Time-flow Linearity

As the fuel in the injector should start to flow first when the injector opens, I'd say that the total flow per injector cycle is not linear with time, because the first moments of injection would not give you full flow. The question is, this time could be small enough to be negligible. And, if not, the flow doesn't need to be linear, its somehow proportional. The controller will compensate for it based on the lambda sensor reading anyway.

Yes Bhrescobar:

I think the same, it should be an acelerating movement but i supose there is also a resistance force proportional to the the scuare of the velocity, i am trying to make mathematical model of it

Regards

Ciro

What kind of injection you consider? Pressure opening? Piloted by controller?

When you use the word gas is it a short for gasoline or is it a gas?

Depending on the type of injector and used pressures the laws are different. give more input and I could suggest you what you can do for a simulation of the behaviour.

If you want use the PM channel.

Nick Name

Thanks for your comment

I am considering an injector for gasoline which is controlled electicaly via a solenoid drive by the car´s computer.

The pressure as far as i know is in the order of 4 psi and because the pressure control i assume it is constant, in other analisis we can make a model taking into acount the effect of the variation in the pressure.

Thanks for your contribution to this matter

What is the PM channel?

Ciro

1- PM= not prime minister but only private message

2- In the configuration you mention the problem is the solenoid and the way it is piloted since due to its impedance the lifting is not linear. I presume you think about a DC solenoid for those you should consider 1st a delay till the field builds up a force high enough to open and then the time to move the poppet mass+spring mass to the stroke end. A similar process occurs when the current is off. 1st a time is required to decrease the force so that the spring accelerates the poppet's mass and moves it on the seat.

3- During all the time the poppet is not on its seat gasoline can flow through the gap at start under laminar conditions since Reynolds is very low then as a turbulent flow. The flow coefficients are not constant. The quantity of injected gasoline is the integral versus time of the instant flow. The relationship is complex and cannot be simulated very easy but you can use for small gaps a coefficient proportional to Re and in the turbulent flow range consider a constant coefficient of 0.6...0.65.

The flow will be proportional to: Q= α*A(t)*(2*Δp/ρ)^0.5 where

α - flow coefficient

A(t)- opening of valve

Δp - pressure drop at valve seat

ρ- specific mass of fluid.

Using either XLS integration programmes "ad-hoc" or Vissim or Simulink you may simulate the valve solenoid+poppet and determine how the flow depends on opening and how the quantity changes versus time.

Such simulations I did years ago for digital (PWM) control of hydraulic (or pneumatic) cylinders in pressure or displacement.

Dear nick name:

This was a wise analisis, but i think the opennig time should be very short (the open time is between 1 to 5 ms and the opening time should be much shorter than that, "i guess") and we perhaps can just forget about that, in that case only we can take into acount the "inercial" and use the newton`s low to solve the problem, but i think we must take the resistance to the flow that i guess is proportional to the scuare of the velocity, when i put all togheter i found a diferencial ecuation that i am trying to solve in this days, but i am still doubt about what i am doing.

What do you think

Thanks for your interest

Regards

Ciro

send me the equation and i shall tell you if it is ok

Dear Nike Name

1.- I made a simplified injector as a cylindrical tube, on one side we have the pressure and in the other is the combustion chamber, it has a section area of "A" and a length "L"

2.- The pressure, supposed constant, is "P"

3.- The force applied in the pressure side is F= p * A

4.- According to the Newton´s law, the force applied to the mass "M" in the tube produce an acceleration "a" so we can write:

F = M*a= M*V′

5.- If we suppose the is a resistance to the flow "R" equal to the velocity squared then:

R = k * V²

6- If there are no other forces F should be equal to R so:

M*V′ = k *V^{2 were k is a constant}

V′ = k * V² / M

This is the differential equation i found and the initial condition is

V(0) = 0

I am not sure all this is OK, it to much time i left the university

Thanks for your interest

Ciro

Make a sketch of the injector.

1-your equation is only partly valid since the "k" factor is variable and you consider it as constant.

2-pressure is NOT only on one side of the fluid column but everywhere inside the volume and it acts as well at the opening area which let the fluid flow to the following space.

3-the fluid volume should not be considered as a solid body as you do this leads to too big errors. Depending on the ration L/d the fluid has or not be considered as a whole body and the compressibility (bulk modulus with or without wall elasticity correction) has or not be put in the equation.

4-are you sure about the pressure? It seems to me too low. Which are the dimensions of the tube to the injector from the pump? Do you have any accumulator? Internal and external diameter of the injector?

5-from your equation one can deduce that: dV/dt= k*V²/M and dV/V²= k/M*dt I let you integrate the equation and verify if it will fit with your initial condition V=0 at t=0.

If I have the injector sketch I can have a better look at the kind of equation one would need to solve your problem but I do not feel the one you wrote as the solution. To your information I too left the college a very long time ago, better I do not say how many years.

I did my best and am sorry that you did not appreciate my answer as good.

Dear nick name:

Perhaps there is a misunderstanding, i do appreciate a lot your answers, thanks you very, very, very much, maybe my English is not good enough to express my gratefulness.

I will try to get an injector sketch to send it to you.

kind regards

Ciro