Question

Air at a temperature of 15 °C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 800°C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 650oC. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/s. calculate a) the rate of heat transfer in the heat exchanger. b) the power output from the turbine assuming no heat loss. c) the velocity at the exit from nozzle, assuming no heat loss. Take enthalpy of air as h-CpT, where Cp is the specific heat equal to 1.005 k]/kg.K and T is the temperature.

Answer 1

Explanation:

The given problem can be easily solved by Steady flow Energy Equation.

Answer 2

Answer:

Rate of heat transfer= 1577.85 kJ/s.

Power output of turbine = 298.8 kW.

velocity at exit from the nozzle = 152.55 m/s.

Explanation:

Temperature of air,     $t_{1} = 15°C$

Velocity of air, $C_{1}$= 30 m/s.

Temperature of air after passing the heat exchanger, $t_{2} = 800 °C$

Velocity of air at entry to the turbine,     $C_{2}= 30 m/s$

Temperature of air after leaving the turbine,       $t_{3} =$650°C

Velocity of air at entry to nozzle,    $C_{3} =$60 m/s

Temperature of air after expansion through the nozzle,  $t_{4} =$ 510°C

Air flow rate,     $m$

= 2 kg/s.

(i) Heat exchanger :

Rate of heat transfer :

Energy equation is given as

$m\left ( h_{1} +\frac{C_{1}^{2}}{2}+Z_{1}g\right )+Q_{1-2}=m\left ( h_{2}+\frac{C_{2}^{2}}{2}+Z_{2} g\right )+W_{1-2}$

Here,       $Z_{1} = Z_{2}, C_{1}, C_{2} = 0, W_{1-2} = 0$

$\therefore mh_{1} + Q_{1-2} = mh_{2}$

or        $Q_{1-2} = m(h_{2} - h_{1})$

=$mc_{p}(t_{2} - t_{1}) = 2 * 1.005 (800 - 15)$ = 1577.85

Hence, rate of heat transfer = 1577.85 kJ/s.

(ii) Turbine :

Power output of turbine :

Energy equation for turbine gives

$m\left ( h_{2}+\frac{C_{2}^{2}}{2} \right )=m\left ( h_{3}+\frac{C_{3}^{2}}{2} \right )+w_{2-3}$

              $[\because$ $Q_{2-3} = 0, Z_{1} = Z_{2}$]

∴ $W_{2-3}=m\left ( h_{2}+\frac{C_{2}^{2}}{2} \right )-m\left ( h_{3} +\frac{C_{3}^{2}}{2}\right )W_{2-3}$

$=m\left [ \left ( h_{2}-h_{3} \right )+\left ( \frac{C_{2}^{2}-C_{3}^{2}}{2} \right ) \right ]$

$=m\left [c_{p} \left ( t_{2}-t_{3} \right )+\left ( \frac{C_{2}^{2}-C_{3}^{2}}{2} \right ) \right ]$

$2\left [ 1.005\left ( 800-650 \right )+\frac{\left (30 \right )^{2}-\left ( 60 \right )^{2}}{2\times 1000} \right ]$

$2[1.005(150)-1.35]$

$= 2 [ 150.75-1.35] = 298.8 kJ/s$  or$298.8 kW$

Hence, power output of turbine = 298.8 kW.

(iii) Nozzle :

Velocity at exit from the nozzle :

Energy equation for nozzle gives,

$h_{3}+\frac{C_{3}^{2}}{2}=h_{4}+\frac{C_{4}^{2}}{2}​$

$[\because W_{3-4} = 0, Q_{3-4}= 0, Z_{1} = Z_{2}]$

$\frac{C_{4}^{2}}{2}=\left ( h_{3} -h_{4}\right )+\frac{C_{3}^{2}}{2}=c_{p}\left ( t_{3}-t_{4} \right )+\frac{C_{3}^{2}}{2}$

$= 1.005(650 - 500) + \frac{60^{2}}{2\times 1000}$

$= 1.005(150)+1.8$

$=150.75+1.8=152.55$

∴ $C_{4}= 152.55$

Hence, velocity at exit from the nozzle = 152.55 m/s.

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