Answer:
P2 = 3.9 MPa
Explanation:
Given that
T₁ = 290 K
P₁ = 95 KPa
Power P = 5.5 KW
mass flow rate = 0.01 kg/s
solution
with the help of table A5
here air specific heat and adiabatic exponent is
Cp = 1.004 kJ/kg K
and k = 1.4
so
work rate will be
W = m × Cp × (T2 - T1) ..........................1
here T2 = W ÷ ( m × Cp) + T1
so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290
T2 = 838 k
so final pressure will be here
P2 = P1 × [tex](\frac{T2}{T1})^\frac{k}{k-1}[/tex] ..............2
P2 = 95 × [tex](\frac{838}{290})^\frac{1.4}{1.4-1}[/tex]
P2 = 3.9 MPa