Engine Finite Heat Release Model (With Heat Transfer) Help

In summary, Jason has trouble with the Finite Heat Release Equation used to simulate combustion and expansion processes in an internal combustion engine. The equation is as follows: P = Cylinder Pressure (kPa), θ = Crank Angle (Deg), k = Specific Heat Ratio, V = Cylinder Volume (m3), Qin = Heat Input (J), ƒ = mass fraction burned (Wiebe Function, %), h = heat transfer coefficient (W/m2-K), A = Cylinder Wall Surface Area (m2), ω = Engine Speed (rad/s..?), Tg = Cylinder Gas Temperature (K), Tw = Cylinder Wall Temperature (K),
  • #1
Jason Louison
70
6
Hello Physics Forum Users! I have an annoying situation with the Finite Heat Release Equation used to simulate combustion and expansion processes in an internal combustion engine. The equation is as follows:
Screen Shot 2018-02-05 at 3.11.48 PM.png

Nomenclature:

P = Cylinder Pressure (kPa)
θ = Crank Angle (Deg)
k = Specific Heat Ratio
V = Cylinder Volume (m3)
Qin = Heat Input (J)
ƒ = mass fraction burned (Wiebe Function, %)
h = heat transfer coefficient (W/m2-K)
A = Cylinder Wall Surface Area (m2)
ω = Engine Speed (rad/s..?)
Tg = Cylinder Gas Temperature (K)
Tw = Cylinder Wall Temperature (K)
I do not know why π/180 is there, maybe someone can explain it to me?
of course, df/dθ and dV/dθ are rate of change equations for Cylinder Volume and Mass Fraction Burned.

So, here's where things get a little weird. To obtain cylinder pressure, I Calculate dP/dθ, then multiply it by (1/720), and then add it to the initial pressure, p(θ-1). Yeah I know, it doesn't make much sense, in fact that right there may be my biggest mistake. But If I were to multiply dP/dθ by dθ, my data would be very inconclusive, as dθ for every cell is 1. multiplying the equation by 1 and adding it to the the initial pressure generates very weird data, and, unfortunately, some errors as well. The second issue is that when the engine speed is low, (Below 6500 RPM), The expansion pressure section is very obscure looking:

1000 RPM (Ignition Timing=-28° BTC): Way too much area under curve
Screen Shot 2018-02-05 at 3.04.36 PM.png

8500 RPM (Ignition Timing=-28° BTC): Normal-looking curve
Screen Shot 2018-02-05 at 3.05.21 PM.png


Sources used:
https://www.engr.colostate.edu/~allan/engines.html

I have looked everywhere to try and find out what I am doing wrong, but I found nothing. :(
 

Attachments

  • Screen Shot 2018-02-05 at 3.11.48 PM.png
    Screen Shot 2018-02-05 at 3.11.48 PM.png
    18 KB · Views: 802
  • Screen Shot 2018-02-05 at 3.04.36 PM.png
    Screen Shot 2018-02-05 at 3.04.36 PM.png
    13.7 KB · Views: 413
  • Screen Shot 2018-02-05 at 3.05.21 PM.png
    Screen Shot 2018-02-05 at 3.05.21 PM.png
    12.4 KB · Views: 431
Engineering news on Phys.org
  • #2
Hi, Jason. I just joined this group and spotted an unexpected opportunity to be slightly useful before any chance arises of being a pest later.
I notice that the parameters which you listed are in S.I. units (System Internationale to someone of your ancestry, and thankfully a standard used by a very high proportion of the known universe) with the exceptions of 1): Pressure (It should be the base unit of Pascals) and 2): Crank Angle in Degrees.
Given that Crank Angle is practicably entered as Degrees, the factor of Π/180 converts Degrees to Radians which is the S.I. unit of angle.
The Radian is considered dimensionless, making the units of angular velocity; ω to be [1/seconds] ~ (such that, for example, Torque * angular velocity = Power.
[Newton*metres] x [1/second] = [Joules / second] =[Watts] ).
Perhaps the value in [kPa] which you may have used, together with your mentioned application of the factor: (1 / 720) have caused your unexpected result.
It seems that the "1/720" is a second (partial) adjustment of units to that already made by the "Π/180" (Your question mark after engine speed units suggests doubt).
You might also have to audit further because of the "%" cited in the Mass Fraction. Disrupting factors of 1000 ( [kPa/Pa] ) & (1 / 720 ) & 0.01 ( [ % ] / Ratio ) will certainly produce "very weird data".
 
  • Like
Likes Randy Beikmann
  • #3
DonKiwi said:
Hi, Jason. I just joined this group and spotted an unexpected opportunity to be slightly useful before any chance arises of being a pest later.
I notice that the parameters which you listed are in S.I. units (System Internationale to someone of your ancestry, and thankfully a standard used by a very high proportion of the known universe) with the exceptions of 1): Pressure (It should be the base unit of Pascals) and 2): Crank Angle in Degrees.
Given that Crank Angle is practicably entered as Degrees, the factor of Π/180 converts Degrees to Radians which is the S.I. unit of angle.
The Radian is considered dimensionless, making the units of angular velocity; ω to be [1/seconds] ~ (such that, for example, Torque * angular velocity = Power.
[Newton*metres] x [1/second] = [Joules / second] =[Watts] ).
Perhaps the value in [kPa] which you may have used, together with your mentioned application of the factor: (1 / 720) have caused your unexpected result.
It seems that the "1/720" is a second (partial) adjustment of units to that already made by the "Π/180" (Your question mark after engine speed units suggests doubt).
You might also have to audit further because of the "%" cited in the Mass Fraction. Disrupting factors of 1000 ( [kPa/Pa] ) & (1 / 720 ) & 0.01 ( [ % ] / Ratio ) will certainly produce "very weird data".
Haha, I know very well that MFB has to be actual percent (1.00 instead of 100). That would result in an astronomical error xD. I have been playing around with the Heat loss term a bit, I have yet to come to a conclusion as to what I am doing wrong, but I'm getting there!
 

Related to Engine Finite Heat Release Model (With Heat Transfer) Help

1. What is the Engine Finite Heat Release Model?

The Engine Finite Heat Release Model is a computational model used to simulate the combustion process in internal combustion engines. It takes into account the dynamics of heat transfer and chemical reactions in the combustion chamber.

2. How does the Engine Finite Heat Release Model work?

The model uses mathematical equations to describe the heat release and energy transfer processes in an engine. It takes into account factors such as fuel injection, ignition timing, and air-fuel mixture to predict the heat release and temperature in the combustion chamber.

3. What are the benefits of using the Engine Finite Heat Release Model?

The model allows for accurate prediction of engine performance, emissions, and efficiency. It also allows for the optimization of engine design and operating conditions to improve overall performance.

4. What types of engines can the Engine Finite Heat Release Model be applied to?

The model can be applied to a wide range of engines, including gasoline, diesel, and natural gas engines. It can also be used for different engine configurations, such as spark ignition and compression ignition engines.

5. How is the Engine Finite Heat Release Model validated?

The model is validated by comparing its predictions to experimental data from actual engine tests. This helps to ensure the accuracy and reliability of the model for various engine conditions and configurations.

Similar threads

Replies
7
Views
2K
  • Mechanical Engineering
Replies
1
Views
1K
  • Mechanical Engineering
Replies
1
Views
1K
Replies
7
Views
6K
  • Mechanical Engineering
Replies
5
Views
35K
Replies
20
Views
1K
Replies
5
Views
3K
  • Mechanical Engineering
Replies
11
Views
14K
Replies
14
Views
3K
  • Mechanical Engineering
Replies
5
Views
3K
Back
Top