---
title: "Air flow and fuel flow of an SI engine"
subject: "Automotive Technology"
chapter: 2
key: "semester_5/automotive_technology:fuel-consumption-air-flow-numerical"
source: https://www.bctengineer.com/q/semester_5/automotive_technology/fuel-consumption-air-flow-numerical
site: BCT Engineering
---

# Air flow and fuel flow of an SI engine

Automotive Technology · Chapter 2: Fuel Supply Systems

- Asked in: year not given, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/automotive_technology/fuel-consumption-air-flow-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-2

## Question

A 1.6 litre, four-cylinder, four-stroke petrol engine runs at 3000 rpm with a volumetric efficiency of 85 %. The intake air density is 1.18 kg/m³ and the brake power is 35 kW. Calculate (i) the mass flow rate of air, (ii) the fuel flow rate for a stoichiometric AFR of 14.7, (iii) the brake specific fuel consumption, and (iv) the fuel flow at full load when the AFR is 12.5 and the equivalence ratio $\lambda$.

## Short answer

Air mass flow = ηv ρ Vd N/(2×60) = 0.85×1.18×0.0016×3000/120 = 0.0401 kg/s (144.4 kg/h). Fuel flow at AFR 14.7 = 144.4/14.7 = 9.83 kg/h. BSFC = 9.83/35 = 0.281 kg/kWh = 281 g/kWh. Full load fuel flow at AFR 12.5 = 144.4/12.5 = 11.55 kg/h, λ = 12.5/14.7 = 0.85.

## Answer

**Given:** $V_d = 1.6\times10^{-3}$ m³, $N = 3000$ rpm, $\eta_v = 0.85$, $\rho_a = 1.18$ kg/m³, $BP = 35$ kW.

#### (i) Air flow

A four-stroke engine inhales once every two revolutions.

$$
\dot m_a = \eta_v \rho_a V_d \frac{N}{2\times 60} = 0.85 \times 1.18 \times 1.6\times10^{-3} \times \frac{3000}{120} = 0.0401\ \text{kg/s}
$$

That is $0.0401 \times 3600 = 144.4$ kg/h.

#### (ii) Fuel flow at $AFR = 14.7$

$$
\dot m_f = \frac{\dot m_a}{14.7} = \frac{144.4}{14.7} = 9.83\ \text{kg/h}
$$

For a petrol density of 740 kg/m³, this is $9.83/0.74 = 13.3$ L/h.

#### (iii) Brake specific fuel consumption

$$
bsfc = \frac{\dot m_f}{BP} = \frac{9.83 \times 1000}{35} = 281\ \text{g/kWh}
$$

#### (iv) Full-load mixture

Taking the same air flow:

$$
\dot m_f = \frac{144.4}{12.5} = 11.55\ \text{kg/h}, \qquad \lambda = \frac{AFR_{actual}}{AFR_{stoich}} = \frac{12.5}{14.7} = 0.85
$$

The mixture is rich: it has 15 % less air than stoichiometric, or about 18 % excess fuel ($\phi = 1/\lambda = 1.18$).

**Answer:** $\dot m_a = 0.0401$ kg/s; $\dot m_f = 9.83$ kg/h at $\lambda = 1$; bsfc $= 281$ g/kWh; full-load fuel $= 11.55$ kg/h with $\lambda = 0.85$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is N divided by 2 x 60 in the air flow formula for the four-stroke engine?
   A. Each cylinder inhales once per two revolutions
   B. The engine has two cylinders
   C. Efficiency is 50 %
   D. Air density halves
   Answer: A. A four-stroke engine inhales once per two revolutions, giving 0.85 x 1.18 x 1.6e-3 x 3000/120 = 0.0401 kg/s. (Key point: Air flow formula)

2. With air flow 144.4 kg/h and AFR 14.7, what is the fuel flow?
   A. About 9.83 kg/h
   B. About 11.55 kg/h
   C. About 2122 kg/h
   D. About 0.68 kg/h
   Answer: A. Fuel flow equals air flow divided by AFR: 144.4/14.7 = 9.83 kg/h. (Key point: Fuel flow at stoichiometric)

3. What is bsfc for 9.83 kg/h and 35 kW?
   A. About 281 g/kWh
   B. About 28.1 g/kWh
   C. About 344 g/kWh
   D. About 0.281 g/kWh
   Answer: A. bsfc = 9.83 x 1000/35 = 281 g/kWh. (Key point: bsfc)

4. At full load AFR = 12.5; what is lambda and what does it mean?
   A. 0.85, mixture is rich
   B. 1.18, mixture is lean
   C. 0.85, mixture is lean
   D. 1.18, mixture is rich
   Answer: A. Lambda = 12.5/14.7 = 0.85; below 1 means rich, with fuel flow 144.4/12.5 = 11.55 kg/h. (Key point: Equivalence ratio)

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