Question

. Air at a temperature of 20°C passes through a heat exchanger at a

velocity of 40 m/s where its temperature is raised to 820°C. It then enters a turbine with same

velocity of 40 m/s and expands till the temperature falls to 620°C. On leaving the turbine, the air

is taken at a velocity of 55 m/s to a nozzle where it expands until the temperature has fallen to

510°C. If the air flow rate is 2.5 kg/s, calculate :

(i) Rate of heat transfer to the air in the heat exchanger ;

(ii) The power output from the turbine assuming no heat loss ;

(iii) The velocity at exit from the nozzle, assuming no heat loss.

Take the enthalpy of air as h = cpt, where cp is the specific heat equal to 1.005 kJ/kg°C and

t th

Answer 1

Solution;

Given;

Temperature of air,$t_1=20°c$

Velocity of air, $C_1=40m/s$

Temperature of air after passing the heat exchanger,$t_2=820°c$

Velocity of air at entry to the turbine,$C_2=40m/s$

Temperature of air leaving the turbine,$t_3=620°c$

Velocity of air at entry to nozzle,$C_3=55m/s$

Temperature of air after expansion through the nozzle,$t_4=510°c$

Air flow rate, $\dot{m}=2.5kg/s$

(i) Heat exchanger;

To find the rate of heat transfer;

From Energy equation given by;

$m(h_1+\frac{C_1^2}{2}+z_1g)+Q_{1-2}=m(h_2+\frac{C_2^2}{2}+z_2g)+W_{1-2}$

In which ;

$z_1=z_2$

$C_1,C_2$ $=0$

$W_{1-2}=0$

Therefore ,the equation reduces to;

$mh_1+Q_{1-2}=mh_2$

$Q_{1-2}=m(h_2-h_1)$

$Q_{1-2}=mc_p(t_2-t_1)$

$Q_{1-2}=2.5×1.005×(820-20)$

$Q_{1-2}=2010kJ/s$

Answer;

2010kJ/s

(ii) Turbine;

Power output from Turbine;

Energy equation for turbine will be ;

$m(h_2+\frac{C_2^2}{2})=m(h_3+\frac{C_3^2}{2})+W_{2-3}$

Since;

$Q_{2-3}=0$

$z_1=z_2=z_3$

$W_{2-3}=m[(h_2-h_3)+(\frac{C_2^2-C_3^2}{2})]$

$W_{2-3}=m[c_p(t_2-t_3)+\frac{C_2^2-C_3^2}{2}]$

$W_{2-3}=2.5[1.005(820-620)+(\frac{40^2-55^2}{2×1000})]$

$W_{2-3}=2.5[201+0.7125]$

$W_{2-3}=504.3kJ/s$

Answer;

504.3kW

(iii) Nozzle

Velocity at exit from the nozzle;

Energy Equation for the nozzle will be ;

$h_3+\frac{C_3^2}{2}=h_4+\frac{C_4^2}{2}$

Since;

$z_1=z_2$

$W_{3-4}=0$

$Q_{2-3}=0$

$\frac{C_4^2}{2}=c_p(t_3-t_4)+\frac{C_3^2}{2}$

$\frac{C_4^2}{2}=1.005(620-510)+\frac{55^2}{2×1000}$

$C_4=473m/s$

Answer;

473m/s