Question

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 370 cm^2. 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 kPa·m^3/kg·K. The enthalpies are h1 = 300.19 kJ/kg and h2 = 315.27 kJ/kg. Determine (a) the exit velocity and (b) the exit pressure of the air.

Answer 1

exit velocity = 62 m/s

exit pressure is 98.52332 kPa

Step-by-step explanation:

given data

pressure p1 = 80 kPa

temperature t1 = 27°C = 300 K

velocity v1 = 220 m/s

temperature t2 = 42°C = 315K

area exit = 370 cm²

lose heat at a rate = 18 kJ/s

gas constant of air = 0.287 kPa·m^3/kg·K.

h1 = 300.19 kJ/kg

h2 = 315.27 kJ/kg

to find out

exit velocity and exit pressure

solution

first we apply here energy balance equation to find out outlet velocity

Einlet = Eoutlet

m(h1 + v1²/2) = m (h2+v2²/2) +Q

m(v1²-v2²) /2 = m (h2 -h1) +Q

v2² = v1² - 2Cp Δt - 2Q/m

here Δt = h2 - h1 and put all value

v2² = 220² - 2(1.005)10³ (15) - 2(18)10³ /2.5

exit velocity = 62 m/s

so

now find outlet pressure that is

p2 = mRT / A2 v2

put value

p2 = 2.5 (287) 315 / (370
`10^(-4)` 62)

p2 = 98.52332 kPa

exit pressure is 98.52332 kPa