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

Answer:

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

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W"$_{cv$ / [ (h₁ - h₂ ) -$\frac{V_2^2}{2}$ ]

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 × $\frac{1\frac{k^3}{s} }{1kW}$ ] / [ (932.93 - 503.02  )k³/kg  -$\frac{100^2\frac{m^2}{s^2} }{2}$|$\frac{ln}{kg\frac{m}{s^2} }$||$\frac{1kJ}{10^3N-m}$| ]

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

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²