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3 years ago

Define the hydraulic diameter for a rectangular duct

Engineering

1 answer:

8090 [49]3 years ago

4 0

Answer with Explanation:

Hydraulic diameter is a term analogous to the diameter of the circular sectional pipe but used for the cases when the cross sectional shape of the pipe is non circular.

It serves as an equivalent diameter that is used to calculate the Reynolds number for the flow.

The hydraulic diameter is 4 times the hydraulic radius of any section.

For a rectangular duct as shown in the attached figure

$R_{h}=\frac{Wetted_{Area}}{Wetted_{perimeter}}\\\\R_h=\frac{d\times b}{2(d+b)}\\\\\therefore D_{h}=4\times R_{h}=4\times \frac{db}{2(d+b)}=\frac{2db}{(d+b)}$

Where

$D_{h}$ is the hydraulic diameter of the duct with depth 'd' and width 'b'

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Answer:

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Explanation:

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An escalator in a shopping center is designed to move 50 people, 75 kg each, at a constant speed of 0.6 m/s at 45° slope. Determ

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Answer:

$\dot E = 15602.842\,W$

Explanation:

By an adequate application of the Principle of Energy Conservation, the escalator need energy to elevate from to the bottom to the top. Hence:

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The minimum power is found by substituting known inputs:

$\dot E = (50)\cdot (75\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.6\,\frac{m}{s} )\cdot \sin 45^{\textdegree}$

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A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t

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Work done = -19.7 KJ

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Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, $P_{1}$ = 1 bar

$v_{2}$ = 4 bar

pV¹°¹ = constant

From superheated propane table, at $P_{1}$ = 1 bar and $T_{1}$ = 27⁰C

$v_{1}$ = 0.557 m³/kg

$v_{2}$ = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

$V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1 \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557 \\\\V2 = 0.158 m^3/kg$

At = 4 bar and v = 0.158 m³/kg

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To find work done in the process:

$W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\ \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\ \\W = -19.75KJ$

(b) Heat transfer = ?

$Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ$

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