Question

A pipe of diameter 0.2 m carries oil (specific gravity = 0.85) at the rate of 100 litres per second and the pressure at a point P is 19.62 kN/m2 (gauge). If the point P is 3 m above the datum line, then determine the total energy at point P in metres of oil.?​

Answer 1

Answer:

The continuum idealization is implicit in many. 1.3 DENSITY AND SPECIFIC GRAVITY. Density is defined as mass per unit volume. Density. 3 m. (kg / m ).

Step-by-step explanation:

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Answer 2

Answer:

The answer to the given question is the total energy at point P in metres of oil is 5.8748 m

Step-by-step explanation:

Given :

SG specific gravity = 0.85.

diameter = 0.2 m

pressure at a point p = 19.62 kN/m²

Q= 100 litres per second = 0.10ml/s

gravity = 9.81 m/s².

z= 3 m.

To find :

Total energy at a point P.

Concept :

The concept we should use here is The total energy T of any section of a pipe when a fluid is in motion can be calculated from Bernoulli's equation.

T is

$= \frac{p}{ (rho)g} + \frac{ {v}^{2} }{2g} + z$

Solution:

T is calculated by the formula

$\frac{p}{ (rho)g} + \frac{ {v}^{2} }{2g} + z$

first, we have to find the value of v by the formula.

Q= A v.

$0.10 = \frac{\pi}{4} \times {d}^{2} \times v$

on substituting the values we get the value of v as

$0.10 = \frac{\pi}{4} \times {(0.2)}^{2} \times v \\ v = \frac{0.10 \times 4}{ {(0.2)}^{2} \times \pi }$

The final value of v is

$\frac{0.4}{0.125} = 3.2 \frac{m}{s}$

rho (Density ) = SG * rho (water ). Density of water

$= 0.85 \times 1000 \\ = 850 \frac{kg}{ {m}^{3} }$

Therefore, the value of T will be

$= \frac{19.62 \times 1000}{850 \times 9.81} + \frac{ {(3.2)}^{2} }{2 \times 9.81} + 3$

on solving the above expression, we get the final answer as

$\frac{19620}{8338.5} + \frac{10.24}{19.62} + 3$

on solving each term and on adding we get the value as

$= 2.3529 + 0.5219 + 3 = 5.8748$

Therefore, the total energy at point P in metres of oil is

5.8748 m

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