Question

A 200 mm long pipe slopes down at 1 in 100 and tapers from 0.25m diameter to 0.15 diameter at the lower end. If the pipe carries 100 litres of oil of specific gravity 0.85 find the pressure at the lower end. The upper end gauge reads 50 KPa.

Answer 1

Value of pressure at upper portion is 38.123kPa

Step-by-step explanation:

We shall use Bernoull's equation to solve

According to Bernoulli's equation we have

`(P)/(\gamma )+(8Q^(2))/(\pi ^(2)D^(4)g)+z=constant\\\\(P_(1))/(\gamma )+(8Q^(2))/(\pi ^(2)D_(1)^(4)g)+z_(1)=(P_(2))/(\gamma )+(8Q^(2))/(\pi ^(2)D_(2)^(4)g)+z_(2)`

Taking
`P_(1)` pressure at upper end

`P_(2)` pressure at lower end

Applying values we get

`(50* 10^(3))/(\gamma _(w)* 0.85)+(8* (0.1)^(2))/(\pi ^(2)* (0.25)^(4)* 9.81)+z_(1)=(P_(2))/(\gamma _(w)* 0.85)+(8* (0.1)^(2))/(\pi ^(2)* (0.15)^(4)* 9.81)+z_(2)\\\\\therefore (P_(2))/(\gamma _(w)* 0.85)=(50* 10^(3))/(\gamma _(w)* 0.85)+(8* (0.1)^(2))/(\pi ^(2)* (0.25)^(4)* 9.81)+z_(1)-(8* (0.1)^(2))/(\pi ^(2)* (0.15)^(4)* 9.81)-z_(2)`

Solving for
`P_(2)` we get

`(P_(2))/(0.85* \gamma _(w))=4.575+(z_(1)-z_(2))\\\\\therefore P_(2)=0.85* 9810* (4.575+0.002)\\\\\therefore P_(2)=38.123kPa`