 Also, note that by using Bernoulli's equation, the following Torricelli's theorem is proved. Y2 = /2gh
Flow from a tank. For the tank shown in figure below, compute the velocity of flow from the nozzle and the volume flow rate for a range of depth from 3.0 m to 0.5 m in steps of 0.5 m. The diameter of the jet at the nozzle is 50 mm. Depth h (m) V, (m/s) Q (m/s) HINT Use Bernoulli's equation between points 1 and 2. 3.0 2.5 P,ly + Z, + V,2/2g = P,) y + Z, + V,2/2g 2.0 Assume the datum to pass horizontally through point 2. Therefore, 1.5 h 1.0 PY+ Z, + V,/2g = P/ y+ Z, + V,?/2g 0.5 Also, note that by using Bernoulli's equation, the following Torricelli's theorem is proved. Y2 = /2gh
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Flow from a Tank**
For the tank shown in the figure below, compute the velocity of flow from the nozzle and the volume flow rate for a range of depths from 3.0 m to 0.5 m in steps of 0.5 m. The diameter of the jet at the nozzle is 50 mm.
**Hint:**
Use Bernoulli’s equation between points 1 and 2:
\[
\frac{P_1}{\gamma} + Z_1 + \frac{V_1^2}{2g} = \frac{P_2}{\gamma} + Z_2 + \frac{V_2^2}{2g}
\]
Assume the datum to pass horizontally through point 2. Therefore,
\[
\frac{P_1}{\gamma} + \cancel{Z_1} + \cancel{\frac{V_1^2}{2g}} = \frac{P_2}{\gamma} + \cancel{Z_2} + \frac{V_2^2}{2g}
\]
Also, note that by using Bernoulli’s equation, the following Torricelli’s theorem is proved:
\[
V_2 = \sqrt{2gh}
\]
**Table:**
| Depth h (m) | \( V_2 \) (m/s) | \( Q \) (m³/s) |
|-------------|-----------------|----------------|
| 3.0 | | |
| 2.5 | | |
| 2.0 | | |
| 1.5 | | |
| 1.0 | | |
| 0.5 | | |
**Diagram Explanation:**
The diagram shows a tank with a liquid flowing out through a nozzle at the bottom. The height of the liquid in the tank is denoted by \( h \), which varies from 3.0 m to 0.5 m. The flow is analyzed at a nozzle of 50 mm diameter using Bernoulli’s equation to compute the exit velocity \( V_2 \) and volume flow rate \( Q \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F535a6610-5408-42b9-810f-cca1f66cb656%2F875b3094-82f9-4efc-80b7-6f50660a6d3b%2F1uubhmc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Flow from a Tank**
For the tank shown in the figure below, compute the velocity of flow from the nozzle and the volume flow rate for a range of depths from 3.0 m to 0.5 m in steps of 0.5 m. The diameter of the jet at the nozzle is 50 mm.
**Hint:**
Use Bernoulli’s equation between points 1 and 2:
\[
\frac{P_1}{\gamma} + Z_1 + \frac{V_1^2}{2g} = \frac{P_2}{\gamma} + Z_2 + \frac{V_2^2}{2g}
\]
Assume the datum to pass horizontally through point 2. Therefore,
\[
\frac{P_1}{\gamma} + \cancel{Z_1} + \cancel{\frac{V_1^2}{2g}} = \frac{P_2}{\gamma} + \cancel{Z_2} + \frac{V_2^2}{2g}
\]
Also, note that by using Bernoulli’s equation, the following Torricelli’s theorem is proved:
\[
V_2 = \sqrt{2gh}
\]
**Table:**
| Depth h (m) | \( V_2 \) (m/s) | \( Q \) (m³/s) |
|-------------|-----------------|----------------|
| 3.0 | | |
| 2.5 | | |
| 2.0 | | |
| 1.5 | | |
| 1.0 | | |
| 0.5 | | |
**Diagram Explanation:**
The diagram shows a tank with a liquid flowing out through a nozzle at the bottom. The height of the liquid in the tank is denoted by \( h \), which varies from 3.0 m to 0.5 m. The flow is analyzed at a nozzle of 50 mm diameter using Bernoulli’s equation to compute the exit velocity \( V_2 \) and volume flow rate \( Q \).
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