Engine Performance Parameter MCQ Quiz in मल्याळम - Objective Question with Answer for Engine Performance Parameter - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 11, 2025
Latest Engine Performance Parameter MCQ Objective Questions
Top Engine Performance Parameter MCQ Objective Questions
Engine Performance Parameter Question 1:
Mechanical efficiency of an automobile engine usually varies in the range
Answer (Detailed Solution Below)
Engine Performance Parameter Question 1 Detailed Solution
Explanation:
Mechanical efficiency:
- Mechanical efficiency is defined as the ratio of brake power (delivered power) to the indicated power (power provided to the piston).
\({\eta _m} = \frac{{bp}}{{ip}} = \frac{{bp}}{{bp + fp}}\)
- Mechanical efficiency takes into account the mechanical losses in an engine.
- In general, the mechanical efficiency of the engine varies from 70 – 90 % and it depends on speed and design of the engine and can fluctuate based on:
- Engine Design and Technology: Older engines tend to have lower efficiencies, while newer engines with advanced technologies like direct injection and variable valve timing can achieve higher efficiencies.
- Engine Load and Speed: Engines operate at their peak efficiency at specific load and speed conditions, often around their cruising range. At lower or higher loads and speeds, efficiency typically drops.
- Temperature: Engines operate most efficiently within their normal operating temperature range. Cold starts and overheating can significantly reduce efficiency.
- Maintenance: Poor engine maintenance, like clogged air filters or worn-out spark plugs, can lead to decreased efficiency.
Important PointsEngine Performance Parameter:
Indicated Power:
- It is the power developed inside the engine cylinder.
\(ip = \frac{{{p_m}LAN.n}}{{60}}\)
Brake Power:
- This is the actual power available at the crankshaft and may be called power output of the engine. It is always less than indicated power.
\(BP = \frac{{2\pi NT}}{{60}}\)
Indicated thermal efficiency:
- It is the ratio of IP to fuel power.
\({\eta _i} = \frac{{ip}}{{{{\dot m}_f} \times CV}}\)
Brake thermal efficiency :
- It is the ratio of bp to fuel power.
\({\eta _b} = \frac{{bp}}{{{{\dot m}_f} \times CV}}\)
Engine Performance Parameter Question 2:
An IC engine produces an indicated power of 12 kW. If the mechanical efficiency of the engine is 90%, then loss of power due to friction is given as:
Answer (Detailed Solution Below)
Engine Performance Parameter Question 2 Detailed Solution
Concept:
Mechanical efficiency is given by:
\({\eta _{mech}} = \frac{{Brake\;power}}{{Indicated\;power}}\)
Indicated power = Brake power + Friction power
Calculation:
Given:
Indicated power = 12 kW, ηmech = 90% = 0.9
Mechanical efficiency is:
\({\eta _{mech}} = \frac{{Brake\;power}}{{Indicated\;power}}\)
\(0.90 = \frac{{Brake\;power}}{{12}} \Rightarrow Brake\;power = 0.90 \times 12 = 10.8\;kW\)
Now,
Friction Power = Indicated power – brake power
Friction power = 12 – 10.8 = 1.2 kW
Engine Performance Parameter Question 3:
If stroke volume of engine = 300 cc, clearance volume of engine = 20 cc. The compression ratio of engine will be:
Answer (Detailed Solution Below)
Engine Performance Parameter Question 3 Detailed Solution
Concept:
The compression ratio is the ratio of volume before compression and after compression.
