A 250 kilo ohms and a 750 kilo ohms resistor are connected in series across a 75-volt source. Determine the error in measuring t
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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
The exit temperature is 586.18K and compressor input power is 14973.53kW
Data;
- Mass = 50kg/s
- T = 288.2K
- P1 = 1atm
- P2 = 12 atm
The exit temperature of the gas can be calculated isentropically as
Let's substitute the values into the formula
The exit temperature is 586.18K
<h3>The Compressor input power</h3>The compressor input power is calculated as
The compressor input power is 14973.53kW
Learn more on exit temperature and compressor input power here;
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