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ryzh [129]
4 years ago
12

For a gearbox power and efficiency test apparatus that accommodates interchangeable gearboxes, drive up to 0.5kW. How to select

the motor for the test apparatus?
Engineering
1 answer:
BigorU [14]4 years ago
4 0

Answer:

First we must know the efficiency of the gearbox in which energy is lost by friction between its components, we divide this efficiency by the nominal power and find the real power.

With this value knowing the speed of rotation of the gearbox we select an engine that has a power just above.

Also keep in mind that the motor torque must be greater than the resistance offered by the gearbox and the 0.5Kw machine

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
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Answer:

i want coins sorry use a calculator or sum

Explanation:

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3 years ago
Does anybody know how to take a screenshot on a HP pavilion computer?
Setler79 [48]

Answer:

I do i do it everyday

Explanation:

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6 0
3 years ago
A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?
seraphim [82]

Answer:

No.

Explanation:

P_r = Power rating = 1 W

R = Resistance = 350\ \Omega

V = Voltage = 24\ \text{V}

Power is given by

P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}

1.65\ \text{W}>1\ \text{W}

So

P>P_r

Hence, the resistor is not operating within its power rating.

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Compute the number of games using the formula of “SINGLE ROUND ROBIN” and plot the games in a table:
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Answer:

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