Design of 4 stroke IC engine from few parameters (bmep, B.P, piston speed)

[marellasunny]

(I have trouble in finding the stroke length and fuel flow rate)
From Heywood,E2.11
You are designing a 4 stroke CI to provide a brake power of 300kW,Naturally aspirated at its max rated speed.Based on typical values for brake mean eff.pressure(bmep){700kPa to 900kPa} and max mean piston speed,estimate the required engine displacement,and the bore and stroke for sensible cylinder geometry and number of engine cylinders. What is the maximum rated engine speed(rpm) for your design? What would be the brake torque and the fuel flow rate(g/h) at this max speed?Assume a max mean piston speed 12 m/s is typical of good engine designs.

I use the usual equations which relate bmep to max power(i.e max rpm) and bmep to max torque(i.e max bmep)

bmep=Pb(max)∗nR∗10^3/Vd∗N(max) ; nR=2 for 4 strokes

bmep(max)=6.28∗nR∗T(max)/Vd

where, Vd is the displacement volume=number of cylinders*3.14/4*B^2*L

where B-bore,L-stroke

The problem I have is that once I have found the bore dimension,I have no other way of finding the stroke length i.e I just have to assume the stroke=bore and continue the problem. To solve for fuel flow rate,I have to assume that the sfc for NA diesels is usually 200(g/kWh)

 

[SteamKing]

This is one of those things where finding out the bore:stroke ratio of similar CI engines can help. A B:L ratio of 1 is OK, but there have been quite a few CI designs where B:L ratios have been less than one.

 

[marellasunny]

I have attached a scanned note of my calculations on paper. I finally arrive at a realistic value for stroke by assuming it as half of the bore diameter. L=B/2
I conclude this assumption realistic because the max rpm is 1300 for the consideration that it is 6 cylinder and having a 300kW horsepower. Is my methodology correct?
I am clueless to find the stroke length any other way than assuming values for the number of cylinders and max rpm range.

Please offer corrections!

 

[Ranger Mike]

Stroke length

L=B x .75 is good start
classic 302 CID V8 was 4 inch bore with 3 inch stroke. Very good performer.

 

[SteamKing]

I have attached a scanned note of my calculations on paper. I finally arrive at a realistic value for stroke by assuming it as half of the bore diameter. L=B/2
I conclude this assumption realistic because the max rpm is 1300 for the consideration that it is 6 cylinder and having a 300kW horsepower. Is my methodology correct?
I am clueless to find the stroke length any other way than assuming values for the number of cylinders and max rpm range.

Please offer corrections!

If you take a look at the attached brochure for Caterpillar Marine Diesel engines (p.3), you will see that all of these engines have B:L ratios below 1. The reason for this is to maximize torque production at low speeds. Diesels typically operate at lower RPMs than SI engines, thus the longer strokes give a longer time for combustion to take place. While Ford SI engines give good performance, they produce their max. power above 3000 RPM, which is typically outside the operating range of most industrial diesel engines.

In my experience, it is unusual to find a diesel engine with a stroke less than the bore.

 

[SteamKing]

If you take a look at the attached brochure for Caterpillar Marine Diesel engines (p.3), you will see that all of these engines have B:L ratios below 1. The reason for this is to maximize torque production at low speeds. Diesels typically operate at lower RPMs than SI engines, thus the longer strokes give a longer time for combustion to take place. While Ford SI engines give good performance, they produce their max. power above 3000 RPM, which is typically outside the operating range of most industrial diesel engines.

In my experience, it is unusual to find a diesel engine with a stroke less than the bore.

Sorry: I omitted the brochure:

 

[marellasunny]

Thanks.I realize what I did wrong the first time. I took the bmep at max Power too low i.e I used the value of 7bar instead of something like 8 bar(which is realistic for NA diesels). But,what makes my design unrealistic is that the bore dia I obtain is 364mm for a 12 cylinder(an engine as large as this doesn't even exist for marine diesels!). Anyway,I proceed further to state that this would be a undersquare engine(with stroke greater than the bore dia).I state this design because having a oversquare engine with a rated power of only 300kW isn't realistic,the power rating given in the question is too low for a oversquare engine. So,the only possibility is to have a 'undersquare' design which means I get a stroke length of >350mm. I am astonished by these values. Especially when undersquare engines are meant for engines that have a need for higher peak torque. I arrive at a max rpm value of 791 rpm,again not realistic.Peak torque is a whopping 3600 Nm or 2655 ft-lb.

Have you ever come across such specs in your experience people?

 

[SteamKing]

It comes down to the operating speed of the engine.

Large 2-stroke, low speed marine diesel engines are gigantic:
https://en.wikipedia.org/wiki/Wärtsilä-Sulzer_RTA96-C

A person would have no problem standing inside one of the cylinders of this engine.

