Question

Air at a temperature of 150C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 8000C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 6500C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle where it expands until the temperature falls to 5000C. If the air flow rate is 2 kg/s, calculate the following.

Answer 1

Please find the attached file for figure

Explanation:

• Fine the Rate of flow of heat exchanger to air

• The power output from turbine assuming no heat loss and the velocity at exit from nozzle where Enthalphy h CpT i.e Cp specific heat = 1.005KJ/KgK

The rate of flow of heat transfer to air exchanger i.e

$w(h1 + \frac{V^{2} 1}{2} + Z1g) + Q_{1-2} = w( h2 + \frac{V2^{2} }{2} + Z1g) + W_{1-2}$

$Wh_{1} + Q_{1-2} = Wh_{2}$

$Q_{1-2} = w(h2 - h1)$

$Q_{1-2} = wc_{p} (t2- t1)$

$Q_{1-2} = 2x1.oo5(8000 - 150)$

$Q_{1-2} = 1577.8 kg j /s$

Given t₁ = 150°C, t₂ = 8000°c

V₁ = 30m/s, V₂ = 30m/s

t₃ = 6500°C , V₃ = 60m/s

Energy equation for turbine given by

$W( \frac{V_{2})^{2}}{2}+ h_{2}) = Wh_{3} + w\frac{(V3)^{2} }{2} + Wt$

$\frac{(V2)^{2}-(V3)^{2} }{2} = (h2 -h3) = Wt/w$

$\frac{(30^{2}-60^{2})x10^{-3} }{2} + ( 1.005)(8000 - 6500) =Wt/w$

Wt = 1.506.15xw

Wt = 1506.15x2 = 3012.3kW