Question

•• Unfiltered olive oil must flow at a minimum speed of 3.0 m/s to prevent settling of debris in a pipe. The oil leaves a pump at a pressure of 88 kPa through a pipe of radius 9.5 mm. It then enters a horizontal pipe at atmospheric pressure. Ignore the effects of viscosity. (a) What is the speed of the oil as it leaves the pump if it flows at 3.0 m/s in the horizontal pipe? (b) What is the radius of the horizontal pipe?

Answer 1

a

The velocity at which the oil leaves the pump is
`v_1 = 5.15 m/s`

b

The radius of the horizontal pipe
`r_2 = 0.01244 m`

Step-by-step explanation:

From the question we are told that

The rate of flow of unfiltered olive oil is
`v = 3.0 m/s`

The pressure on the pump is
`P_1 = 88 kPa = 88 *10^(3) \ Pa`

The radius of the pipe is
`r = 9.5 = (9.5)/(1000) = 0.0095 \ m`

Generally we can define this motion with Bernoulli's Equation as

`P_1 + (1)/(2) \rho v_1 ^2 + \rho g h_1 = P_2 + (1)/(2) \rho v_2 + \rho g h_2`

`P_2` is the pressure inside the pipe which is the atmospheric pressure which has a value of
`P_2 = 1.01 *10^(5) \ Pa`

`\rho` is the density of olive oil which has a value of
`\rho = 980 kg/m^3`

`g` is the acceleration due to gravity

h the height of the pipe since from the question we are not told that it is place on anything hence the height is the diameter of the pipe

`v_1 \ and \ v_2` are the speed at which it leaves the pump and the speed at which it flows in the pipe respectively

Making
`v_2` the subject of the equation

`v_1 = \sqrt{[(2)/(\rho) (P_2 -P_1 - (1)/(2) v^2_2)]}`

Substituting values

`v_1 = \sqrt{[(2)/(980) (1.01 *10^5 -88 *10^(3) - (1)/(2)(3.0)^2)]}`

`v_1 = 5.15 m/s`

Using continuity equation to define the motion of this fluid we have

`A_1 v_1 = A_2 v_2`

Where
`A_1` is the area of the pump which is circular so mathematically it is

`A_1 = \pi r^2_1`

`A_2` is the area of the horizontal pipe which is circular so mathematically it is

`A_2 = \pi r^2_2`

So the continuity equation becomes

`\pi r^2 _1 v_1 = \pi r^2 _2 v_2`

Making
`r_2` subject of formula

`r_2 = \sqrt{(r_1^2 v_1)/(v_2) }`

Substituting values

`r_2 = \sqrt{( 0.0095^2 5.15 )/(3.0) }`

`r_2 = 0.01244 m`