(a) 0.249 (24.9 %)
The maximum efficiency of a heat engine is given by
where
Tc is the low-temperature reservoir
Th is the high-temperature reservoir
For the engine in this problem,
Therefore the maximum efficiency is
(b-c) 0.221 (22.1 %)
The second steam engine operates using the exhaust of the first. So we have:
is the high-temperature reservoir
is the low-temperature reservoir
If we apply again the formula of the efficiency
The maximum efficiency of the second engine is
Answer:
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Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
Answer:
B. Her curiosity got the best of her
Explanation:
Pandora gave in to her curiosity .
