Interactive Lecture
Engine Performance, Torque Curves, and Transmission Matching
by Dr Arnaldo Delli Carri
Thanks to the contribution of Dr Kristaq Hazizi
The performance of an internal combustion (IC) engine is fully characterised by a family of curves plotted against engine speed in revolutions per minute (RPM). Powertrain engineers read these maps to select gear ratios, calibrate fuel injection, design hybrid strategies, and benchmark engines of different sizes. The five key quantities are: brake torque, brake power, Brake Mean Effective Pressure (BMEP), Brake Specific Fuel Consumption (BSFC), and volumetric efficiency.
Brake quantities are measured at the crankshaft output (after internal friction losses), using a dynamometer (brake). Brake torque \(T\) and brake power \(P\) are related by:
where \(N\) is engine speed in RPM and \(\omega\) is angular velocity in rad/s. In the practical units used by engineers:
\[ P\;[\text{kW}] = \frac{T\;[\text{N·m}] \;\times\; N\;[\text{RPM}]}{9549} \]The constant 9549 = 60 000 / (2π).
Brake torque is proportional to the mass of air–fuel mixture burned per cycle. The more charge the cylinder traps, the higher the peak cylinder pressure, and the more work each power stroke delivers. Understanding the torque curve therefore means understanding what controls cylinder filling across the RPM range.
At any given RPM the engine attempts to inhale a fixed swept volume \(V_d\) of fresh charge. How much it actually captures — relative to that ideal — is the volumetric efficiency \(\eta_v\). Because torque \(\propto\eta_v\), the torque curve and the \(\eta_v\) curve are almost identical in shape (you can verify this in the simulation below).
The flame front that ignites the charge takes a nearly fixed time to propagate across the combustion chamber. At higher RPM the crankshaft rotates further during that time, so ignition must be advanced (triggered earlier) to ensure peak cylinder pressure still occurs slightly after top dead centre (TDC) — the crank angle that maximises mechanical advantage. The engine management system maps ignition advance against RPM, but this map can only be optimal at one operating condition; at other speeds, combustion phasing is a compromise that costs torque.
Both friction mean effective pressure (piston rings, bearings, valve train) and pumping mean effective pressure (pushing exhaust out through the valves and ports) increase with RPM. These losses subtract directly from the gross indicated work, pulling the torque curve down at high speed even if combustion were perfect. Together they set the practical rev limit of the engine.
BMEP is the hypothetical constant cylinder pressure that, acting over every power stroke, would produce the same net work per cycle as the real engine. For a 4-stroke engine:
where \(V_d\) is the total swept (displaced) volume in m³ and \(T\) is brake torque in N·m. BMEP is expressed in Pa (typically quoted in kPa).
BMEP is the most useful single metric for comparing engines of different sizes because it normalises torque by displacement, and its formula carries no engine-speed term — it measures only how hard the engine is working per unit of displacement, so two engines of different capacity can be judged on the same scale. Naturally aspirated petrol engines typically achieve 850–1200 kPa; turbocharged diesels can exceed 2500 kPa. Since BMEP ∝ Torque (for a given engine), the BMEP curve has exactly the same shape as the torque curve, simply scaled by \(4\pi / V_d\).
where \(\dot{m}_f\) is the fuel mass flow rate [g/s] and \(P\) is brake power [kW]. BSFC is expressed in g/kWh. A lower BSFC indicates a more fuel-efficient engine.
BSFC forms a characteristic U-shaped curve against RPM. At low RPM, mechanical friction losses consume a relatively large fraction of the indicated work, raising BSFC. At very high RPM, pumping losses and reduced time for complete combustion increase fuel consumption. The minimum BSFC — the engine's “sweet spot” — occurs near the RPM of peak BMEP. Modern diesel engines achieve BSFC below 210 g/kWh; petrol engines typically range from 250–350 g/kWh at best efficiency.
