# Help with engine displacement volume, fuel flow rate and crank angle

## Homework Statement

An automobile has a 2.5-litre, four-cylinder, homogeneously charged, direct injected, four-stroke diesel engine. The maximum power output of this naturally aspirated engine is 200 HP at 4000 rpm. By turbocharging, the intake pressure is boosted twice as that for the naturally aspirated design; the volumetric efficiency is increased to 110%. The air-fuel ratio is 17:1 for both turbocharged and naturally aspirated designs. What is the displacement volume required for the turbo engine to produce the same power at same rated rpm?

For this engine at 4000 rpm, if the fuel injection started 12 CAD bTDC and lasted for 600μs, at what crank angle did the injection end? Determine the fuel flow rate (cc/min) per cylinder per cycle for the turbocharged engine given that the density of diesel is 850 kg/m3.

## Homework Equations

P=((ηvola*Vsw*N*ηf*Qf)/2)*(1/AFR)

Mod note: Added the following explanation for the variables in this equation.
ηvol=Volumetric efficiency
ρa=density of air
Vsw=Swept Volume
N=engine speed
ηf=combustion of fuel efficiency
Qf=Heating value of fuel
AFR=Air fuel ratio

## The Attempt at a Solution

For the first question I used the equation above. Since AFR ratio, Qf, N, ηf are the same for both they can be crossed out. That leaves a simple equation. Converted the HP into kw so 200HP=149.1KW multiplied by 2 gives 298kw=ηvola*Vsw. Plugged in the other known values and got a swept volume of 135.45. Seems a little off might need a little help here.

For the second and third question I'm not even sure where to begin or what equation to use. My professor isn't that great so I'm not left with much and the internet is of little help. I need help with these.

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Mark44
Mentor

## Homework Statement

An automobile has a 2.5-litre, four-cylinder, homogeneously charged, direct injected, four-stroke diesel engine. The maximum power output of this naturally aspirated engine is 200 HP at 4000 rpm. By turbocharging, the intake pressure is boosted twice as that for the naturally aspirated design; the volumetric efficiency is increased to 110%. The air-fuel ratio is 17:1 for both turbocharged and naturally aspirated designs. What is the displacement volume required for the turbo engine to produce the same power at same rated rpm?

For this engine at 4000 rpm, if the fuel injection started 12 CAD bTDC and lasted for 600μs, at what crank angle did the injection end? Determine the fuel flow rate (cc/min) per cylinder per cycle for the turbocharged engine given that the density of diesel is 850 kg/m3.

## Homework Equations

P=((ηvola*Vsw*N*ηf*Qf)/2)*(1/AFR)

## The Attempt at a Solution

For the first question I used the equation above. Since AFR ratio, Qf, N, ηf are the same for both they can be crossed out. That leaves a simple equation. Converted the HP into kw so 200HP=149.1KW multiplied by 2 gives 298kw=ηvola*Vsw. Plugged in the other known values and got a swept volume of 135.45. Seems a little off might need a little help here.

For the second and third question I'm not even sure where to begin or what equation to use. My professor isn't that great so I'm not left with much and the internet is of little help. I need help with these.

I don't know what the variables in your equation represent, so can't offer any help on the first question.

For question 2, the crank is spinning a 4000 revolutions per minute, or 4000/60 = 400/6 = 200/3RPS. This would be 360°/revolution * 200/3 revolutions/sec. For a given cylinder, injection starts at 12 ° bTDC (I'm assuming that CAD = crank angle degrees) and lasts for 600 μsec = 600/1,000,000 seconds. Knowing the crank speed, in °/sec, you can figure out how many degrees the crank will spin while the injector is on, which you can use to determine the crank angle when the injector shuts off.

For question 3, the idea is that the engine is spinning at 4000 RPM, and on each intake stroke (every fourth stroke) you know the volume of air that is being packed into the cylinder and hence the volume of fuel (17:1).

BTW, maybe that ratio is OK for diesels, but it's way too lean for gasoline engines, which typically run around 14.7:1 or so. Running a gasoline engine that lean could lead to serious problems, such as burnt valves and maybe pistons.

Hope this helps.

