This set of Farm Tractor Multiple Choice Questions & Answers (MCQs) focuses on “Engine Performances – Set 2”.

1. How to calculate the brake power of an engine in watts?

a) 2πNT / 60

b) 2πNT

c) 2 * 10^{3} * πNT / 60

d) πNT / 60

View Answer

Explanation: It is a formula to calculate the brake power of a tractor engine. In the formula, N represents revolutions per minute, T represents torque in Nm. 1 Nm/s = 1 watt. The 60 in the denominator represents the conversion of revolutions per minute into revolutions per second.

2. Find the torque of the engine having the tangential force of 569.5 Kg-m/s^{2} and crank radius of 30cm.

a) 170.8 Nm

b) 171.8 Nm

c) 173 Nm

d) 172 Nm

View Answer

Explanation: T = F * r

Where T – torque of the engine, F – tangential force on the crankshaft, r – crank radius

T = 569.5N * 0.3m

= 170.8 Nm

1 Kg-m/s

^{2}= 1N and 30cm = 0.3m

3. Find the brake horsepower of a tractor engine having engine torque 170 Nm, crankshaft rotation 2500rpm.

a) 60

b) 61

c) 62

d) 63

View Answer

Explanation: BP = 2πNT/60

= (2π*2500*170)/60

= 44505.8 watts

= 44505.8/745.6 (1 Hp = 745.6 watts)

= 59.6 ≈ 60 Hp

4. Find indicated power of a two-cylinder four-stroke engine with mean effective pressure of 686700 N/m^{2}, stroke length 15 cm, bore 12.5 cm and rpm as 1200.

a) 33 Hp

b) 32 Hp

c) 32.5 Hp

d) 35 Hp

View Answer

Explanation: Indicated power = k.P

_{m}.L.A.n/60

= (2*686700 N/m

^{2}*0.15m*(π*(0.125m)

^{2}/4)*1200/2)/60

= 25281.23 watts

= 33.90 Hp

5. Find the specific output of a tractor engine of stroke length 17 cm, bore 15 cm with brake power 30 Hp.

a) 9 Hp/l

b) 12 Hp/l

c) 14 Hp/l

d) 8 Hp/l

View Answer

Explanation: Specific output = Brake power/A*L where A – Area and L – stroke length

= 30/((π*0.15

^{2})/4)*0.17

= 9986.19 Hp/m

^{3}

= 9.98619 Hp/l

6. What is the purpose of calculation of specific output of an engine?

a) Engine performance

b) Engine input

c) Engine torque

d) Engine pressure

View Answer

Explanation: Specific output is a measure of engine performance. It is the power produced by the engine per unit of piston displacement. It is measured in the units of horsepower/m

^{3}or horsepower/litre or Kilowatts/m

^{3}.

7. Find the brake power of a four-stroke 4 cylinder engine having brake mean effective pressure 9 Kg/cm^{2}, piston displacement 1178.09 cm^{3} and the crankshaft of the engine rotates at 2000 rpm.

a) 94 Hp

b) 96 Hp

c) 97 Hp

d) 98 Hp

View Answer

Explanation: Brake power = P

_{mb}*(L*A)*k*n/60

= (9 Kg/cm

^{2}*1178.09 cm

^{3}*4*(2000/2))/60

= 706854 kg-cm/s

= 7068.54 kg-m/s

= 94.24 hp

Where P

_{mb}= Brake mean effective pressure

8. Find the friction mean effective pressure of an engine having brake mean effective pressure 25 Hp and indicated mean effective pressure 30 Hp.

a) 5.0 Hp

b) 5.5 Hp

c) 5.3 Hp

d) 5.4 Hp

View Answer

Explanation: FMEP = IMEP – BMEP

= 30 – 25 = 5 Hp

Where FMEP – friction mean effective pressure

BMEP – Brake mean effective pressure

IMEP – Indicated mean effective pressure

9. How to calculate the fuel consumption of an engine in kg/hr theoretically?

a) (M_{f} = V_{f} * ρ_{f} * 3600) / ∆t

b) (M_{f} = V_{f} * ρ_{f})

c) (M_{f} = V_{f} / ∆t)

d) (M_{f} = L * A)

View Answer

Explanation: Actual fuel consumption of an engine is the amount of fuel consumed per unit distance. It is also calculated using the formula in a theoretical manner. Where M

_{f}– mass flow rate, V

_{f}– Volume of fuel, ρ

_{f}– density of fuel and ∆t – time in which the tank empties itself while supplying fuel to a running engine.

10. Find the indicated specific fuel consumption of an engine having mass flow rate 5 kg/hr and indicated power 30 Hp.

a) 0.2235 kg/kw-hour

b) 0.009 kg/kw-hour

c) 1.023 kg/kw-hour

d) 2.003 kg/kw-hour

View Answer

Explanation: Indicated specific fuel consumption = M

_{f}/ I.P = 5 / 30

= 5 / (30 * 745.6)

= 0.2235 kg/Kw-hour

11. Find the brake specific fuel consumption of an engine having mass flow rate 6 kg/hr, indicated power 30 Hp and friction power 5 Hp.

a) 0.321 kg/kw-hr

b) 0.5 kg/kw-hr

c) 0.66 kg/kw-hr

d) 0.88 kg/kw-hr

View Answer

Explanation: Brake specific fuel consumption = M

_{f}/ B.P = 6 / B.P

= I.P – F.P = B.P = 30 – 5 = 25

= 6 / (25 * 745.6)

= 6 / 18.64

= 0.321 kg/kw-hr

Where B.P – Brake power

**Sanfoundry Global Education & Learning Series – Farm Tractor.**

To practice all areas of Farm Tractor, __here is complete set of Multiple Choice Questions and Answers__.

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