 

[jack action]

I have attached a scanned note of my calculations on paper. I finally arrive at a realistic value for stroke by assuming it as half of the bore diameter. L=B/2
I conclude this assumption realistic because the max rpm is 1300 for the consideration that it is 6 cylinder and having a 300kW horsepower. Is my methodology correct?
I am clueless to find the stroke length any other way than assuming values for the number of cylinders and max rpm range.

Please offer corrections!

Thanks.I realize what I did wrong the first time. I took the bmep at max Power too low i.e I used the value of 7bar instead of something like 8 bar(which is realistic for NA diesels). But,what makes my design unrealistic is that the bore dia I obtain is 364mm for a 12 cylinder(an engine as large as this doesn't even exist for marine diesels!).


Anyway,I proceed further to state that this would be a undersquare engine(with stroke greater than the bore dia).I state this design because having a oversquare engine with a rated power of only 300kW isn't realistic,the power rating given in the question is too low for a oversquare engine. So,the only possibility is to have a 'undersquare' design which means I get a stroke length of >350mm. I am astonished by these values. Especially when undersquare engines are meant for engines that have a need for higher peak torque. I arrive at a max rpm value of 791 rpm,again not realistic.Peak torque is a whopping 3600 Nm or 2655 ft-lb.

Have you ever come across such specs in your experience people?

Your problem in your calculations is the 10³ you are introducing. You shouldn't have it. This way your answer will be in m² instead of dm² (which is a wrong assumption by the way). You have to actually multiply your bore in mm by 0.31623 to get the right answer [which corresponds to 1000 / 100 / sqrt(1000)]. So instead of 364 mm, the proper answer would be 115 mm for a 12-cylinder and 174 mm instead of 550 mm for a 6-cylinder.

By using this calculator, you can find the answer in the proper units easily. For a 12-cylinder with BMEP = 8 bar and MPS = 12 m/s, I find a torque of 845 lb.ft @ 2500 rpm, a bore X stroke of 115 mm X 144 mm (assuming B/S = 0.8) and a total displacement of 18 L.

 

[marellasunny]

You have to actually multiply your bore in mm by 0.31623 to get the right answer [which corresponds to 1000 / 100 / sqrt(1000)].

Why did you multiply by 0.31623? Is this formula of 1000/100/sqrt(1000) designed to serve a particular principle of engine design?

 

[jack action]

Why did you multiply by 0.31623? Is this formula of 1000/100/sqrt(1000) designed to serve a particular principle of engine design?

No, it is just to «convert» it to proper units.

To convert your «wrong» answer (B_wrong) to the «right» answer (B_right), starting from your final answer, going backward:

1. Your bore B of 364 mm must be converted to your assumed 3,64 dm ( = B_wrong/100 dm)
2. Then, it has to be squared to go back into the B² X n equation. So it becomes (B_wrong/100)²
3. This is where you've introduced incorrectly a 1000 multiplier that you have to remove. So it becomes (B_wrong/100)²/1000. At this point your displacement will be in the correct unit of m³.
4. Now to find again your bore dimension, you will need to go from B² to B, so it becomes sqrt((B_wrong/100)²/1000) or B_wrong/100/sqrt(1000).
5. Finally, you multiply it by 1000 to convert your m into mm. So you get B_wrong*1000/100/sqrt(1000).

Hence B_right = B_wrong*1000/100/sqrt(1000)

Or B_right = B_wrong* 0.31623

Improper unit conversion can make weird and unexpected results.

 

[Aybi]

No, it is just to «convert» it to proper units.

To convert your «wrong» answer (B_wrong) to the «right» answer (B_right), starting from your final answer, going backward:

1. Your bore B of 364 mm must be converted to your assumed 3,64 dm ( = B_wrong/100 dm)
2. Then, it has to be squared to go back into the B² X n equation. So it becomes (B_wrong/100)²
3. This is where you've introduced incorrectly a 1000 multiplier that you have to remove. So it becomes (B_wrong/100)²/1000. At this point your displacement will be in the correct unit of m³.
4. Now to find again your bore dimension, you will need to go from B² to B, so it becomes sqrt((B_wrong/100)²/1000) or B_wrong/100/sqrt(1000).
5. Finally, you multiply it by 1000 to convert your m into mm. So you get B_wrong*1000/100/sqrt(1000).

Hence B_right = B_wrong*1000/100/sqrt(1000)

Or B_right = B_wrong* 0.31623

Improper unit conversion can make weird and unexpected results.

Hi can you share the final corrected document?

 

[berkeman]

Welcome to PF.

Hi can you share the final corrected document?

Do you have this schoolwork assignment as well? If so, we probably should not just give you the answer. Can you post what you think the corrected version should be?