Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust
Question Statement:
A tractor engine developing 30 kW rejects heat at the rate of 0.58 kW per kW of engine output. A water cooling system is to be installed in the tractor. The expected temperature rise as air moves through the radiator is 20 K. The frontal area of the radiator is limited to 0.16 m². If density of air is 1.29 kg m⁻³ and specific heat of air is 1.0 kJ kg⁻¹K⁻¹, the amount of air to be blown per unit time through the radiator frontal area is
Options:
(A) 0.674 m³s⁻¹
(B) 0.870 m³s⁻¹
(C) 1.162 m³s⁻¹
(D) 1.502 m³s⁻¹
Official Answer Key:
A
Step-by-Step Solution & Explanation
Heat to be rejected via radiator, Q = 30 × 0.58 = 17.4 kW. Air mass flow: ṁ = Q/(c_p·ΔT) = 17.4/(1.0×20) = 0.870 kg/s. Volumetric flow: V̇ = ṁ/ρ = 0.870/1.29 = 0.674 m³/s. (B) is the mass flow value mistaken for volumetric flow.
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