VFR= AxV ; VFR= 3.11ft^2(1.4578ft/sec)=4.51cfmAir Flow Rate: 4.51cfm

[Windseaker]

Homework Statement

(Static)
Measure air flow in a duct 28"x16" w/ a Pitot tube is .4 inches w.c.
What is air flow rate?

Homework Equations

P=dxh , VFR= AxV , P=(Hs1-Hs2)+(V2^2-V1^2/2g) + (He2-He1) + Hf or Hv=(V/4000)^2

The Attempt at a Solution

28X16= 448in^2(1/144in^2)=3.11 ft^2
.4"w.c.(1/12")=.0333ft of w.c.
P=dxh : p= 62.4lb/ft^3(.0333ft w.c.)= 2.08 lb/ft^2 ( 1ft^2/144in^2)=.0144psi
MMMMM- lost now!
How about
Hv=(V/4000)^2 ,,,no

P= 0+(V2^2-V1^2/2g) + 0 + 0
(V2^2-V1^2)=P(2g)
(V2-V1)=(P(2g))1/2
(V2-V1)=(.033ftw.c.(64.4))1/2
(V2-V1)=1.4578ft/sec

 

[mark726]

VFR=AxV ,,, no

VFR=(3.11ft^2)(1.4578ft/sec)=4.53ft^3/sec I would first clarify the units and variables used in the problem. It seems that the given pressure is in inches of water column (w.c.) and the dimensions of the duct are in inches. However, the equations used in the solution attempt are using feet and pounds, which can lead to errors in the calculation. Therefore, I would convert all units to a consistent system, either all in inches or all in feet.

Next, I would review the equations used in the solution attempt and make sure they are applicable to the problem. For example, the equation P=dxh is typically used to calculate the pressure at a certain depth in a fluid, but it may not be relevant to this problem. Similarly, the equation Hv=(V/4000)^2 may not be applicable here as it is used to calculate the head loss in a pipe.

To calculate the air flow rate, I would use the equation VFR=AxV, where A is the cross-sectional area of the duct and V is the air velocity. The given dimensions of the duct can be used to calculate the cross-sectional area, which is 3.11 ft^2. To find the air velocity, I would use the equation (V2-V1)=(P(2g))1/2, where V2 and V1 are the air velocities at two different points in the duct and P is the pressure difference between those points. In this case, V1 can be assumed to be 0 as the air is entering the duct and V2 can be calculated using the given pressure of 0.4 inches w.c.

Therefore, the final calculation would be VFR=(3.11 ft^2)(1.4578 ft/sec)=4.53 ft^3/sec.