A. Imagine the white car in the left lane is moving more slowly than the surrounding traffic. How is this a violation of a state
2 answers:
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The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W =
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
(a). The value of temperature at the end of heat addition process = 2042.56 K
(b). The value of pressure at the end of heat addition process = 1555.46 k pa
(c). The thermal efficiency of an Otto cycle = 0.4478
(d). The value of mean effective pressure of the cycle = 1506.41
Explanation:
Compression ratio = 8
Initial pressure = 95 k pa
Initial temperature = 278 °c = 551 K
Final pressure = 8 ×
= 8 × 95 = 760 k pa
Final temperature =
×
Final temperature = 551 ×
Final temperature = 998 K
Heat transferred at constant volume Q = 750
(a). We know that Heat transferred at constant volume =
⇒ 1 × 0.718 × ( - 998 ) = 750
⇒ = 2042.56 K
This is the value of temperature at the end of heat addition process.
Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.
⇒ P ∝ T
⇒ =
⇒ =
× 760
⇒ = 1555.46 k pa
This is the value of pressure at the end of heat addition process.
(b). Heat rejected from the cycle
For the compression and expansion process,
⇒
⇒
⇒ = 1127.7 K
Heat rejected = 1 × 0.718 × ( 1127.7 - 551)
⇒ = 414.07
Net heat interaction from the cycle =
-
Put the values of &
we get,
⇒ = 750 - 414.07
⇒ = 335.93
We know that for a cyclic process net heat interaction is equal to net work transfer.
⇒ =
⇒ = 335.93
This is the net work output from the cycle.
(c). Thermal efficiency of an Otto cycle is given by
Put the values of &
in the above formula we get,
⇒ = 0.4478
This is the thermal efficiency of an Otto cycle.
(d). Mean effective pressure :-
We know that mean effective pressure of the Otto cycle is given by
=
---------- (1)
where is the swept volume.
=
---------- ( 2 )
From ideal gas equation
= m × R ×
Put all the values in above formula we get,
⇒ 95 × = 1 × 0.287 × 551
⇒ = 0.6
From the same ideal gas equation
= m × R ×
⇒ 760 × = 1 × 0.287 × 998
⇒ = 0.377
Thus swept volume = 0.6 - 0.377
⇒ = 0.223
Thus from equation 1 the mean effective pressure
⇒ =
⇒ = 1506.41
This is the value of mean effective pressure of the cycle.