---
title: "Jet diameter of a simple carburettor"
subject: "Internal Combustion Engines"
chapter: 4
key: "semester_6/internal_combustion_engines:simple-carburettor-jet-size-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/simple-carburettor-jet-size-numerical
site: BCT Engineering
---

# Jet diameter of a simple carburettor

Internal Combustion Engines · Chapter 4: Carburetor & Fuel Injection systems

- Asked in: year not given, 6 marks
- Video: [Problem 1 on Carburetor,  IC engine,  Simple jet carburetor, throat diameter, orifice diameter](https://www.youtube.com/watch?v=jBMfM7v1W5k) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/simple-carburettor-jet-size-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-4

## Question

A simple carburettor has a venturi throat of diameter 32 mm. Air at 101.3 kPa and 300 K flows through it and the pressure depression at the throat is 5 kPa. The coefficient of discharge for air is 0.85 and for the fuel jet 0.65. Petrol density is 740 kg/m$^3$. Find the air flow rate and the diameter of the fuel jet needed to give an air-fuel ratio of 15:1. Neglect the nozzle lip and the compressibility of air.

## Short answer

Air flow = 0.0742 kg/s (267 kg/h). Required fuel jet diameter = 1.89 mm.

## Answer

Given: $d_t = 0.032$ m, $p = 101.3$ kPa, $T = 300$ K, $\Delta p = 5000$ Pa, $C_{da} = 0.85$, $C_{df} = 0.65$, $\rho_f = 740$ kg/m$^3$.

#### Air density and throat area
$$
\begin{aligned}
\rho_a &= \frac{p}{RT} = \frac{101\,300}{287 \times 300} = 1.177\ \text{kg/m}^3 \\
A_t &= \frac{\pi}{4}(0.032)^2 = 8.042 \times 10^{-4}\ \text{m}^2
\end{aligned}
$$

#### Air flow
$$
\begin{aligned}
\dot{m}_a &= C_{da} A_t \sqrt{2 \rho_a \Delta p} \\
&= 0.85 \times 8.042 \times 10^{-4} \times \sqrt{2 \times 1.177 \times 5000} \\
&= 0.85 \times 8.042 \times 10^{-4} \times 108.48 = 0.0742\ \text{kg/s}
\end{aligned}
$$

This is 267 kg/h. The throat velocity is $\sqrt{2\Delta p/\rho_a} = 92.2$ m/s, which is well below sonic, so the incompressible assumption is acceptable.

#### Fuel flow for A/F = 15
$$
\dot{m}_f = \frac{0.0742}{15} = 4.944 \times 10^{-3}\ \text{kg/s}
$$

#### Jet area and diameter
$$
\begin{aligned}
A_f &= \frac{\dot{m}_f}{C_{df}\sqrt{2\rho_f \Delta p}} = \frac{4.944 \times 10^{-3}}{0.65 \times \sqrt{2 \times 740 \times 5000}} \\
&= \frac{4.944 \times 10^{-3}}{0.65 \times 2720.3} = 2.796 \times 10^{-6}\ \text{m}^2 \\
d_f &= \sqrt{\frac{4 A_f}{\pi}} = 1.887 \times 10^{-3}\ \text{m}
\end{aligned}
$$

**Answer:** air flow $= 0.0742$ kg/s (267 kg/h); fuel flow $= 0.00494$ kg/s (17.8 kg/h); jet diameter $\approx 1.89$ mm.

## Check yourself

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

1. How is air density found for the carburettor air flow?
   A. rho = p/(RT)
   B. rho = RT/p
   C. rho = p x T
   D. rho = p/R
   Answer: A. Ideal gas: 101325/(287 x 300) = 1.177 kg/m^3. (Key point: Air density)

2. Which equation gives air mass flow?
   A. Cda At sqrt(dp / rho_a)
   B. Cda At sqrt(2 rho_a dp)
   C. Cda At rho_a dp
   D. At sqrt(2 dp)
   Answer: B. Neglecting compressibility, m_a = Cd A sqrt(2 rho dp) = 0.0742 kg/s. (Key point: Air flow equation)

3. How is the fuel flow rate found for A/F = 15?
   A. m_f = 15 x m_a
   B. m_f = m_a x 1.2
   C. m_f = m_a / 15
   D. m_f = m_a / 740
   Answer: C. A/F is air mass over fuel mass, so m_f = 0.0742/15 = 0.00494 kg/s. (Key point: Fuel flow)

4. Why does the jet area formula use petrol density 740 rather than air density?
   A. The air density is unknown
   B. The jet carries air
   C. Petrol is compressible
   D. The fluid through the jet is fuel, with the same pressure depression
   Answer: D. A_f = m_f /(Cdf sqrt(2 rho_f dp)) with rho_f = 740 gives d about 1.89 mm. (Key point: Jet area)

## Related in Internal Combustion Engines

- [Simple carburettor and its limitations](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Asimple-carburettor-working-compensation) — Internal Combustion Engines, chapter 4
- [Volumetric efficiency from orifice air measurement](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Avolumetric-efficiency-orifice-air-consumption) — Internal Combustion Engines, chapter 8
- [IP, BP and efficiencies of a petrol engine](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Aindicated-brake-power-efficiency-numerical) — Internal Combustion Engines, chapter 8
- [Fuel-air cycle and actual cycle losses](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Afuel-air-cycle-assumptions-losses) — Internal Combustion Engines, chapter 1
- [Efficiency ratio and specific fuel consumption](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Arelative-efficiency-sfc-numerical) — Internal Combustion Engines, chapter 8
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