Let VS = Stroke volume
VC = Clearance volume
\(Compression\ ratio\left( {{r_c}} \right) = \frac{{Effective\ Swept\ Volume}}{{Clearance\ Volume}}\)
\({r_c} = \frac{{{V_S} + {V_C}}}{{{V_C}}}\)
\({r_c} = \frac{{{V_S} + {V_C}}}{{{V_C}}}=1+\frac{{{V_S} }}{{{V_C}}}\)
\({r_c} = 1 + \frac{{300}}{{20}} = 16\)
Engine Performance Parameter Question 4:
The following data recorded during a test on oil engine:
Speed of Engine = 1000 rpm, Load on the brake = 1000 N, Length of brake = 750 mm,
Then brake power is:Answer (Detailed Solution Below)
Engine Performance Parameter Question 4 Detailed Solution
Concept:
In Prony Brake Dynamometer,
Torque T = Wl in N-m.
Brake Power \(B.P = \frac{{2\pi NT}}{{60}}\) in kW
where l is the length of the lever
Calculation:
Given:
Speed of engine N = 1000 rpm, load on brake W = 1000 N and length L = 750 mm = 0.75 m.
T = 1000 × 0.75 = 750 N
\(B.P = \frac{{2 \times \pi \times 1000 \times 750}}{{60}} = 78539.81\;W\)
Thus B.P = 78.54 kW
Engine Performance Parameter Question 5:
The torque developed by the engine is maximum
Answer (Detailed Solution Below)
Engine Performance Parameter Question 5 Detailed Solution
Explanation:
Peak torque appears earlier in the RPM range because of the engine's volumetric efficiency characteristic.
Volumetric efficiency = \(\frac{Actual\;mass\;of\;air\;intake\;per\;stroke}{Theoretical\;mass\;of\;air\;intake\;at\;that\;temperature}\)
Theoretical mass intake = Displacement Volume × Air density at that temperature
Volumetric efficiency:
It is a measure of the engine breathing capacity - varies from low (during low speeds) to a peak at about 2/3rd of the speed range and again to a low value at higher speeds.
The reason for this is the engine piston speed and valve timing.
At low speeds, the valves stay open for a longer time but the suction in the engine is less because the piston moves slowly. Because of this, the air intake into the engine is lesser than what is maximum possible.
As the engine picks up speed, the higher piston speed creates more suction which makes the engine intake more and more air with each stroke. This increases the volumetric efficiency which peaks at a speed for a given engine and thus increasing the Torque.
Engine Performance Parameter Question 6:
A spark ignition engine has a compression ratio of 8 and the volume before compression is 0.9 m3/kg. Net heat interaction per cycle is 1575 kJ/kg. What is the mean effective pressure?
Answer (Detailed Solution Below)
Engine Performance Parameter Question 6 Detailed Solution
Concept:
Compression ratio = \(r = \frac{{volume\;before\;compression\;}}{{volume\;after\;compression}}\)
Work done = Mean effective pressure × swept volume
swept volume = (v1 – v2)
Calculation:
Given Compression ratio (r) = 8, v1 = 0.9 m3/kg, Net heat interaction for cycle = 1575 kJ/kg
For any cycle, the net heat interaction = Net work interaction
\(r = \frac{{{v_1}}}{{{v_2}}}\)
⇒ \({v_2} = \frac{{0.9}}{8} = 0.1125\frac{{{m^3}}}{{kg}}\)
Work done = Mean effective pressure × swept volume (v1 – v2)
∴ 1575 × 103 = pm × (0.9 – 0.1125)
pm = 20 bar
Engine Performance Parameter Question 7:
Which of the following is a transmission dynamometer?
Answer (Detailed Solution Below)
Engine Performance Parameter Question 7 Detailed Solution
Explanation
Dynamometer – The work output or power output from the piston of the engine is found by a brake mechanism attached to the shaft of the engine, this mechanism is known as Dynamometer.
Types of dynamometer
Absorption dynamometer |
Transmission dynamometer |
The entire energy or power produced by the engine is absorbed by the friction resistance of the brakes of dynamometer and is transformed into heat during the process of measurement.
|
In this dynamometer, energy is not wasted in the friction but is used for doing work. The energy produced by the engine is transmitted to some other machines through dynamometer where power developed is measured.
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Heenan and Froude dynamometer is a hydraulic dynamometer invented by William froude in 1877.
