Question

Carbon dioxide flows at a rate of 1.5 ft3 /s from a 3-in. pipe in which the pressure and temperature are 20 psi (gage) and 120 °F into a 1.5-in. pipe. If viscous effects are neglected and incompressible conditions are assumed, determine the pressure in the smaller pipe

Answer 1

the absolute pressure in the smaller pipe = 19.63 psi

Step-by-step explanation:

Let A be the diameter of the first pipe = 3 inches

Let B be the diameter of the second pipe. = 1.5 inches

To feet (ft) ; we have

Diameter of the pipe A
`D_1 = ((3)/(12))ft = 0.25 \ ft`

Diameter of pipe B
`D_1 = ((1.5)/(12))ft = 0.125 \ ft`

Temperature T = 120° F = (120+ 460)°R

= 580 ° R

The pressure gage to atmospheric pressure ; we have:

`P_(Absolute )=P _(Atm) + P_(guage)`

where;

atmospheric pressure = 1.47 psi

pressure gage = 20 psi

`P_(Absolute )=(1.47+20)psi`

`P_(Absolute )=34.7 \ psi`

To lb/ft²; we have:

`P_(Absolute )=(34.7 *144 ) lb/ft^2`

`P_(Absolute )=` 4.998.6 fb/ft²

The density of carbon dioxide can be calculated by using the relation

`\rho = (P)/(RT)`

`\rho = \frac{4996.8}{(1130 \ lb /slug ^0 R)*(580{^0} R)}`

`\rho = 7.64*10^(-3)\ slug /ft^3`

Formula for calculating cross sectional area is

`A = (\pi)/(4)D`

For diameter of pipe
`D_1 = 0.025`

A₁ =
`(\pi)/(4)*0.25^2`

A₁ = 0.04909 ft²

For diameter of pipe
`D_2 - 0.0125`

A₂
`=(\pi)/(4)*0.125^2`

A₂ = 0.012227 ft²

Using the continuity equation to determine the velocities V₁ and V₂ respectively.

For V₁

Q = A₁V₁

V₁ = Q₁/ A₁

V₁ = 1.5/0.04909

V₁ = 30.557 ft/s

For V₂

Q = A₂V₂

V₂= Q₂/ A₂

V₂ = 1.5/0.04909

V₂ = 30.557 ft/s

Finally; using Bernoulli's Equation to the flow of the carbon dioxide from the larger pipe to the smaller pipe ; we have:

`p_1 + (\rho V_1^2)/(2)+\gamma Z_1= p_2 + (\rho V_2^2)/(2)+\gamma Z_2`

Since the pipe is horizontal then;

`\gamma Z_1= \gamma Z_2`

So;

`p_1 + (\rho V_1^2)/(2)= p_2 + (\rho V_2^2)/(2)`

`p_2 =p_1 +(1)/(2) \rho(V_1^2-V_2^2)`

`p_2 =4996.8+(1)/(2) *7.624*10^(-3)(30.557^2-122.23^2)`

`p_2 =4943.41 \ lb/ft^2`

To psi;

`p_2 =(4943.41 )/(144)psi`

`p_2 =34.33 \ psi` gage

The absolute pressure in the smaller pipe can be calculated as:

`p_2 _(absolute) = 34.33 - 14.7`

`p_2 _(absolute) = 19.63 \ \ absolute`

Hence, the absolute pressure in the smaller pipe = 19.63 psi