Answer to Question #197914 in Molecular Physics | Thermodynamics for asim

Question

The power developed by a gas turbine is 15000 kW. Gases flow through the turbine of the gas turbine unit at 20 kg/s. The specific enthalpies of the gases at the inlet and the outlet of the turbine are 1300 kJ/kg and 400 kJ/kg respectively. The velocities of the gases at the inlet and the outlet are 65 m/s and 155 m/s respectively.

(a) Draw a schematic diagram of the turbine and label the flows in and out of it. (2 marks)

(b) What is the purpose of a gas turbine? How does irreversibility affect the performance of a gas turbine? (4 marks)

(c) Explain the equation which can be used to assess changes in energy in the turbine. (3 marks)

(d) Calculate the rate at which heat is rejected from the turbine. (11 marks)

(e) Calculate the area of the turbine inlet pipe if the specific volume of the gases at inlet is 0.6 m3 /kg. (5 marks)

Answer 1

Answer:-

m=20 kg/s

V=0.6m³/kg

Power developed by the turbine,   P = 15000 kW

∴ Work done, $W=\frac{15000}{20}=750\frac{kJ}{kg}$  

Enthalpy of gases at the inlet,   h₁ = 1300 KJ/kg

Enthalpy of gases at the oulet, h₂=400 KJ/kg

Velocity of gases at the inlet, C₁=  65 m/s

Velocity of gases at the outlet, C₂=155 m/s

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(a)

(b)Irreversibility is the Heat losses to the surroundings and the frictional losses as the fluid moves through the turbine and theMechanical losses due to friction in the turbine mechanism.

(c)Steady-Flow Energy Equation

$Q-W=m(h_2-h_1)+\frac{m}{2000}(C_2^2-C_1^2)+ \frac{mg}{1000}(Z_2-Z_1)$

Q = heat transfer, kW

w = work transfer, kW

 m= mass flow rate, kg/s

h = specific enthalpy, kJ/kg

C = velocity, m/s

g = gravitational acceleration, m/s²

z = elevation,

subscripts 1 and 2 = inlet and outlet conditions

(d)

 Heat rejected,  Q :

 

Using the flow equation,

$h_1+\frac{C_1^2}{2}+Q=h_2+\frac{C_2^2}{2}+W$

putting values

$1300+\frac{(65m/s)^2}{2}+Q=400+\frac{(155)^2}{2}+750\frac{kJ}{kg}$

1300 KJ/kg +2112.5 Nm/kg + Q=400 KJ/kg + 12012.4 Nm/kg + 750 KJ/kg

$\implies$

1300+2.1125+ Q = 400 +12.0124+750

Q= -140.1 KJ/kg

i.e.  Heat rejected =  140.1 KJ/kg $\times 20 kg/s$ = 2802 Kw

(e)

 Inlet area, A : 

Using the relation,

 $m=\frac{CA}{V}$

$\because$ $A=\frac{mV}{C}={20\times 0.6\over 65}=0.18 m^2$