Question

 derive expression Cd=Cv xCc
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Answer 1

Answer:

We are asked to derive the the given expression that is- $C_{d} =C_{v} \times C_{c}$

Explanation:

Here, $C_{d}$ is coefficient of discharge

$C_{v}$ is coefficient of flow/ velocity

$C_{c}$ is coefficient of contraction

1. Co- efficient of contraction $C_{c}$ is defined as the ratio of the area of the jet at vena- contracts to the area of the orifice

It is denoted by symbol $C_{c}$

a = area of orifice

ac= area of jet at vena- contracta

Then $C_{c}=\frac{area of jet at vena- contracta}{area of orifice}\\ C_{c}=\frac{ac}{a}$

the value of $C_{c}$ varies from 0.61 to 0.69 depending on shape and size.

in generally, the value of $C_{c}$ may be taken as 0.64

2. Co- efficient of velocity: is defined as the ratio between the actual velocity of a jet at vena contracta to the theoretical velocity of jet.

it is denoted by symbol $C_{v}$

$C_{v}=\frac{actua; velocity of jet at vena- contraccta}{theoretical velocity}\\ C_{v}=\frac{v}{\sqrt{2gh} }$

where, v is actual velocity and root 2gh is theoretical velocity

The value of $C_{v}$ varies from 0.95 to 0.99 depending on the shape and size. generally the value of $C_{v}$ is 0.98 is taken for sharp edged orifices.

3. Co- efficient of discharge: is defined as the ratio of the actual discharge from the orifice

It is denoted by $C_{d}$. If Q is actual discharge and $Q_{th}$ is the theoretical discharge

$C_{d}=\frac{actual velocity \times actual area}{theoretical velocity \times theoretical area}\\ C_{d}= C_{v} \times C_{v}$

The values of $C_{d}$ varies from 0.61 to 0.65, generally the value of $C_{d}$ is taken as 0.62.

Hence it is proved.

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