Engine Testing and Performance Questions and Answers

Answer: B

The brake thermal efficiency is about 19.13%.

The calorific value of diesel, 35280 MJ m⁻³, is 35.28 MJ per litre. The hourly energy input from both fuels is:

\[ \begin{aligned} \text{diesel} &= 0.46 \times 35.28 = 16.23\ \text{MJ/h} \\ \text{gas} &= 12.5 \times 3.97 = 49.63\ \text{MJ/h} \\ \text{input} &= 65.85\ \text{MJ/h} = 18.29\ \text{kW} \\ \eta &= 3.5/18.29 = 19.13\% \end{aligned} \]

Counting only the producer gas gives 3.5/13.78 = 25.4%, close to the 25.32 distractor. In dual fuel mode the pilot diesel is part of the energy input and must be included.

Answer: C

The maximum electricity produced is 3.15 kW.

The engine is coupled directly to the generator, so the generator receives the full brake power of 3.5 kW. Its electrical output is that shaft input times the generator efficiency:

\[ \begin{aligned} P_{e} &= \eta_{g} \times P_{b} \\ &= 0.90 \times 3.5 = 3.15\ \text{kW} \end{aligned} \]

3.50 kW ignores the generator losses. 3.00 and 2.85 kW would correspond to generator efficiencies of about 86% and 81%, not the 90% given. The engine's thermal efficiency from Q48 plays no part here, because 3.5 kW is already the brake (shaft) output.

Answer: B

Specific fuel consumption is inversely proportional to the thermal efficiency of an engine. Thermal efficiency is the work output divided by the energy of the fuel burnt, and specific fuel consumption is the fuel used per unit of power output, so when more fuel is needed for the same power the efficiency is lower. Mean piston speed, specific power output and mean effective pressure are not inversely proportional to thermal efficiency in this sense.

Answer: D

The ratio of the thermal efficiency of the actual cycle to that of the ideal cycle is called the relative efficiency. Mechanical efficiency is the ratio of brake power to indicated power, indicated thermal efficiency is the ratio of the indicated power to the fuel energy, and volumetric efficiency is the ratio of the volume of air actually drawn to the swept volume, so none of them is the ratio of actual to ideal thermal efficiency.

Relative efficiency is always below 1, since the actual cycle has losses that the ideal cycle does not have.

Answer: C

The mechanical efficiency is 67.21%.

Brake power from the torque:

\[ \begin{aligned} BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 2200 \times 55}{60} \\ &\approx 12.67\ \text{kW} \end{aligned} \]

Indicated power from the mean effective pressure, with one power stroke every two revolutions in a 4-stroke engine:

\[ \begin{aligned} A &= \tfrac{\pi}{4}(0.073)^{2} = 4.185 \times 10^{-3}\ \text{m}^{2} \\ IP &= p_{i} L A \cdot \tfrac{N}{2 \times 60} \cdot n \\ &= 1050 \times 0.078 \times 4.185 \times 10^{-3} \\ &\quad \times 18.33 \times 3 \approx 18.85\ \text{kW} \end{aligned} \]

\[ \eta_{m} = \frac{12.67}{18.85} = 0.6721 = 67.21\% \]

Forgetting the factor 2 for a 4-stroke engine doubles the indicated power and halves the efficiency, and the other listed values do not follow from consistent working.

Answer: A

The power developed in the cylinder is the indicated horsepower (IHP).

Indicated horsepower is the power developed by the burning gases on the piston, worked out from the indicator diagram (pressure against volume) of the cylinder, hence the name. It is the highest of the engine's power figures, since nothing has yet been lost to friction.

BHP is what remains at the flywheel after friction, FHP is the power lost to friction (IHP − BHP), and DBHP is the power left at the tractor's drawbar. Mechanical efficiency is BHP/IHP.