Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure
ratio of the cycle is 9. The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K. Accounting for the variation of specific heats with temperature, determine the minimum mass flow rate of air needed to develop a net power output of 105 MW.
1 answer:
Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
Explanation:
from the T-S diagram, we get the overall pressure ratio of the cycle is 9
Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3
P5/P6 = P7/P8 = √9 =3
get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.
At temperature T1 =300K
Specific enthalpy of air h1 = 300.19 kJ/kg
Relative pressure pr1 = 1.3860
At temperature T5 = 1200 K
Specific enthalpy h5 = 1277.79 kJ/kg
Relative pressure pr5 = 238
Calculate the relative pressure at state 2
Pr2 = (P2/P1) Pr5
Pr2 =3 x 1.3860 = 4.158
get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure pr = 4.153
The corresponding specific enthalpy h = 411.12 kJ/kg
Relative pressure pr = 4.522
The corresponding specific enthalpy h = 421.26 kJ/kg
Find the specific enthalpy of state 2 by the method of interpolation
(h2 - 411.12) / ( 421.26 - 411.12) =
(4.158 - 4.153) / (4.522 - 4.153 )
h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))
h2 - 411.12 = 0.137
h2 = 411.257kJ/kg
Calculate the relative pressure at state 6.
Pr6 = (P6/P5) Pr5
Pr6 = 1/3 x 238 = 79.33
Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure Pr = 75.29
The corresponding specific enthalpy h = 932.93 kJ/kg
Relative pressure pr = 82.05
The corresponding specific enthalpy h = 955.38 kJ/kg
Find the specific enthalpy of state 6 by the method of interpolation.
(h6 - 932.93) / ( 955.38 - 932.93) =
(79.33 - 75.29) / ( 82.05 - 75.29 )
(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )
h6 - 932.93 = 13.427
h6 = 946.357 kJ/kg
Calculate the total work input of the first and second stage compressors
(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )
= 222.134 kJ/kg
Calculate the total work output of the first and second stage turbines.
(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )
= 662.866 kJ/kg
Calculate the net work done
Wnet = (Wturb)out - (Wcomp)in
= 662.866 - 222.134
= 440.732 kJ/kg
Calculate the minimum mass flow rate of air required to generate a power output of 105 MW
W = m × Wnet
(105 x 10³) kW = m(440.732 kJ/kg)
m = (105 x 10³) / 440.732
m = 238.2 kg/s
therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
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If the heat engine operates for one hour:
a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.
b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.
In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.
Explanation:
The Carnot efficiency is obtained as:
Where is the atmospheric temperature and
is the maximum burn temperature.
For the case (B), the efficiency we will use is:
The work done by the engine can be calculated as:
where Hv is the heat value.
If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.
If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
