A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kj/kg, repectively, and exit specific enthalpy of 297.31 kj/kg. For negligible heat transfer with the surroundings, the exit velocity is

a) 223 m/s

b) 197 m/x

c) 90 m/s

d) 70 m/s

Answer: c) 90 m/s

Explanation:

Given

Invest velocity, v1 = 250 m/s

Inlet specific enthalpy, h1 = 270.11 kJ/kg = 270110 J/kg

Outlet specific enthalpy, h2 = 297.31 kJ/kg = 297310 J/kg

Outlet velocity, v2 = ?

0 = Q (cv) - W (cv) + m[ (h1 - h2) + 1/2 (v1² - v2²) + g (z1 - z2) ]

0 = Q (cv) + m[ (h1 - h2) + 1/2 (v1² - v2²) ]

0 = [ (h1 - h2) + 1/2 (v1² - v2²) ]

Substituting the values of the above, we get

0 = [ (270110 - 297310) + 1/2 (250² - v²)

0 = [-27200 + 1/2 (62500 - v²) ]

27200 = 1/2 (62500 - v²)

54400 = 62500 - v²

v² = 62500 - 54400

v² = 8100

v = √8100

v = 90 m/s