Question

Water is pumped from a lake through an 9-in-diameter pipe at a rate of 10 ft3/s. If viscous effects are negligible, what is the gauge pressure in the suction pipe (the pipe between the lake and the pump) at an elevation of 7 ft above the lake?

Answer 1

The gauge pressure in the suction pipe at an elevation of 7 ft above the lake is the same as the atmospheric pressure.

Step-by-step explanation:

To find the gauge pressure in the suction pipe, we can use Bernoulli's equation which relates the pressure, velocity, and height of a fluid. The equation is:

P + 1/2 ρv^2 + ρgh = constant

Since the suction pipe is at the same horizontal level as the lake, the height component can be ignored. Also, because the viscous effects are negligible, we can ignore the velocity component. Therefore, we can simplify the equation to:

P = constant

This means that the gauge pressure in the suction pipe at an elevation of 7 ft above the lake is the same as the atmospheric pressure, which is approximately 14.7 pounds per square inch (psi).

Answer 2

Step-by-step explanation:

Expression to calculate gauge pressure is as follows.

`(p_(1))/(\gamma) + (v^(2)_(1))/(2g) + z_(1) = (p_(2))/(\gamma) + (v^(2)_(2))/(2g) + z_(2)`

where,
`(p)/(\gamma)` = pressure head

`(v^(2))/(2g)` = velocity head z

where,
`p_(1)` = 0,
`v_(1)` = 0,
`z_(1)` = 0, and
`z_(2)` = 6.0 ft

`V_(2) = (Q)/(A_(2)) = (4Q)/(\pi * D^(2)_(2))`

=
`(4 * 10 ft^(3)/s)/(3.14 * ((9)/(12)ft)^(2)))`

= 22.64 ft/s

Therefore,

`p_(2) = -\gamma z_(2) - (1)/(2) \rho V^(2)_(2)`

=
`-62.4 lb/ft^(3) * 7 ft - (1)/(2) * 1.94 slugs/ft^(3) * 22.64 ft/s`

= 414.9
`lb/ft^(2)`

As
`1 lbf/ft^(2) = 0.00694 psi`

So, 414.9
`lb/ft^(2)` =
`414.9 * 0.00694`

= 2.879 psi

Thus, we can conclude that the gauge pressure in the suction pipe is 2.879 psi.