Question

An 8" ID pipe carries air at 80.0 ft/sec, 21.5 psi absolute and 80°F. How many pounds of air are flowing? The 8" pipe reduces to a 4" ID pipe and the pressure and temperature in the 4" pipe are 19.0 psi absolute and 48°F respectively. Find the velocity in the 4" pipe and compare the flows in ft3/sec in the two pipes.

Answer 1

ID = 8 inch = 2/3 feet

Area A = pi x D²/4 = 3.14 x (2/3)² / 4 = 0.34889 ft²

Volumetric flow rate Q = Area x velocity = 0.34889 ft² x 80 ft/s = 27.911 ft³/s

We need density to get Mass flow rate. By Ideal Gas Law, we have

density = PM/RT

P = 21.5 psia

M = 28.96 lb/lb-mol

R = 10.73 (ft³)(psia)/lb-mol)(°R)

T = 80 F = 80 + 460 = 540 R

Substituting, we get

density d = 21.5 x 28.96 /( 10.73 x 540) = 0.10746 lb/ft³

Thus, mass flow rate = Volumetric flow rate (Q) x density (d)

                                = 27.911 ft³/s x 0.10746 lb/ft³

                                = 3 lb/s

3 pounds of air flow each second through the pipe.

New density with

P = 19 psia

T = 48 F = 48 + 460 = 508 R

density d₁ = 19 x 28.96 / ( 10.73 x 508) = 0.100946 lb/ft³

diameter = 4 inch = 1/3 feet

New area A₂ = pi x (1/3)²/4 = 0.0872225 ft²

We have,

Mass flow rate = constant = Area x velocity x density

We calculated mass flow rate in the previous part as 3 lb/s.

Thus, 3 lb/s = 0.0872225 ft² x velocity x 0.100946 lb/ft³

velocity = 340.715 ft/s - Velocity in the smaller pipe.

Since mass flow rate is same,

Volumetric flow rate in pipe 1 / Volumetric flow rate in pipe 2 = density₂/density₁ = 0.100946/0.10746 = 0.9394

Flow rate in ft³/s in pipe 2 (smaller) is 0.9394 time that in pipe 1 ( larger).

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