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Water is flowing through two horizontal pipes of different diameters which are connected together. In the first pipe the speed of water 4 m/s and the pressure is `2.0xx10^(4) N//m^(2)`. Calculate the speed and pressure of water in the second pipe. The diameter of the pipes the diameter of the pipes are 3 cm and 6 cm respectively.

Text Solution

Verified by Experts

The correct Answer is:
`2.75`
According to equation of continuity
`a_(1) v_(a) = a_(2) v_(2)`
or `pi _(1)^(2) v_(1) = pi r_(2) ^(2) v_(2) or v_(2) = (( r_(1))/( r_(2)))^(2) v_(1)`
Here `r_(1) = 1.5 cm = 1.5 xx 10 ^(-2) m`
`r_(2) = 3 cm = 3 xx 10^(-2) m and v_(1) = 4 ms^(-1)`
`v_(2) = (( 1. 5 xx 10 ^(-2))/( 3 xx 10^(-2))) xx 4 = 1 ms ^(-1)`
Applying Bernouilli s theorem for horizontal flow
`p_(1) + (1)/(2) rho v_(1)^(2) = p_(2) + (1)/(2) rho v_(2)^(2) or p_(2) = p_(1) + (1)/( 2) rho ( v_(1) ^(2) - v_(2) ^(2))`
But ` rho _(omega ) (water) = 10 ^(3) kgm ^(-3) v_(1) = 4 ms^(-1) v_(2) = 1 ms^(-1) , p_(1) = 2 . 0 xx 10 ^(4) Nm ^(-2)`
`:. p_(2) = 2.0 xx 10 ^(4) + (1)/(2) xx 10^(3) ( 4^(2) - 1^(2))`
` = 2.75 xx 10 ^(4) N m ^(-2)`
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