Air is heated at steady state as it flows through a constant area duct. At the inlet, the pressure is 2 bar, the temperature is 300 K and the velocity is 87 m/s. At the exit, the pressure is 1.45 bar and the velocity is 360 m/s. Neglecting potential energy effects and assuming ideal gas behavior for the air, determine the heat transfer per unit mass of air (qcv) flowing through the duct, in kJ/kg.

Heat transfer  q = 664015.5 J/kg (Approx)

Explanation:

Given:

Area = a

Velocity flow rate Q1 = 87a m³/s

Velocity flow rate Q2 = 360a m³/s

Find:

Heat transfer = q

Computation:

[P1][v1]/T1 = [P2][v2]/T2

[2][(87a]/300 = [1.45][360a] / T2

T2 = 900 K

Heat transfer  q = Cp(T2 - T1) + (V2² /2 - V1² /2 )

Heat transfer  q = 1,005(900 - 300) + (360² /2 - 87² /2)

Heat transfer  q = 664015.5 J/kg (Approx)

the inner core.

good luck: )

klicker