Question

Problem 1 - Gas Turbine Engine. Assume the gas turbine engine operates on basic non-ideal Brayton cycle with the following specifications Inlet Condition: Dry air at 537 R and 14.7 psia Pressure Ratio: 24.2 Actual Exhaust Temperature: 950 F Compressor and Turbine Efficiency: 85% Actual Power Output 34 MW Using given/assumed values, determine: a) firing temperature, b) cycle efficiency, c) air mass flow rate. Problem 2- HRSG. Assume the HRSG is used to produce saturated steam at 600 psia using combination of heat from the gas-turbine engine exhaust, and supplemental natural gas firing. Additional natural gas firing is used to superheat the steam to 750 F. Liquid water supply: 75 F and 14.7 psia. Liquid water pump: Ideal compression to 600 psia. Turbine Exhaust Inlet Temperature: 950 F. Turbine Exhaust Exit Temperature: Assume equal to saturation temperature of steam. HRSG Steam Output Rate: 210,000 lbm/hour Based on given/assumed values, determine: a) heat transfer from gas turbine exhaust (Btu/sec), b) fraction of steam mass flow produced by the gas turbine exhaust (%), c) amount of supplemental firing required to provide total steam mass flow (Btu/sec), d) addition firing heat required to superheat steam (Btu/sec). Problem 3-Steam Turbine. A portion of the superheated steam produced by the HRSG + superheater is used to drive a steam turbine. The steam turbine has a two-stage expansion process first to 60 psia, then to 20 psia. We will assume the following data: Inlet Steam Condition: 600 psia and 750 F Fraction of steam diverted out of turbine after first stage expansion: 5% Turbine efficiency (both stages): 85% Turbine Power Output: 11 MW Based on given/assumed values, determine: a) outlet state after first expansion (T,P and/or quality), b) outlet state of second expansion (T,P and/or quality), c) fraction of HRSG steam needed for the power turbine.

Answer 1

Given Information :

• Gas turbine engine working on basic non ideal brayton cycle
• Inlet condition of air :
  T1 298.33 K
    
  P14.7 psi
  
• Pressure ratio :
  P3 24.2 PA P2 p Pi
  
• Actual exhaust temperature :
  T783.15 K
  
• Compressor and turbine efficiency :
  0.85 isen
  
• Actual power output :
  34 MW
  

Assumptions :

• Air is taken as ideal gas with constant specific heats :
• 
• 
  C,=0.718 kJ/kg - K
  
• 
  R 0.287 kJ/kg - K
  
• 
  = 1.4
  

Solution :

Calculating all the temperatures at every stage of the non ideal brayton cycle :

Initially considering an ideal brayton cycle where the compression and the expansion processes are isentropic in nature :

Considering Isentropic Compression, the relation between the temperature and the pressure between the inlet and exit of the compressor can be written as follows :

$\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{\gamma-1}{\gamma}}$

Substituting the values in the above equation and rearranging :

$T_2=T_1(24.2)^{\frac{1.4-1}{1.4}}$

$T_2=298.33*2.4852$

Now considering the isentropic efficiency of the compressor and calculating the actual temperature of air at the exit of the compressor :

12 11 nisen T Ti

Where
$T_2'=$ Actual temperature of air after the compression considering the efficiency of the compressor

Rearranging and substituting the values :

12 11 T2 Ti Tisen

$T_2'=298.33+\frac{741.44-298.33}{0.85}$

Considering Isentropic Expansion, the relation between the temperature and the pressure between the inlet and exit of the turbine can be written as follows :

$\frac{T_3}{T_4}=(\frac{P_3}{P_4})^{\frac{\gamma-1}{\gamma}}$

Since we know that

$\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{\gamma-1}{\gamma}}$ and  $\frac{P_2}{P_1}=\frac{P_3}{P_4}$

Equating the above equations :

$\frac{T_3}{T_4}=\frac{T_2}{T_1}$

Substituting the values :

$\frac{T_4}{T_3}=\frac{298.33}{741.43}$

$\frac{T_4}{T_3}=0.4023$

Now considering the isentropic efficiency of the turbine and calculating the actual temperature of air at the exit of the turbine:

$\eta_{isen}=\frac{T_3-T_4'}{T_3-T_4}$
Where
$T_4'=$ Actual temperature of air after the expansion considering the efficiency of the turbine

$T_3=$ Maximum temperature that the air attains after combustion.

Substituting the values in the above equation :

$\eta_{isen}=\frac{1-\frac{T_4'}{T_3}}{1-\frac{T_4}{T_3}}$

$0.85=\frac{1-\frac{783.15}{T_3}}{1-0.4023}$

$1-\frac{783.15}{T_3}=0.5079$

783.15 1-0.5079 T3

783.15 T3 0.4920

Calculating the value of temperature if the expansion in the turbine was isentropic :

$\frac{T_4}{T_3}=0.4023$

Substitute the values in the above equation :

$\frac{T_4}{1591.718}=0.4023$

Tabulating all the temperatures at each point in the non ideal basic rankine cycle

Point	Temperature
1	298.33 K
2	741.44 K
2'	819.63 K
3	1591.718 K
4	640.34 K
4'	783.15 K

The firing temperature is equal to :

$\mathbf{T_3=T_{firing }=1591.718\ k}$

Calculating the air mass flow rate from the actual power output :

The actual power output can be defined as the difference between the actual turbine work output and the actual compressor work input.

Calculating the actual turbine work output :

$P_{turbine}=\dot H_3-\dot H_4'$

$P_{turbine}=\dot m_f *C_p*T_3-\dot m_f *C_p*T_4'$

Prurbine C (T3 - T;)

Calculating the actual compressor work input :

$P_{compressor}=\dot H_2'-\dot H_1$

Pcompressor = m* Cp T- m*Cp* T

$P_{compressor}=\dot m_f *C_p*(T_2'-T_1)$

Where
$\dot m_f=$ Air Mass Flow rate

Actual work output is given by :

Pturbine Peo net= compressor

Substituting the values :

$34*10^6=\dot m_f *C_p*(T_3-T_4')-\dot m_f *C_p*(T_2'-T_1)$

$34*10^6=\dot m_f *C_p*(T_3-T_4'-T_2'+T_1)$

$34*10^6=\dot m_f *1.005*10^3*(1591.718-783.15-819.63+298.33)$

$34*10^6=\dot m_f *1.005*10^3*287.268$

$34*10^6=\dot m_f *288704.34$

$\dot m_f= \frac{34*10^6 }{288704.34}$

$\dot m_f= 117.76\ kg/s$

$\therefore$ The air mass flow rate is given by  $\mathbf{\dot m_f= 117.76\ kg/s}$

Calculating the cycle efficiency :

Cycle efficiency is given by the formula :

Pnet TeHsupplied nc

where

Hsupplied
Heat supplied during the combustion
$P_{net}=$ Net Power output of the cycle

Heat supplied during the combustion can be found out from the following formula :

$H_{supplied}=\dot H_3-\dot H_2'$

$H_{supplied}=\dot m_f*C_p*T_3-\dot m_f*C_p*T_2'$

$H_{supplied}=\dot m_f*C_p*(T_3-T_2')$

Substituting the values :

$H_{supplied}=117.76*1.005*10^3(1591.718-819.63)$

H supplied = 91375688.29 W

Now calculating the cycle efficiency

Pnet TeHsupplied nc

$\eta_c=\frac{34}{91.38}$

$\eta_c=0.372$

$\eta_c=37.2$%

$\therefore$ The cycle efficiency of the basic non ideal brayton cycle is 37.2%