4-6. A vertical cylindrical storage vessel is 10 m high and 2 m in diameter. The vessel contains liquid cyclohexane currently at
a liquid level of 8 m. The vessel is vented to the atmosphere. A 1-cm-diameter hole develops in the bottom of the vessel. Calculate and plot the discharge rate of cyclohexane as a function of time until the vessel is completely empty. Show that the time to empty is the same as the time calculated using Equation 4-21.
1 answer:
Answer:
See attachment for step by step procedure into getting answer.
Explanation:
Given that;
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4-6. A vertical cylindrical storage vessel is 10 m high and 2 m in diameter. The vessel contains liquid cyclohexane currently at a liquid level of 8 m. The vessel is vented to the atmosphere. A 1-cm-diameter hole develops in the bottom of the vessel. Calculate and plot the discharge rate of cyclohexane as a function of time until the vessel is completely empty. Show that the time to empty is the same as the time calculated using Equation 4-21.
See attachment for completed steps.
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Answer:
Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.
AC Signal = An electrical signal that alternates between positive and negative voltage.
DC Signal = An electrical signal that only has positive voltage.
Rectify = A fancy word for converting something.
Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.
The load resistor is just there to prevent a short circuit from happening.
Answer:
radius = 9.1 × m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = .................1
here F is applied load and is length
put here value and we get
radius =
solve it we get
radius = 9.1 × m
Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>
<em>i) Engineering stress at maximum load = P/ A </em>
= P / = 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
<em>ii) True Fracture stress = P/A </em>
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2