--- title: "Engine displacement, compression ratio and power" subject: "Automotive Technology" chapter: 1 key: "semester_5/automotive_technology:engine-displacement-bmep-numerical" source: https://www.bctengineer.com/q/semester_5/automotive_technology/engine-displacement-bmep-numerical site: BCT Engineering --- # Engine displacement, compression ratio and power Automotive Technology · Chapter 1: Introduction - Asked in: year not given, 8 marks - Video: [IC Engine Terms, Bore, Stroke, Compression Ratio, gtu #bme #engine](https://www.youtube.com/watch?v=2gmiLU6SXQs) — Mechanical Engineering Management - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/automotive_technology/engine-displacement-bmep-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-1 ## Question A four-cylinder, four-stroke petrol engine has a bore of 80 mm and a stroke of 90 mm. The clearance volume of each cylinder is 55 cc. At 4000 rpm the engine develops a brake torque of 120 N m. Calculate (i) the swept volume per cylinder and the total engine displacement, (ii) the compression ratio, (iii) the brake power, (iv) the brake mean effective pressure and (v) the mean piston speed. ## Short answer Swept volume per cylinder Vs = (π/4)×(0.08)²×0.09 = 452.4 cm³, total displacement Vd = 1809.6 cm³ (1.81 L). Compression ratio r = (452.4+55)/55 = 9.23. Brake power BP = 2π×4000×120/60 = 50.27 kW. BMEP = 8.33 bar. Mean piston speed = 2×0.09×4000/60 = 12 m/s. ## Answer **Given:** $D = 80$ mm, $L = 90$ mm, $n = 4$ cylinders, $V_c = 55$ cc, $N = 4000$ rpm, $T = 120$ N m. #### (i) Swept volume and displacement $$ \begin{aligned} V_s &= \frac{\pi}{4} D^2 L = \frac{\pi}{4}(8)^2(9)\ \text{cm}^3 = 452.4\ \text{cm}^3 \\ V_d &= n V_s = 4 \times 452.4 = 1809.6\ \text{cm}^3 \approx 1.81\ \text{L} \end{aligned} $$ #### (ii) Compression ratio $$ r = \frac{V_s + V_c}{V_c} = \frac{452.4 + 55}{55} = 9.23 $$ #### (iii) Brake power $$ BP = \frac{2\pi N T}{60} = \frac{2\pi \times 4000 \times 120}{60} = 50\,265\ \text{W} \approx 50.27\ \text{kW} $$ #### (iv) Brake mean effective pressure For a four-stroke engine, $BP = \dfrac{p_{mb} V_d N}{2 \times 60}$ (one power stroke per two revolutions). $$ p_{mb} = \frac{BP \times 120}{V_d N} = \frac{50\,265 \times 120}{1.8096\times10^{-3} \times 4000} = 8.33\times10^{5}\ \text{Pa} = 8.33\ \text{bar} $$ #### (v) Mean piston speed $$ \bar{S}_p = 2LN/60 = \frac{2 \times 0.09 \times 4000}{60} = 12\ \text{m/s} $$ **Answer:** $V_s = 452.4$ cc, $V_d = 1809.6$ cc, $r = 9.23$, $BP = 50.27$ kW, bmep $= 8.33$ bar, $\bar S_p = 12$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the swept volume per cylinder for D = 80 mm, L = 90 mm? A. About 226 cm^3 B. About 452 cm^3 C. About 1810 cm^3 D. About 905 cm^3 Answer: B. Vs = (pi/4)(8 cm)^2(9 cm) = 452.4 cm^3; times 4 cylinders gives 1809.6 cm^3. (Key point: Swept volume) 2. With Vs = 452.4 cc and Vc = 55 cc, what is the compression ratio and which formula is used? A. r = Vs/Vc = 8.2 B. r = (Vs + Vc)/Vc = 9.23 C. r = (Vs + Vc)/Vs = 1.12 D. r = Vc/(Vs + Vc) = 0.108 Answer: B. Compression ratio is total volume over clearance volume: (452.4 + 55)/55 = 9.23. (Key point: Compression ratio) 3. What is the brake power at 4000 rpm and 120 N m? A. About 50.3 kW B. About 25.1 kW C. About 14.4 kW D. About 100.5 kW Answer: A. BP = 2 pi N T / 60 = 2 pi x 4000 x 120 / 60 = 50 265 W = 50.27 kW. (Key point: Brake power) 4. Why does the bmep formula for a four-stroke engine use N/(2 x 60)? A. There are two cylinders firing B. One power stroke occurs per two revolutions C. The torque is halved D. The piston speed is halved Answer: B. A four-stroke cylinder gives one power stroke every two revolutions, so BP = pmb Vd N/120, giving 8.33 bar. 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