Choke

During cold starting period, at low cranking speed and before the engine has warmed up, a mixture much richer than usual mixtures (almost five to 10 times more fuel) must be supplied because a large fraction of the fuel will remain liquid even in the cylinder, and only the vapour fraction can provide a combustible mixture with the air. The most common means of obtaining this rich mixture is by the use of a choke, which is a butterfly-type valve placed between the entrance to the carburettor and the venturi throat as shown in Fig. 10.33. By partially closing the choke, a large pressure drop can be produced at the venturi throat that would normally result from the amount of air flowing through the venturi. This strong suction at the throat will draw large quantities of fuel from the main nozzle and supply a sufficiently rich mixture so that the ratio of evaporated fuel to air in the cylinders is within combustible limits. Choke valves are sometimes made with a spring loaded so that high-pressure drops and excessive choking will not result after the engine has started and has attained a higher speed.

images

Figure 10.32 Acceleration pump

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Figure 10.33 Choke valve with spring-loaded bypass

Some manufacturers make the choke operate automatically by means of a thermostat such that when the engine is cold, the choke is closed by a bimetallic element. After starting and as the engine warms up, the bimetallic element gradually opens the choke to its fully open position.

An alternative to the choke is the provision of auxiliary fuel jets that are opened manually or automatically only as required.

Example 10.1

A single jet carburettor has to supply 5 kg of air per minute. The air is at a pressure of 1.013 bar and at a temperature of 27°C. Calculate the throat diameter of the choke for air flow velocity of 90 m/s. Take the velocity coefficient as 0.8. Assume the flow to be incompressible and isentropic.

Solution

Given that ṁ_a = 5 kg/min, p₁ = 1.013 bar, T₁ = 273 + 27 = 300 K, c₂ = 90 m/s, C_da = 0.8

Applying Bernoulli’s theorem between inlet and throat, we have

Density of air at inlet, ρ_a₁ = images

Density of air at throat, ρ_a₂ = images

Mass flow rate at throat, ṁ_a = ρ_a₂ A₂c₂

throat dia, d₂ = 0.0325 m or 32.5 mm

Example 10.2

A six-cylinder, four-stroke engine, 80 mm in diameter and 120 mm stroke length runs at 3600 rpm. The volumetric efficiency of the engine is 0.8. If the maximum head causing the flow is limited to 11.765 cm of mercury, find the throat diameter of the venturi required. Find the diameter of the nozzle orifice if the desired A: F ratio is 15:1. Take C_da = 0.9, C_df = 0.7, ρ_a = 1.3 kg/m³ and ρ_f = 720 kg/m³.

Solution

Given that i = 6, d = 0.08 m, L = 0.120 m, N = 3600 rpm, η_vol. = 0.8, h_Hg = 11.765 cm

Volume flow rate of air,

Mass flow rate of air, ṁ_a = ρ_aV̇_a = 1.3 × 0.0868 = 0.113 kg/s

h_w = h_Hg × γ_Hg = 11.765 13.6 = 160 of water

Example 10.3

The throat diameter of a carburettor is 80 mm and the nozzle diameter is 5.5 mm.The nozzle lip is 6 mm. The pressure difference causing the flow is 0.1 bar. Find (a) the air-fuel (A:F) ratio supplied by the carburettor neglecting nozzle lip. (b) A:F ratio considering nozzle lip and (c) the minimum velocity of air required to start the fluid flow. Neglect air compressibility. Take C_da = 0.85, C_df = 0.7, ρ_a = 1.2 kg/m³ and ρ_f =750 kg/m³.

Solution

Given that d_a = 80 mm, d_f = 5.5 mm, z = 6 mm, ∆p = 0.1 bar, C_da = 0.85, C_df = 0.7, ρ_a = 1.2 kg/m³, ρ_f = 750 kg/m³

Example 10.4

A 4-stroke petrol engine of 2 litres capacity is to develop maximum power at 4000 rpm. The volumetric efficiency at this speed is 0.75 and the air-fuel ratio is 14:1. The venturi throat diameter is 28 mm. The coefficient of discharge of venturi is 0.85 and that for fuel jet is 0.65. Calculate (a) the diameter of the fuel jet and (b) the air velocity at the throat.

The specific gravity of petrol is 0.76. Atmospheric pressure and temperature are 1 bar and 17°C respectively.

Solution

Given: V_s = 2 litres or 2 × 10−³ m³, N = 4000 rpm, n = 2, η_vol = 0.75, A/F = 14/1, d₂ = 28 mm, C_da = 0.85, C_df = 0.65, ρ_f = 760 kg/m3, p₁ = 1 bar, T₁ = 17 + 273 = 290 K

or d_f = 28 × 0.32733 = 9.165 mm∴ Diameter of fuel jet = 9.165 mm
Volume of air supplied per cycle = stroke volume × η_volV₁ = 2 × 10⁻³ × 0.75 = 1.5 × 10⁻³ m³Mass of this air per cycle at suction condition, = 1.8 × 10⁻³ kg/sMass of air supplied, ṁ_a = A₂ C₂ρ_a2 (ρ_a1 = ρ_a2)or c₂ = 81.2 m/s

Example 10.5

The venturi of a simple carburettor has a throat diameter of 20 mm and the fuel orifice has a diameter of 1.12 mm. The level of petrol surface in the float chamber is 6.0 mm below the throat venturi. Coefficient of discharge for venturi and fuel orifice are 0.85 and 0.78 respectively. Specific gravity of petrol is 0.75. Calculate (a) the air-fuel ratio for a pressure drop of 0.08 bar, (b) petrol consumption in kg/hr and (c) the critical air velocity. The intake conditions are 1.0 bar and 17°C. For air c_p = 1.005 and c_v = 0.718 kJ/kg-K.

