An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t
he heat addition the volume triples. Find the two specific heat transfers (q) in the cycle and the overall cycle efficiency. Solution:
1 answer:
Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
= R*
in(
/
)
The specific heat capacity:
=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
=1 - (
/
)
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67
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Answer:
a)6.7m/S
b)6.8m/s
Explanation:
Hello ! To solve the point b you must follow the steps below
1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)
2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)
3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)
4)
the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation
where
Vfy= final verticaly speed
Vy=initial verticaly speed=0.59m/S
g=gravity=9.81m/S^2
y=height of slide=2.31m
solving
The last step is to add the velocity components vectorally at the end of the route with the following equation
point A
taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.
vy=0
Vfy==6.7m/s
