Question

Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 100 m/s. The turbine operates at steady state and develops a power output of 3200 kW. Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Calculate the mass flow rate of air, in kg/s, and the exit area, in m2 .

Answer 1

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Step-by-step explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W"
`_{cv` / [ (h₁ - h₂ ) -
`(V_2^2)/(2)` ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW ×
`(1(k^3)/(s) )/(1kW)` ] / [ (932.93 - 503.02 )k³/kg -
`(100^2(m^2)/(s^2) )/(2)`|
`(ln)/(kg(m)/(s^2) )`||
`(1kJ)/(10^3N-m)`| ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

so

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[
`((8.314)/(28.97)(k^3)/(kg.K))`×
`500 K`×
`(7.54 kg/s)` ] / [(1 bar)(100 m/s )]} |
`(1 bar)/(10N/m^2)`||
`10^3N.m/1k^3`

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²