The denominator is the theoretical mass of air that would fill the displacement at ambient density \(\rho_{air} \approx 1.225\;\text{kg/m}^3\), once every two crankshaft revolutions (one complete cycle for a 4-stroke engine). \(\eta_v\) is dimensionless, typically expressed as a percentage.
Volumetric efficiency peaks at an intermediate RPM where intake tuning (Helmholtz resonance and pressure wave effects) enhances cylinder filling. At low speeds, residual exhaust gas dilutes the fresh charge; at very high speeds, flow inertia and throttling losses restrict inflow. Turbocharging forces additional air mass into the cylinder, allowing \(\eta_v\) (relative to ambient air) to exceed 100%.
Select an engine type and adjust its parameters. The top panel shows the Torque and Power curves. The lower panel shows BMEP, BSFC, and ηv on fixed axes, so the absolute magnitudes stay comparable as you change engine type and settings — a weaker engine's curves sit visibly lower instead of being rescaled to fill the panel. Tick the Auto-scale axes box to rescale to the current curves. Use the RPM Cursor for actual values in the infobox.
Compare a turbocharged engine against a naturally aspirated baseline on the same fixed axes. Solid red/blue lines show the NA engine; dashed orange/cyan lines show the turbocharged engine. The key trade-off: a smaller turbocharged engine can match or exceed the peak torque of a much larger NA unit — this is the principle behind modern engine downsizing.
A turbocharger recovers waste exhaust enthalpy and uses it to drive a compressor, forcing additional air mass into the cylinders. Both petrol and diesel engines benefit, but in very different ways and with different challenges.
| Concept | Equation | Notes |
|---|---|---|
| Brake power from torque | \(\displaystyle P = \frac{T \cdot N}{9549}\) | \(P\) in kW, \(T\) in N·m, \(N\) in RPM |
| BMEP (4-stroke engine) | \(\displaystyle \text{BMEP} = \frac{4\pi T}{V_d}\) | Result in Pa; \(V_d\) in m³ |
| Torque from BMEP | \(\displaystyle T = \frac{\text{BMEP} \cdot V_d}{4\pi}\) | Useful for sizing engines to a target BMEP |
| Brake Specific Fuel Consumption | \(\displaystyle \text{BSFC} = \frac{\dot{m}_f}{P}\) | Units: g/kWh; minimum near peak BMEP |
| Volumetric efficiency | \(\displaystyle \eta_v = \frac{\dot{m}_{air}}{\rho_{air}\, V_d\, N/(2 \times 60)}\) | 4-stroke; \(\rho_{air} \approx 1.225\;\text{kg/m}^3\) |
| Specific power | \(\displaystyle P_{sp} = \frac{P_{\max}}{V_d}\) | Units: kW/L; used to compare engine designs |
A 4-stroke petrol engine has a displacement of 2.0 L. At 4000 RPM it operates at a BMEP of 1100 kPa.
Find: (a) the brake torque at this operating point; (b) the brake power.
Show Solution
Given: \(V_d = 2.0\;\text{L} = 2.0 \times 10^{-3}\;\text{m}^3\), \(\text{BMEP} = 1100\;\text{kPa} = 1.100 \times 10^6\;\text{Pa}\), \(N = 4000\;\text{RPM}\)
(a) Brake Torque:
\[ T = \frac{\text{BMEP} \cdot V_d}{4\pi} = \frac{1.100 \times 10^6 \times 2.0 \times 10^{-3}}{4\pi} = \frac{2200}{12.566} = 175.1\;\text{N·m} \](b) Brake Power:
\[ P = \frac{T \cdot N}{9549} = \frac{175.1 \times 4000}{9549} = \frac{700\,400}{9549} = 73.3\;\text{kW} \]Verification: \(P = \text{BMEP} \cdot V_d \cdot \frac{N}{2\times60} = 1.1\times10^6 \times 2\times10^{-3} \times \frac{4000}{120} = 2200 \times 33.33 = 73{,}333\;\text{W} \approx 73.3\;\text{kW}\) ✓
Answer: (a) \(T = 175.1\;\text{N·m}\); (b) \(P = 73.3\;\text{kW}\)
An engine produces 85 kW of brake power while consuming fuel at a mass flow rate of 6.5 g/s. Calculate the BSFC at this operating point.
