Question

A 0.25 m diameter pipe carries oil of specific gravity
0.8 at the rate of 120 Litres per second and the
pressure at a point A is 19.62 kN/m^2 ( gauge). If
the point A is 3.5 m above the datum line , calculate
the total energy at point A in meters of oil ?​

Answer 1

Concept

If the fluid is in motion, the total head $T$ at any section of pipe can be calculated from Bernoulli's equation , $T=\dfrac{P}{\rho g}+\dfrac{v^2}{2g}+z$

Given

$SG=$$0.8$, d=$0.25m,$ $P=19.62KN/m^2$,$z=3.5m$, $Q=120 l/s=0.12m^3/s$,$g=9.81m/s^2$

To find

Total energy $T$, at point $A$

Solution

Total head $T=\dfrac{P}{\rho g}+\dfrac{v^2}{2g}+z$

$Q=Av$

$0.12=\dfrac{\pi }{4} \times d^2*v\\0.12=\dfrac{\pi }{4} \times (0.25)^2*v\\v=\dfrac{0.12\times 4}{\pi \times (0.25)^2} \\v=2.445m/s$

$\rho=SG\times \rho _{water}\\ \Rightarrow \rho =0.8\times 1000\\\Rightarrow \rho =800 kg/m^3$

Therefore,

$T=\dfrac{19.62\times 1000}{800\times 9.81} +\dfrac{2.445^2}{2\times 9.81} +3.5\\T=2.5+0.305+3.5\\T=6.305m$

As a result, Total head and Total energy,  $T$ at point $A$ is 6.305m

Answer 2

Answer:

Step-by-step explanation: