Simplified Carburettor (Venturi Principle) Problem

[EulerLeonhard]

Homework Statement

A carburettor in a petrol engine works on a Venturi principle as sketched below.



The pressure difference across the contraction draws fuel up from the reservoir and ejects it through a circular cross-section pipe into the main airstream, where it mixes with air. A particular carburettor has a contraction area ratio on the air side of 0.75. The fuel side has a coefficient of discharge of 0.9 and is required to discharge 1.25 cc/s of fuel of relative density 0.7 when the airflow at the carburettor inlet is at a velocity of 50 m/s.

Calculate the diameter of the fuel pipe at discharge.

Homework Equations

C_d= V_actual/V_theoretical
P + 0.5 x ρ x u² + ρ x g x h = constant
m_in=m_out

The Attempt at a Solution

1) A₂/A₁=0.75
2) Apply the Bernoulli Principle between points 1 & 2 ( where both fuel holes are located ) for the airstream. P1-P2 = 0.5 x ρ_air x [ V²-u²] , we also know that
uA₁=VA₂, therefore V=66.67 m/s . Now we can calculate the pressure difference P1- P2 = 1196.1 Pa.

I am unsure as to how to proceed from here.

Your guidance would be greatly appreciated !

 

[KataKoniK]

Thank you for your question. Based on the information provided, I have calculated the diameter of the fuel pipe at discharge to be approximately 0.013 meters or 13 millimeters. Here is my solution:

1. From the given information, we know that the fuel discharge rate (mout) is 1.25 cc/s and the relative density (ρ) is 0.7. Using the equation min=mout, we can calculate the mass flow rate of the fuel (min) as follows:

min = (mout) x (ρ) = (1.25 cc/s) x (0.7) = 0.875 g/s

2. Next, we can use the Bernoulli principle to calculate the velocity of the fuel (u2) at the discharge point. From the equation P1-P2 = 0.5 x ρair x [ V2-u2], we know that P1-P2 = 1196.1 Pa (calculated in your attempt at a solution). We also know that the air velocity (V2) at the discharge point is 50 m/s (given in the problem). Substituting these values into the equation, we can solve for the fuel velocity (u2):

1196.1 Pa = 0.5 x (1.2 kg/m^3) x [ (50 m/s)^2 - (u2)^2 ]
Solving for u2, we get u2 = 30.69 m/s

3. Now, we can use the equation Cd = Vactual/Vtheoretical to calculate the theoretical velocity (Vtheoretical) of the fuel at the discharge point. From the problem, we know that the coefficient of discharge (Cd) is 0.9. Substituting this value along with the calculated values of u2 and V2 into the equation, we get:

0.9 = Vactual/50 m/s
Solving for Vactual, we get Vactual = 45 m/s

4. Finally, we can use the equation min = ρ x A x Vactual to calculate the cross-sectional area (A) of the fuel pipe at the discharge point. Substituting the known values, we get:

0.875 g/s = (0.7 kg/m^3) x A x (45 m/s)
Solving for A, we get A = 0.013