Show Solution
Given: \(P = 85\;\text{kW}\), \(\dot{m}_f = 6.5\;\text{g/s}\)
Formula:
\[ \text{BSFC} = \frac{\dot{m}_f}{P} \]Converting \(\dot{m}_f\) to g/h: \(6.5\;\text{g/s} \times 3600\;\text{s/h} = 23{,}400\;\text{g/h}\)
\[ \text{BSFC} = \frac{23{,}400\;\text{g/h}}{85\;\text{kW}} = 275\;\text{g/kWh} \]This is a typical value for a well-optimised naturally aspirated petrol engine operating near its peak efficiency island.
Answer: \(\text{BSFC} = 275\;\text{g/kWh}\)
Two engines are being evaluated for a sports car application:
Engine B produces more torque, but which engine uses its displacement more effectively? Calculate the peak BMEP for each.
Show Solution
Engine A: \(V_{d,A} = 1.6\;\text{L} = 1.6 \times 10^{-3}\;\text{m}^3\), \(T_A = 200\;\text{N·m}\)
\[ \text{BMEP}_A = \frac{4\pi \times 200}{1.6 \times 10^{-3}} = \frac{2513.3}{1.6 \times 10^{-3}} = 1{,}570{,}796\;\text{Pa} \] \[ \text{BMEP}_A = 1571\;\text{kPa} \]Engine B: \(V_{d,B} = 2.5\;\text{L} = 2.5 \times 10^{-3}\;\text{m}^3\), \(T_B = 280\;\text{N·m}\)
\[ \text{BMEP}_B = \frac{4\pi \times 280}{2.5 \times 10^{-3}} = \frac{3518.6}{2.5 \times 10^{-3}} = 1{,}407{,}434\;\text{Pa} \] \[ \text{BMEP}_B = 1407\;\text{kPa} \]Conclusion: Despite producing less absolute torque, Engine A achieves a higher BMEP (1571 kPa vs 1407 kPa). It extracts more work per unit of displacement — it is the more intensively developed engine. Engine A is likely turbocharged or achieves higher volumetric efficiency.
Answer: \(\text{BMEP}_A = 1571\;\text{kPa}\), \(\text{BMEP}_B = 1407\;\text{kPa}\) — Engine A uses its displacement more effectively.
A 4-stroke, 2.0 L engine running at 3000 RPM draws in air measured at 50 g/s. Ambient air density is 1.225 kg/m³. Calculate the volumetric efficiency.
Show Solution
Given: \(\dot{m}_{air} = 50\;\text{g/s} = 0.050\;\text{kg/s}\), \(N = 3000\;\text{RPM}\), \(V_d = 2.0\;\text{L} = 2.0 \times 10^{-3}\;\text{m}^3\), \(\rho_{air} = 1.225\;\text{kg/m}^3\)
Theoretical air mass flow rate (4-stroke: one induction stroke per 2 revolutions):
\[ \dot{m}_{theory} = \rho_{air} \cdot V_d \cdot \frac{N}{2 \times 60} = 1.225 \times 2.0 \times 10^{-3} \times \frac{3000}{120} \] \[ \dot{m}_{theory} = 1.225 \times 2.0 \times 10^{-3} \times 25 = 0.06125\;\text{kg/s} \]Volumetric Efficiency:
\[ \eta_v = \frac{\dot{m}_{air,\,\text{actual}}}{\dot{m}_{theory}} = \frac{0.050}{0.06125} = 0.816 = 81.6\% \]This is a reasonable value for a naturally aspirated engine at moderate speed; peak volumetric efficiency typically occurs at a higher RPM where intake tuning is most effective.
Answer: \(\eta_v = 0.816 = 81.6\%\)