Engine Performance Parameter Question 8:
The compressor that increase the pressure of air supplied to the inlet port by drawing the power from crankshaft is ________.
Answer (Detailed Solution Below)
Engine Performance Parameter Question 8 Detailed Solution
Explanation:
The most efficient method of increasing the power of an engine is by increasing the flow of air into the engine to enable more fuel to be burnt, which can be done by two devices supercharger and turbocharger. Some of the key points are mentioned for both the devices in the tabular form below.
Turbocharging |
Supercharging |
Turbocharging is a forced induction system that compresses the atmospheric gases i.e. increases the pressure and density and sends it to the engine cylinder. |
Supercharging is also a forced induction system. It compresses the atmospheric gases and sends it to the engine cylinder. |
It uses exhaust gases for its energy. |
It is connected to the crankshaft of the engine for its energy. |
It is not directly connected to the engine. |
It is directly connected to the engine through a belt. |
The compressor is rotated by the turbine. |
The compressor is rotated by the engine through a belt. |
It has a lag problem due to a discontinuous supply of energy. |
Negligible lag problem because of the continuous supply of energy by the crankshaft. |
It has smog alternating equipment, which helps in lowering the carbon emission. |
Does not have a wastegate, so the smog emits from the supercharger. |
The compressed air in the turbocharger has a high temperature and therefore requires an intercooler to lower its temperature. |
The compressed air in the supercharger has less temperature, therefore may or may not require an intercooler. |
Engine Performance Parameter Question 9:
The main object of Morse test is to find out
Answer (Detailed Solution Below)
Engine Performance Parameter Question 9 Detailed Solution
- The Morse test is a test conducted to determine the power developed in each cylinder in a multi-cylinder IC engine
- In this test first of all the power developed by all the cylinders together is determined experimentally
- Then the power of the individual cylinders is determined by cutting off the power supply to the spark plug of the cylinder under test
- Then the power developed by the engine with the remaining cylinders is determined experimentally and this value is subtracted from the first value and this gives you the power developed in the cylinder, whose spark plug was cut off
- Similarly, this test is carried out on all the cylinders of the engine individually
Thus Morse test is used to find out the frictional power of the multi-cylinder engine.
The methods are used to find out the frictional power are -
(a) Willan's line method
(b) Morse test
(c) Motoring test
(d) Retardation test
First indicated power is found out by cut out one cylinder at a time simultaneously. When all cylinders are working then it is possible to find out the Break Power of the engine. From Indicated Power and Break Power we can find out the Frictional power.
But the main point is that the Morse test is only applicable to multi-cylinder engines.
Here in the options, only one option has a multi-cylinder engine and that is option (3) and hence it is correct.
Engine Performance Parameter Question 10:
In four-stroke engines, the camshaft gear has
Answer (Detailed Solution Below)
Engine Performance Parameter Question 10 Detailed Solution
Explanation:
- Let’s start with the piston at the TDC.
- 1st stroke the piston moves down, and the intake valve opens.
- 2nd stroke, the intake valve closes, and the piston moves up.
- The crankshaft has rotated once, and one valve has opened.
- 3rd stroke, the spark plug fires the mixture; the piston moves down.
- 4th stroke, the exhaust valve opens, and the piston moves up.
- The engine has completed 2 revolutions and each valve has opened once.
- The camshaft travels at half the crankshaft speed because each valve opens only once on every two revolutions of the crankshaft.
- Therefore, In four-stroke engines, the camshaft gear has twice the number of teeth as on crankshaft gear.
Additional Information
- A two-stroke engine requires only two strokes of the piston to complete one full cycle.
- Therefore, it requires only one rotation of the crankshaft to complete a cycle.
- This means several events must occur during each stroke for all four events to be completed in two strokes.
- In a two-stroke engine, the camshaft is geared so that it rotates at the same speed as the crankshaft (1: 1).