Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl
1 answer:
Answer:
exergy = 33.39 kW
Explanation:
given data
thermal energy reservoir T2 = 1500 K
heat at a rate = 150,000 kJ/h = kW = 41.67 kW
environment temperature T1 = 25°C = 298 K
solution
we get here maximum efficiency that is reversible efficiency is express as
reversible efficiency = 1 - ...............1
reversible efficiency = 1 -
reversible efficiency = 0.80133
and
the exergy of this supplied energy that is
exergy = efficiency × hat supply ................2
exergy = 0.80133 × 41.67 kW
exergy = 33.39 kW
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Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
304.13 mph
Explanation:
Data provided in the question :
The Speed of the flying aircraft = 300 mph
Tailwind of the true airspeed = 50 mph
Now,
The ground speed will be calculated as:
ground speed =
or
The ground speed =
or
The ground speed = 304.13 mph
Hence, the ground speed is 304.13 mph