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

Answer:

the absolute pressure in the smaller pipe = 19.63 psi

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 $D_1 = (\dfrac{3}{12})ft = 0.25 \ ft$

Diameter of pipe B  $D_1 = (\dfrac{1.5}{12})ft = 0.125 \ ft$

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

$\rho = \dfrac{P}{RT}$

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

$\rho = 7.64*10^{-3}\ slug /ft^3$

Formula for calculating cross sectional area is

$A = \dfrac{\pi}{4}D$

For diameter of pipe $D_1 = 0.025$

A₁ = $\dfrac{\pi}{4}*0.25^2$

A₁ = 0.04909 ft²

For diameter of pipe $D_2 - 0.0125$

A₂ $=\dfrac{\pi}{4}*0.125^2$

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:

$p_1 + \dfrac{\rho V_1^2}{2}+\gamma Z_1= p_2 + \dfrac{\rho V_2^2}{2}+\gamma Z_2$

Since the pipe is horizontal then;

$\gamma Z_1= \gamma Z_2$

So;

$p_1 + \dfrac{\rho V_1^2}{2}= p_2 + \dfrac{\rho V_2^2}{2}$

$p_2 =p_1 +\dfrac{1}{2} \rho(V_1^2-V_2^2)$

$p_2 =4996.8+\dfrac{1}{2} *7.624*10^{-3}(30.557^2-122.23^2)$

$p_2 =4943.41 \ lb/ft^2$

To psi;

$p_2 =\dfrac{4943.41 }{144}psi$

$p_2 =34.33 \ psi$ 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