A cylindrical tank has diameter D=1.20 m. It starts at h0=0.30 m and must reach h=0.80 m. The inlet volumetric flow rate is Qin=0.0026 m^3/s and the outlet volumetric flow rate is Qout=0.0009 m^3/s. Find the time required.单项选择题
A
270 s
B
330 s
C
360 s
D
300 s

Question

A cylindrical tank has diameter D=1.20 m. It starts at h0=0.30 m and must reach h=0.80 m. The inlet volumetric flow rate is Qin=0.0026 m^3/s and the outlet volumetric flow rate is Qout=0.0009 m^3/s. Find the time required.单项选择题

A

270 s

B

330 s

C

360 s

D

300 s

BlackTom AI · Accepted Answer

Net volumetric accumulation rate equals Qin − Qout, and the rise in tank height comes from that net volume filling the cylindrical cross-sectional area.

1) Relevant concepts
- Conservation of mass (volume balance for an incompressible liquid): dV/dt = Qin − Qout.
- For a vertical right circular cylinder, volume V = A·h where A = πD^2/4 is the tank cross-sectional area.
- Time to change height by Δh: t = ΔV / (Qin − Qout) = A·Δh / (Qin − Qout).

2) Step-by-step reasoning
- Compute cross-sectional area A from the diameter D.
- Compute the required volume increase ΔV = A·(h − h0).
- Compute net volumetric rate Qnet = Qin − Qout.
- Compute t = ΔV / Qnet and round to the nearest available choice.

3) Calculations (all intermediate steps)
- D = 1.20 m → A = π·D^2/4 = π·(1.20^2)/4 = π·1.44/4 = π·0.36 = 0.36π m^2 ≈ 1.1309734 m^2.
- Height change Δh = 0.80 − 0.30 = 0.50 m.
- Volume needed ΔV = A·Δh = 0.36π · 0.50 = 0.18π m^3 ≈ 0.5654867 m^3.
- Qin = 0.0026 m^3/s, Qout = 0.0009 m^3/s → Qnet = 0.0026 − 0.0009 = 0.0017 m^3/s.
- Time t = ΔV / Qnet = 0.5654867 / 0.0017 ≈ 332.64 s.

4) Why the correct answer is correct
The computed time is about 332.6 s. Among the multiple-choice options, the closest value is 330 s. Standard practice is to choose the option nearest the calculated value; rounding 332.6 s to the nearest ten seconds gives 330 s, which matches the provided correct answer.

5) Why the other options are incorrect
- 270 s, 300 s, and 360 s deviate substantially (by ≈63 s, 33 s, and 27 s respectively) from the computed 332.6 s and are not the nearest available choices.

6) Conclusion
Therefore, the required time is approximately 330 s, which agrees with the provided correct answer.

A cylindrical tank has diameter D=1.20 m. It starts at h0=0.30 m and must reach h=0.80 m. The inlet volumetric flow rate is Qin=0.0026 m^3/s and the outlet volumetric flow rate is Qout=0.0009 m^3/s. Find the time required.单项选择题

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A cylindrical tank has diameter D=1.20 m. It starts at h0​=0.30 m and must reach h=0.80 m. The inlet volumetric flow rate is Qin​=0.0026 m^3/s and the outlet volumetric flow rate is Qout​=0.0009 m^3/s. Find the time required.单项选择题

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A river 935 m wide and 3.5 m deep, flowing at 1.5 m sec-1 enters narrow canyon 55 m wide and 15 m deep. What is the average river speed in the canyon assuming its cross section is a elliptical? The area of an ellipse is 𝜋 a b, where a and b are the two axes of the ellipse.

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更多留学生实用工具

希望你的学习变得更简单

A cylindrical tank has diameter D=1.20 m. It starts at h0=0.30 m and must reach h=0.80 m. The inlet volumetric flow rate is Qin=0.0026 m^3/s and the outlet volumetric flow rate is Qout=0.0009 m^3/s. Find the time required.单项选择题