Energy transfer in mechanical systems: During steady-state operation, a mechanical gearbox receives 70 KW of input power through
the input shaft and delivers power through the output shaft. Considering the gearbox as a thermodynamic system, the rate of convection heat transfer generated during its operation is defined by: where h 0.171 kW/m2-K is the convection heat transfer coefficient, A 1.0 m is the outer surface area of the gearbox, T 300 K is the temperature of the outer surface, and T 293 K is the temperature of the surrounding air. For the gearbox, evaluate the heat transfer rate, 2, and the power delivered through the output shaft, W.
1 answer:
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Answer:
i). Compression ratio = 3.678
ii). fuel consumption = 0.4947 kg/hr
Explanation:
Given :
Fuel calorific value = 45 MJ/kg
We know, engine efficiency is given by,
where is compression ratio =
=
where is compression volume
is swept volume
Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.
Then,
and at 70% compression, 30% of the swept volume remains
∴
We know,
∴
Therefore, compression ratio is = 3.678
Now efficiency,
, this is the ideal efficiency
Therefore actual efficiency,
Therefore total power required = 1 kW x 3600 J
= 3600 kJ
∴ we know efficiency,
Therefore fuel required =
= 0.4947 kg/hr
Answer:
the pressure at the end of the combustion is 2.68 MPa
Solution:
As per the question:
Initial Pressure, P = 0.95\ MPa
Temperature before combustion, = 273 + 425 = 698 K
Temperature after combustion, = 1973 K
Now,
To calculate the pressure at the end of combustion, P':
By using the Pressure-Temperature relation from Gay- Lussac's law: