Question

A pipeline carrying oil of specific gravity 0.87, changes in diameter from 200 mm diameter

at a position A to 500 mm Diameter at a Position B which is 4 meters at a higher level. If the

pressures at A and B are 9.81 N/cm2 and 5.886 N/cm2 respectively and the discharge is 200

litres/s determine the loss of head and direction of flow.​

Answer 1

Answer:

The head loss is 22.54Kpa and

The direction of liquid flow will be from point A to point B

Explanation:

Given data is,

Diameter at point A=200mm

Diameter at point A=500mm

Discharge of oil=200L/sec

Using the equation of continuity at point A and point B we have

$A_{1}V_{1}=A_{2}V_{2} \\ = > \frac{\pi}{4} (0.2) {}^{2} V_{1} = \frac{\pi}{4} (0.5) {}^{2} V_{2} = 200 \times 10 {}^{ - 3} m {}^{3} s {}^{ - 1} \\ = > V_{1} =6.36ms {}^{ - 1} \\ V_{2} =1.01ms {}^{ - 1}$

Also applying bernoulli's equation at both the points we get

$P_{1}+\frac{1}{2}pv _{1} ^2+pgh_{1} \\ = P_{2}+\frac{1}{2}pv _{2}^2+pgh_{2}+∆h_{f}$

Substituting given values we get

$9.81 \times 10 {}^{4} + \frac{1}{2} \times 870 \times 6.36 {}^{2} + 870 \times 9.81 \times 0 \\ = 5.886 \times 10 {}^{4} + 0.5 \times 870 \times 1.01 {}^{2} + 870 \times 9.81 \times 4 +∆h_{f}$

On calculation,we get the head loss as

$∆h_{f} = 22543.21Pa = 22.54kPa$

The direction of liquid flow will be from point A to point B

#SPJ3

Answer 2

A pipeline carrying oil of specific gravity 0.87, changes in diameter from 200 mm diameter at a position A to 500 mm Diameter at a Position B which is 4 meters at a higher level. If the pressures at A and B are 9.81 N/cm2 and 5.886 N/cm2 respectively and the discharge is 200 litres/s determine the loss of head and direction of flow.​

Solution -

Given ,

Diameter at point A =  200mm

Diameter at point B = 500 mm

Discharge of oil = 200/sec

equation of continuity at point A and B

$A_{1} V_{1} = A_{2}V_2} \\= \frac{\pi }{4} (0.2)^{2} V_{1} = \frac{\pi }{4} (0.5)^{2} \\=200 * 10^{-3} m^{3} s^{-1} \\= V_{1} = 6.36 ms^{-1}$

apply bernoulli's theorem

$P_{1} + \frac{1}{2} pv_{1} ^{2} + pgh_{1} \\= P_{2} + pv_{2} ^{2} + pgh_{2} +ahf$

Substituting given values we get

9.81 ×$10^{4}$ + $\frac{1}{2} * 870 * 6.36^{2} + 870 * 9.81 *0\\= 5.886 * 10^{4} +0.5* 870 * 1.01^{2} +870 * 9.81 * 4 + ahf$

On calculation , we get the head loss as

$ahf = 22543.21 Pa = 22.54kPa$

The direction of liquid flow will be from point A to B.

#SPJ2