Question

3.A pipe of diameter 400mm carries water at a velocity of 25m/s.the pressure at the points and a and b are given as 29.43N/m'and 22.563N/m' respectively while the datum head at a & bare 28m & 30m. Find the loss of head between a and b
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Answer 1

the attached is the answer

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Answer 2

Given: The diameter of a pipe $=400mm=0.4m$

The velocity of water flowing through the pipe $=25m/s$

The pressure at points A and B is $=29.43N/cm^2$ and $22.563N/cm^2$

Lengths of datum head $=28m$ and $30m$

To find: The loss of head between a and b.

Solution:

According to the formula of Bernoulli's equation for real fluids,

$\frac{P_1}{d_1}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{d_2}+\frac{v_2^2}{2g}+z_2+h_L$

where, $P_1,P_2$ are the pressure at the points A and B.

$v_1,v_2$ is the velocity of the fluid

$z_1,z_2$ are the lengths of the datum

$h_L$ is the loss of head between the two points.

At point A,

$P_A=29.43N/cm^2=29.43*10^4N/m^2$

$z_A=28m$

The velocity for both the points will remain the same that is,

$v_1=v=25m/s$

The total energy at point A will be:

$E_A=\frac{P_1}{d_1}+\frac{v_A^2}{2g}+z_A$

$E_A=\frac{29.43*10^4}{1000*9.81}+\frac{25*25}{2*9.81}+28$

$E_A=30+31.85+28=89.85J$

At point B,

$P_B=22.563N/cm^2=22.563*10^4N/m^2$

$z_B=30m$

The velocity remains the same which is $=25m/s$

∴ The total energy at B,

$E_B=\frac{P_2}{d_2}+\frac{v_2^2}{2g}+z_B$

$E_B=\frac{22.563*10^4}{1000*9.81}+\frac{25*25}{2*9.81}+30$

$E_B=23+31.85+30=84.85J$

∴ The loss of head between point A and B will be,

$=E_A-E_B\\=89.85-84.85\\=5.0J$

Final answer:

The loss of head between A and B will be 5.0J.