Question 3 of 6: Internal Combustion Engine

Question 3: Internal Combustion Engine (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. From the page-9 attachment and the air-fuel characteristics stated in the question:

Find. The swept and clearance volumes, the trapped charge and fuel masses, the four cycle temperatures, the indicated work per cylinder per cycle, and the ideal, actual and useful power at 2600 rpm compared with the manufacturer's own torque figure.

Approach. Geometry fixes the two volumes, the gas law fixes the trapped mass and hence the fuel available, and the four temperatures follow from isentropic compression, constant-volume heat release and isentropic expansion using the charge specific heats. Work comes from the energy balance, and the power chain is then simply work per cycle times firing frequency, degraded twice.

1. Part (b) — cylinder volumes. The page-9 bore and stroke reproduce the printed displacement, $$V_s = \frac{\pi D^2}{4}L = \frac{\pi (8.25)^2}{4}\times 9.28 = 496.1\ \text{cm}^3 \ \text{per cylinder} \quad (\times 4 = 1984\ \text{cm}^3)$$ The compression ratio relates swept and clearance volume, $r = (V_s + V_c)/V_c$, so $$V_c = \frac{V_s}{r-1} = \frac{496.0}{9} = 55.11\ \text{cm}^3, \qquad V_1 = V_s + V_c = \boxed{551.1\ \text{cm}^3}, \qquad V_2 = V_c = \boxed{55.11\ \text{cm}^3}$$ $V_1$ is the volume at bottom dead centre where compression begins and $V_2$ that at top dead centre where it ends.
2. Part (c) — trapped charge and fuel. The charge is air plus fuel together, so its gas constant is $R = c_p - c_v = 1.42 - 1.14 = 0.28\ \text{kJ/kg}\cdot\text{K}$ and its specific-heat ratio is $k = c_p/c_v = 1.2456$. Filling the full cylinder volume at inlet conditions, $$m = \frac{p_1 V_1}{R T_1} = \frac{100 \times 551.1\times 10^{-6}}{0.28 \times 303.15} = \boxed{0.6493\ \text{g}}$$ Of that, one part in $(1 + r_f) = 16$ is fuel: $$m_{fuel} = \frac{m}{1 + r_f} = \frac{0.6493}{16} = \boxed{0.04058\ \text{g}}$$ leaving 0.6087 g of air, the correct 15:1 proportion.
3. Part (d) — temperature after compression. Compression is isentropic over the volume ratio $r = 10$, so from the page-17 isentropic relations $$T_2 = T_1\, r^{\,k-1} = 303.15 \times 10^{0.2456} = \boxed{533.7\ \text{K}}$$ (260.5 °C). The exponent is $k - 1 = 0.2456$ rather than the 0.4 of pure air, because the fuel vapour and the products raise $c_v$ and soften the compression.
4. Part (d) — peak and exhaust temperatures. All the fuel releases its calorific value at constant volume into the whole charge: $$Q_{in} = m_{fuel}\,\mathrm{CV} = 0.04058\times10^{-3} \times 45\,000 = 1.8261\ \text{kJ}$$ $$T_3 = T_2 + \frac{Q_{in}}{m\,c_v} = 533.7 + \frac{1.8261}{0.000649\times 1.14} = \boxed{3000.8\ \text{K}}$$ and the expansion returns over the same volume ratio, $$T_4 = \frac{T_3}{r^{\,k-1}} = \frac{3000.8}{1.7604} = \boxed{1704.6\ \text{K}}$$ (1431 °C), a realistic blow-down temperature for a spark-ignition engine.
5. Part (e) — net work per cylinder per cycle. The cycle rejects heat at constant volume between 4 and 1, $$Q_{out} = m c_v (T_4 - T_1) = 0.000649 \times 1.14 \times 1401.4 = 1.0373\ \text{kJ}$$ $$W_{net} = Q_{in} - Q_{out} = 1.8261 - 1.0373 = \boxed{0.7888\ \text{kJ}}$$ As a check this equals the air-standard efficiency times the heat input, $\eta = 1 - r^{1-k} = 0.4320$, giving $0.4320 \times 1.8261 = 0.7888$ kJ — the same number to four figures.
6. Part (f) — ideal power for the whole engine. A four-stroke cylinder fires once every two revolutions, so at 2600 rpm each cylinder completes $2600/(2\times 60) = 21.667$ cycles per second and $$P_{ideal} = W_{net}\,\frac{N}{2\times 60}\,n_{cyl} = 0.7888 \times 21.667 \times 4 = \boxed{68.36\ \text{kW}}$$
7. Part (g) — actual cycle power. Time loss (finite burn duration), blowdown loss (the exhaust valve opening before bottom dead centre) and heat loss to the walls together remove 20 % of the indicated work: $$P_{actual} = 0.80\,P_{ideal} = 0.80 \times 68.36 = \boxed{54.69\ \text{kW}}$$
8. Part (h) — useful (brake) power. Pumping and friction remove a further 18 % at the crankshaft: $$P_{useful} = 0.82\,P_{actual} = 0.82 \times 54.69 = \boxed{44.84\ \text{kW}}$$ The two stages together pass $0.80 \times 0.82 = 0.656$ of the indicated work, so only 66 % of it reaches the flywheel.
9. Part (i) — comparison with the specification. The page-9 attachment lists 165 N·m at 2600 rpm, and the page-17 internal-combustion relation gives $$P = \frac{2\pi N \tau}{60} = \frac{2\pi \times 2600 \times 165}{60} = \boxed{44.92\ \text{kW}}$$ against the 44.84 kW computed in (h) — agreement to 0.2 %. That is far closer than the modelling deserves and should not be read as proof of the individual loss splits; what it does confirm is that the assumed loss chain is of the right total magnitude and that the fuel-air cycle with modified specific heats is a sound basis for estimating a real engine's output from its geometry alone. The corresponding brake mean effective pressure, $W_{useful}/V_s = 1043$ kPa, sits in the 9–10 bar band expected of a naturally aspirated gasoline engine at peak torque.