Question

An engine consumes 30 kg of coal per hour and does the work at the rate of 45H.P. Find the efficiency of heat engine. The heat of combustion of coal is 2 x l0⁶cal per kg.

Solution

Power developed = 45 HP = 33570 J/s
Coal consumed per hour = 30kg
Heat produced per hour = $30 cross times 2 cross times 10 to the power of 6 space cal divided by hr$
Heat produced per sec = $space 30 cross times 2 cross times 10 to the power of 6 fraction numerator 4.2 over denominator 3600 end fraction straight J divided by straight s$
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Efficiency, $space space space straight eta equals 33570 over 70000 cross times 100 almost equal to 48 %$

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Mechanical Properties Of Fluids

An engine consumes 30 kg of coal per hour and does the work at the rate of 45H.P. Find the efficiency of heat engine. The heat of combustion of coal is 2 x l0⁶cal per kg.

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For Daily Free Study Material Join wiredfaculty

Mechanical Properties Of Fluids

An engine consumes 30 kg of coal per hour and does the work at the rate of 45H.P. Find the efficiency of heat engine. The heat of combustion of coal is 2 x l06cal per kg.

Some More Questions From Mechanical Properties of Fluids Chapter

Who explained the correct variation of specific heat with temperature in the lower range of temperature?

What is the SI unit of molar specific heat?

Define heat capacity.

What are the units of heat capacity?

Define Dulong and Pettit’s law of specific heat.

Who explained the correct variation of specific heat with temperature in the lower range of temperature?

Do the gases have only one value of specific heat?

What is thermometry?

Define temperature.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

An engine consumes 30 kg of coal per hour and does the work at the rate of 45H.P. Find the efficiency of heat engine. The heat of combustion of coal is 2 x l0⁶cal per kg.