Solution

Given: d₂ = 20 mm, d_f = 1.12 mm, z = 6 m, C_da = 0.85, C_df = 0.78, s_f = 0.75, ∆ p = 0.08 bar, p₁ = 1.0 bar, T₁= 290 K, c_p = 1.005 kJ/kg.K, c_v = 0.718 kJ/kg.K

Fuel-air ratio,
Petrol consumption,
Critical air velocity

Example 10.6

The air-fuel ratio of a mixture supplied to an engine by a carburettor is 15. The fuel consumption of the engine is 7.5 kg/h. The diameter of the venturi is 2.2 cm. Find the diameter of fuel nozzle if the lip of the nozzle is 4 mm. Take ρ_f = 750 kg/m³, C_da = 0.82, C_df = 0.7, atmospheric pressure = 1.013 bar and temperature = 25°C. Neglect compressibility effect of air.

Solution

Given that A: F = 15:1, ṁ_f = 7.5 kg/h, d_a = 2.2 cm, z = 4 mm, ρ_f = 750 kg/m³, C_da = 0.82, C_df = 0.7, p = 1.013 bar, T = 273 + 25 = 298 K

Density of air,

Mass flow rate of air,

Example 10.7

A four-stroke petrol engine of 1710 cm³ capacity is to be designed to develop maximum power at 5400 rpm, when the mixture A:F ratio supplied to the engine is 13:1. Two carburettors are to be fixed. The maximum velocity of air is limited to 107 m/s. Find the diameters of venturi and fuel nozzle jet, if η_vol (engine) = 0.7, C_da = 0.85, C_df = 0.66, nozzle lip = 6 mm and ρ_f = 750 kg/m³. Atmospheric pressure and temperature are 1.013 bar and 300 K. Take c_p = 1 kJ/kgK.

Solution

Given that V_e = 1710 cm³, N = 5400 rpm, A:F = 13, c_a = 107 m/s, η_vol = 0.7, C_da = 0.85, C_df = 0.66, z = 6 mm, ρ_f = 750 kg/m³, p = 1.013 bar, T = 300 K, c_p = 1 kJ/kgK

Volume of air supplied to engine at STP,

Air flow through each carburettor,

Mass flow rate of air,

Air velocity through venturi,

Assuming the flow to be isentropic,

Volume of air passing through each carburettor,

V̇₂ = A_ac_aC_da

or d_a = 2 cm

Mass flow rate of fuel per carburettor,

Example 10.8

A 4-stroke petrol engine has a swept volume of 2.0 litres and is running at 4000 rpm. The volumetric efficiency at this speed is 75% and the air-fuel ratio is 14:1. The venturi throat diameter of the carburettor fitted to the engine is 30 mm. Estimate the air velocity at the throat if the discharge coefficient for air is 0.9. The ambient conditions are, pressure = 1.0 bar, temperature = 20°C. Calculate the diameter of the fuel jet if the fuel density is 760 kg/m3. For air c_p= 1.005 kJ/kgK and R = 287 J/kgK. Assume C_df = 1.0.

Solution

Given: n = 2, V_s = 2 litres or 2 × 10⁻³ m³, N = 4000 rpm, η_vol= 0.75, A/F = 14, d₂ = 30 mm, C_da = 0.9, C_df = 1.0, p₁ = 1.0 bar, T₁= 293 K, ρ_f = 760 kg/m³, c_p = 1.005 kJ/kgK, R = 287 J/kgK

Actual volume sucked by the engine is, V₁ =

Air velocity at venturi throat, images

or p₂ = 0.964 bar

∆p = p₁ − p₂ = 1 − 0.964 = 0.036 bar

Mass of air flowing through the venturi/s,

∴ ṁ_f = 0.0408/14 = 2.916 × 10⁻³ kg/s

or d_f = 1.26 × 10⁻³ m or 1.26 mm

Example 10.9

An engine fitted with a single jet carburettor having a jet diameter of 1.25 mm has a fuel consumption of 6 kg/hr. The specific gravity of fuel is 0.7. The level of fuel in the float chamber is 5 mm below the top of the jet when the engine is not running. Ambient conditions are 1 bar and 17°C. The fuel jet diameter is 0.6 mm. The discharge coefficient of air is 0.85. Air-fuel ratio is 15. Determine the critical velocity of flow at throat and the throat diameter. Express the pressure at throat in mm of water column. Neglect compressibility effect. Assume discharge coefficient of fuel flow is 0.60.

Solution

Given: d_j = 1.25 mm, ṁ_j = 6 kg/hr, S_f = 0.7, z = 5 mm, p₁ = 1 bar, T₁ = 273 + 17 = 290 K, d_f = 0.6 mm, C_da = 0.85, A:F = 15, C_df = 0.60.