During steady state operation, a gear box receives 60kW through the input shaft and delivers power through the output shaft. For the gearbox as the system, the rate of energy transfer by convection is given by Q=-hA(Tb - T) Where h 0.171kW/m2.k is the heat transfer coefficient, A=1.0m2 is the outer surface are of the gearbox. Tb 300k is the temperature at the outer surface and T 293 K is the temperature of the surrounding air. For the gearbox, evaluate the heat transfer rate and the power delivered through the output shaft, in kW.

- the heat transfer rate is -1.197 kW

- the power delivered through the output shaft is 58.8 kW

Explanation:

Given the data in the question;

A gear box receives 60kW through the input shaft and delivers power through the output shaft

Heat received by gear box W₁ = - 60 kW

heat transfer coefficient h = 0.171k W/m².k

Area A = 1.0 m²

Outer surface temperature T_b = 300 K

Surrounding temperature T_f = 293 K

Now,

Heat transfer rate

Q = -hA( T_b - T_f )

we substitute

Q = -( 0.171 ) × 1.0 × ( 300 - 293 )

Q = -( 0.171 ) × 1.0 × 7

Q = -1.197 kW

Hence, the heat transfer rate is -1.197 kW

Now, from energy balance equation;

Heat transfer rate = Heat received by gearbox + Power delivered by output shaft

i.e

Q = W₁ + W₂

so we substitute

-1.197 = -60 + W₂

W₂ = 60 - 1.197

W₂ = 58.8 kW

Therefore, the power delivered through the output shaft is 58.8 kW

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