
If the intake of an I.C engine increases, its efficiency will
(A) increase
(B) decrease
(C) remains the same
(D) depend on other factors
Answer
572.1k+ views
Hint: The engine efficiency is of three types, they brake thermal efficiency, indicated thermal efficiency, and mechanical efficiency. If the intake of an I.C engine is increased it is related to thermal efficiency. By using the brake thermal efficiency and the indicated thermal efficiency the solution is determined.
Formula used:
Brake thermal efficiency,
${\eta _{bt}} = \dfrac{{BP}}{{{m_f} \times CV}}$
Where, ${\eta _{bt}}$ is the brake thermal efficiency, $BP$ is the brake power, ${m_f}$ is the mass of the fuel supplied and $CV$ is the calorific value.
Indicated thermal efficiency,
${\eta _{it}} = \dfrac{{IP}}{{{m_f} \times CV}}$
Where, ${\eta _{it}}$ is the indicated thermal efficiency, $IP$ is the indicated power, ${m_f}$ is the mass of the fuel supplied and $CV$ is the calorific value.
Complete step by step answer:
Brake thermal efficiency,
${\eta _{bt}} = \dfrac{{BP}}{{{m_f} \times CV}}\,..................\left( 1 \right)$
If the intake of the IC engine increases, a huge amount of air is intake by the engine, if more amount of the air is taken, proportionally the more amount of fuel is supplied, so by equation (1), the brake thermal efficiency and the mass flow of the fuel is inversely proportional. So, efficiency will decrease.
Indicated thermal efficiency,
${\eta _{it}} = \dfrac{{IP}}{{{m_f} \times CV}}\,...................\left( 2 \right)$
If the intake of the IC engine increases, a huge amount of air is intake by the engine, if more amount of the air is taken, proportionally the more amount of fuel is supplied, so by equation (2), the indicated thermal efficiency and the mass flow of the fuel is inversely proportional. So, efficiency will decrease.
By equation (1) and equation (2), the efficiency will decrease.
Hence, the option (B) is the correct answer.
Note:
In relation to the efficiency and temperature, the efficiency of the engine decreases as the temperature of the inlet air increases. As the air becomes thinner at high temperatures so it does not expend much so doing less work and thus the engine efficiency will decrease.
Formula used:
Brake thermal efficiency,
${\eta _{bt}} = \dfrac{{BP}}{{{m_f} \times CV}}$
Where, ${\eta _{bt}}$ is the brake thermal efficiency, $BP$ is the brake power, ${m_f}$ is the mass of the fuel supplied and $CV$ is the calorific value.
Indicated thermal efficiency,
${\eta _{it}} = \dfrac{{IP}}{{{m_f} \times CV}}$
Where, ${\eta _{it}}$ is the indicated thermal efficiency, $IP$ is the indicated power, ${m_f}$ is the mass of the fuel supplied and $CV$ is the calorific value.
Complete step by step answer:
Brake thermal efficiency,
${\eta _{bt}} = \dfrac{{BP}}{{{m_f} \times CV}}\,..................\left( 1 \right)$
If the intake of the IC engine increases, a huge amount of air is intake by the engine, if more amount of the air is taken, proportionally the more amount of fuel is supplied, so by equation (1), the brake thermal efficiency and the mass flow of the fuel is inversely proportional. So, efficiency will decrease.
Indicated thermal efficiency,
${\eta _{it}} = \dfrac{{IP}}{{{m_f} \times CV}}\,...................\left( 2 \right)$
If the intake of the IC engine increases, a huge amount of air is intake by the engine, if more amount of the air is taken, proportionally the more amount of fuel is supplied, so by equation (2), the indicated thermal efficiency and the mass flow of the fuel is inversely proportional. So, efficiency will decrease.
By equation (1) and equation (2), the efficiency will decrease.
Hence, the option (B) is the correct answer.
Note:
In relation to the efficiency and temperature, the efficiency of the engine decreases as the temperature of the inlet air increases. As the air becomes thinner at high temperatures so it does not expend much so doing less work and thus the engine efficiency will decrease.
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