The power supply converts the wall outlet AC power into DC power. T or F
1 answer:
You might be interested in
Answer:
The answers to the question are
(1) Process 1 to 2
W = 295.16 kJ/kg
Q = -73.79 kJ/kg
(2) Process 2 to 3
W = 0
Q = 1135.376 kJ/kg
(3) Process 3 to 4
W = -1049.835 kJ/kg
Q = 262.459 kJ/kg
(4) Process 4 to 3
W=0
Q = -569.09 kJ/kg
(b) The thermal efficiency = 49.9 %
(c) The mean effective pressure is 9.44 bar
Explanation:
(a) Volume compression ratio
Initial pressure p₁ = 1 bar
Initial temperature, T₁ = 310 K
cp = 1.005 kJ/kg⋅K
Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle
For a polytropic process we have
Therefore p₂ = p₁ ÷
= (1 bar) ÷
= 19.953 bar
Similarly for a polytropic process we have
or T₂ = T₁ ÷
=
= 618.531 K
The molar mass of air is 28.9628 g/mol.
Therefore R = = 0.287 kJ/kg⋅K
cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K
1). For process 1 to 2 which is polytropic process we have
W = =
= 295.16 kJ/kg
Q = =
= -73.79 kJ/kg
W = 295.16 kJ/kg
Q = -73.79 kJ/kg
2). For process 2 to 3 which is reversible constant volume heating we have
W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg
W = 0
Q = 1135.376 kJ/kg
3). For process 3 to 4 which is polytropic process we have
W = = Where T₄ is given by
or T₄ = T₃ ×
= 2200 × T₄ = 1102.611 K
W = = -1049.835 kJ/kg
and Q = 262.459 kJ/kg
W = -1049.835 kJ/kg
Q = 262.459 kJ/kg
4). For process 4 to 1 which is reversible constant volume cooling we have
W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg
W=0
Q = -569.09 kJ/kg
(b) The thermal efficiency is given by
=
= 0.499 or 49.9 % Efficient
(c) The mean effective pressure is given by
where r = compression ratio and
=
However p₃ = =
=70.97 atm
=
=
Therefore = 9.44 bar
Please find attached generalized diagrams of the Otto cycle
Answer:
0.31
126.23 kg/s
Explanation:
Given:-
- Fluid: Water
- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%
- Pump: Isentropic
- Net cycle-work output, Wnet = 100 MW
Find:-
- The thermal efficiency of the cycle
- The mass flow rate of steam
Solution:-
- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.
First process: Isentropic compression by pump
P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )
h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )
s1 = s-P1 = 0.6492 KJ/kg.K
v1 = v-P1 = 0.001010 m^3 / kg
P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )
s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )
- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:
- From the following relation we can determine ( h2 ) as follows:
h2 = h1 + wp
h2 = 191.81 + 8.0699
h2 = 199.88 KJ/kg
Second Process: Boiler supplies heat to the fluid and vaporize
- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).
- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).
P3 = 8 MPa
T3 = ? ( assume fluid exist in the saturated vapor phase )
h3 = hg-P3 = 2758.7 KJ/kg
s3 = sg-P3 = 5.7450 KJ/kg.K
- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:
Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).
- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.
- Under the isentropic conditions the steam exits the turbine at the following conditions:
P4 = 10 KPa
s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )
- Compute the quality of the mixture at condenser inlet by the following relation:
- Determine the isentropic ( h4s ) at this state as follows:
- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:
- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.
- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.
Answer: The mass flow rate of the steam would be 126.23 kg/s
- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):
Answer: The thermal efficiency of the cycle is 0.31