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NikAS [45]
4 years ago
11

A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl

aced within a channel to maintain an upstream depth of 2.5 m for a flow of 0.5 m3/s
Physics
1 answer:
Alex4 years ago
7 0

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

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Serga [27]

Answer:

Option C is the correct answer.

Explanation:

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Option C is the correct answer.

3 0
3 years ago
Read 2 more answers
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

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Where R_o is the initial resistance

Making the change in temperature the subject of the formula

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Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

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