Air is drawn in from the outside through a blower heated and

Air is drawn in from the outside through a blower, heated, and then flows steadily through a turbine to generate electricity as shown. If the turbine generates 100 kW of electric power, calculate the outlet velocity, u_2, and temperature, T_2, of the air. Assume that the air can be treated as a perfect and calorically-perfect gas and that the flow properties are uniform across the inlet and outlet cross-sections. Away from the inlet, the ambient outdoor conditions are T_0 = 10 degree C and p_0 = 101.3 kPa. The outlet flow is at pressure, p_2 = 0.9p_0. The blower work and heat flux are 8 kW and 18 kW, respectively, and the mass flow rate through the system is 15 kg/s. The inlet cross-sectional area is A_1 = 3.0 m^2, while the outlet cross-section is A_2 = 0.4 m^2. The total drop in height from inlet to outlet is z_1 - z_2 = 6 m. Ignore all head losses.

Solution

solution:

1)here steady flow energy equation are for turbine

energy content at inlet=energy content at outlet

Q\'+m\'(h1+c1^2/2+9.81z1)=Wsf\'+m\'(h2+c2^2/2+g*z2)

here Q\'=18 kw+0=18 kw

Wsf\'=8+100=108 kw

here density=Po/RoTo=1.01325*10^5/287*283=1.2475 kg/m3

m=density*Q

Q=15/1.2475=12.0238 m3/s

hence c1=Q/A1=4.0074 m/s

c2=Q/A2=30.05958 m/s

on putting value we get that

10-108=15[(h2-h1)+443.761+58.86]

we get that

h1-h2=508.621

h=m\'cpT1

so we get that

T1=283 K

h1-h2=m\'Cp(T1-T2)=508.621

T2=249.22 K

hence velocity oultet are

c2=30.059m/s and t2=249.22 k