Calculating Mass Flow Rate of Nitrogen in a Circular Pipe at 100 psia and 300 F

[xzibition8612]

Homework Statement

Nitrogen at a pressure of 100 psia and a temperature of 300 F flows through a circular pipe with a diameter of 6 inches. The velocity profile is parabolic with V= [1-(r/3)^2]
where r is the local radius in inches and V is the velocity in ft/s. Determine the mass flow rate of nitrogen through the pipe.

Homework Equations

m = pVA

The Attempt at a Solution

I suppose this would be a steady flow. Then plug in r (3 inches) into V, get V=0. And mass flow rate is 0. Which makes sense because it's a steady flow. But this answer seems to be too trivial. So I must've screwed up somewhere. Any help would be appreciated. Thanks!

 

[KataKoniK]

Hello! Thank you for posting this problem. Let's take a closer look at the given information and equations to determine the mass flow rate of nitrogen through the pipe.

First, let's define the variables given in the problem:

P = 100 psia (pressure)
T = 300 F (temperature)
d = 6 inches (diameter of pipe)
r = local radius in inches
V = velocity in ft/s

The equation given, V = [1-(r/3)^2], is the velocity profile of the nitrogen flow. This means that the velocity of the nitrogen is not constant throughout the pipe, but rather varies with the radius.

To determine the mass flow rate, we can use the equation m = pVA, where m is the mass flow rate, p is the density of the nitrogen, V is the velocity, and A is the cross-sectional area of the pipe.

To find the density of nitrogen, we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Rearranging this equation to solve for density, we get p = (nRT)/V.

Since we are dealing with a steady flow, the number of moles (n) remains constant. Therefore, we can rewrite the equation as p = (mRT)/V, where m is the mass of the nitrogen.

Now, let's plug in the given values and solve for the density:

p = (mRT)/V
= (m*R*300)/(100*144)
= (m*R*3)/144

Next, we need to find the cross-sectional area of the pipe. The area of a circle is given by A = πr^2. Since the diameter of the pipe is given as 6 inches, the radius is half of that, or 3 inches. Therefore, the area of the pipe is A = π*(3)^2 = 9π square inches.

Now, we can plug all of our values into the equation m = pVA to solve for the mass flow rate:

m = pVA
= (m*R*3)/144 * 9π * [1-(r/3)^2]
= (m*R*3)/144 * 9π * [1-(3/3)^2]
= (m*R*3