Question 2 of 8 1 current attempt in progress as shown in...

Question 2 of 8 < > -/1 Current Attempt in Progress As shown in the figure, air with a volumetric flow rate of 16,000 ft³/min enters an air-handling unit at 80°F, 1 atm. The air-handling unit delivers air at 45°F, 1 atm to a duct system with three branches consisting of two 26-in.-diameter ducts and one 50-in. duct. The velocity in each 26-in. duct is 13 ft/s. Assume steady state operation and ideal gas behavior for the air. 4D-50 in. T2 T3 T45°F D₂=D3=26 in. V2=V3 Duct system Air-handling unit 1+==P4 =1 atm = 80°F (AV)1 Determine the mass flow rate of air entering the air-handling unit, in lb/s, the volumetric flow rate in each 26-in. duct, in ft³/min, and the velocity in the 50-in. duct, in ft/s.

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that air enters the dock of an air conditioning system at 15 psa and 50 fahrenheit degrees at a volume flow that is given that of 450 cubic feet per minute.

00:18 Now, the diameter of the dock is 10 inches and the heat is transferred to the earth in the dot from the surroundings at a rate of 2 btu.

00:30 Per second.

00:31 So for part a of this problem, we are asked to determine the velocity of the earth at the dot.

00:37 So to calculate that, we know that we can use the inlet velocity of the earth through the dook, and that is determined, let's call that big one, that is the volume flow one.

00:54 This divided by the area.

00:56 So that volume flow that we are given, that is 450, the cubic feet per minute and this divided by the area that that will be pi times 5 divided by 12 feet this to the square and then from this we obtain a value of so the value that we obtained from this is equal to 825 feet per minute so the solution for par a of this problem.

01:33 Now for par b, we are asked about the temperature of the earth at the exit.

01:46 So in this case, first we calculate the volume 1.

01:53 That volume 1 is equal to the gas constant times the temperature 1 that we are given for this case, this divided by the pressure.

02:03 1.

02:04 So we substitute the values in here.

02:07 So the constant, the gas constant, it has about 0 .3704 psa times cubic feet per pounds and this the temperature that we are given, 510 in units of art.

02:33 This divided by the pressure that is 15 ,000 that is 15 ,000, psa.

02:38 So from this we obtain a value of 12 .6 cubic feet per minute.

02:45 Oh sorry, per pounds...

Question

Please give Ace some feedback