Question

Required information Problem 05.038. Diffuser flow DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air at 80 kPa, 27'C, and 220 m/s enters a diffuser at a rate of 2.5 kg/s and leaves at 42'C. The exit area of the diffuser is 360 cm2. The air is estimated to lose heat at a rate of 18 kJ/s during this process. The gas constant of air is 0.287 kPam®/kg K. The enthalples are 7 = 300.19 kJ/kg and in = 315.27 kJ/kg. Problem 05.038.b - Diffuser flow Determine the exit pressure of the air. The exit pressure of the air is kPa.

Answer 1

Sol 05.038.b) The pressure at the exit of the diffuser can be calculated from the mass flow rate equation as shown below:

.....................(1)

m = pz V2 P2 A202 RI2 MRT2 P2 A202

So, The exit pressure depends upon the mass flow rate m = 2.5 kg/s, gas constant R = 0.287 kPam3/kgK , Exit Temperature of Air T2= 420C = 315 K, Area of Cross section at the exit A2 = 360 cm2= 360 x 10-4 m2 and the velocity of air at the Exit v2. Since we know all the required parameters excluding the exit velocity v2.

To find the exit velocity v2 we need apply Conservation of Energy in a fluid flow:

Ein = Ėout (* = v) = n(h2 – hı) +

2.5 x (ví - vž) V=2.5 (315.27 – 300.19) 1000 + 18 x 1000

....................[in Joules]

2.5 x (4-), 5700

ບ: － ບ3 = 44560

= - 44560

เ: = 220 - 44560

$v_{2} = 61.97 \approx 62m/s$.....................(2)

Putting value of v2 from equation (2) in the equation (1)

$P_{2} = \frac{2.5\times .287\times 315}{360 \times 10^{-4}\times 62}$

Sol 05.017 . c) According to the Given data:

Diameter of the pipe d = 28cm = .28m

Inlet pressure P1 = 200 kPa

Inlet Temperature T1 = 200C = 293 K

Inlet Velocity v1 = 4.6 m/s

Specific Volume at inlet V1 = 0.1142 m3/kg

Outlet pressure P2 = 180 kPa

Outlet Temperature T2 = 400C = 313 K

Outlet Velocity = v2 = ?

Specific Volume at Outlet V2 = 0.1374 m3/kg

i) The Volume flow rate at the exit Vf :

The mass flow rate at the inlet is equal to the mass flow rate at the outlet by the law of conservation of mass i.e. m.

Let the Volume flow rate at the inlet be Vi

Volume flow rate at the inlet = Area of Cross section at inlet x Velocity at the inlet

$V_{i}= \tfrac{\Pi }{4}d^{2} \times v_{1}$

$V_{i}= 0.0616 \times 4.6$

Mass flow rate m

V 0.283 m = 0.1142

$m=2.48 \frac{kg}{s}$

And Mass flow rate at exit will be same as that of inlet

$m=\frac{V_{f}}{V_{2}}$

$V_{f}=m\times V_{2}$

$V_{f}= 2.48\times 0.1374$

Vi = 0.3408"

Ans: The Volume flow rate at the exit Vf is 0.3408 m3/s.

ii) The Velocity at the exit v2 :

As we know that volume flow rate at the exit = Area of cross section x outlet velocity

$V_{f} = \frac{\pi }{4}d^{2}\times v_{2}$

$0.3408 = \frac{\pi }{4}\times 0.28^{2}\times v_{2}$

On solving for v2, we get

v2 = 5.53 m/s

Ans: The Velocity at the exit v2 is 5.53 m/s.