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
The A/F ratio for diesels will be different than gasoline engines because there is more energy per unit volume of diesel fuel than gasoline and diesel fuel is also slightly denser than gasoline as well.

W.r.t. engine displacement, you have omitted the units for your calculations. This is a serious mistake for any calculation.

Mark44
Mentor
SteamKing might have been referring to the last line below.
syt0x said:
An automobile has a 2.5-litre, four-cylinder, homogeneously charged, direct injected, four-stroke diesel engine.
.
.
.
Plugged in the other known values and got a swept volume of 135.45.
Swept volume means the same thing as displacement. Did you switch to cubic inches in your calculations? If so, your swept volume figure (assuming cu. in.) is low.

I think an equation for the 3rd part of number 2 might be

Power=(volumetric efficiency)*(heating value of fuel)*(mass flow rate of fuel)

then solve for the mass flow rate of fuel?

Mark44
Mentor
I think an equation for the 3rd part of number 2 might be

Power=(volumetric efficiency)*(heating value of fuel)*(mass flow rate of fuel)

then solve for the mass flow rate of fuel?

Where are the other parts in this formula? You can't just throw them away.

There's something you said in post #1 that I don't quite understand.
syt0x said:
Since AFR ratio, Qf, N, ηf are the same for both they can be crossed out. That leaves a simple equation.
What did you do here? If you are calculating the ratio of the power of the turbocharged engine to the power of the normally aspirated engine, that makes sense, but if that's not what you did, then I'm confused.

It looks to me like you took this statement - "the intake pressure is boosted twice as that for the naturally aspirated design" - and doubled the power. I am almost certain that's wrong.

The problem is, intake pressure is not part of the formula. However, volumetric efficiency is. In the OP it is stated that volumetric efficiency increased to 110%, but I don't see what this efficiency is for a normally aspirated engine.

For part one. I set up a ratio V2=(V1*n1)/n2 V1=2.5litre n1=.9 n2=1.10 This gave me a V2 of 2.045 litres, seem correct?

For part 2 did (4000rev/min)*(1min/60sec)*(360deg/1rev)=24000deg/sec
Then I did 24000deg/sec*(600*10^(-6)=14.4 deg
Finally I took -12 deg+14.4 deg to equal 2.4 CAD ATDC Seem correct?

For the third part I'm still stuck. I took the 2.5 litre of air divided that by 4 to get .625 litre of air per cylinder. Divided that by 17(the AFR) to get .03676 litre of fuel per cylinder. Converted that to cm^3 to get 36.76cm^3. 4000rpm/4(because only 1/4 is intake)=1000
So 1000*36.76cm^3=36,760cc/m I didn't use density which the problem specified. What am I doing wrong? Thanks!!

Mark44
Mentor
For part one. I set up a ratio V2=(V1*n1)/n2 V1=2.5litre n1=.9 n2=1.10 This gave me a V2 of 2.045 litres, seem correct?
Can't be. V1 seems to be the displacement of the engine, which doesn't change, so V2 = V1.

I don't have time right now to look over the stuff below, but I'll take another look later on this evening.
For part 2 did (4000rev/min)*(1min/60sec)*(360deg/1rev)=24000deg/sec
Then I did 24000deg/sec*(600*10^(-6)=14.4 deg
Finally I took -12 deg+14.4 deg to equal 2.4 CAD ATDC Seem correct?

For the third part I'm still stuck. I took the 2.5 litre of air divided that by 4 to get .625 litre of air per cylinder. Divided that by 17(the AFR) to get .03676 litre of fuel per cylinder. Converted that to cm^3 to get 36.76cm^3. 4000rpm/4(because only 1/4 is intake)=1000
So 1000*36.76cm^3=36,760cc/m I didn't use density which the problem specified. What am I doing wrong? Thanks!!

And part 1 is asking what is the displacement volume of the turbo engine to produce the same power. What I did isn't correct? The displacement would be lower but produce the same power due to the turbo which is the point of downsizing an engine and adding a turbo to save fuel.

Mark44
Mentor
Now that I understand the problem better, I agree with your answer for part 1 - 2.045L. for the turbo engine. I didn't bother converting the 200 HP to KW - just set the two expressions for power equal, and solve for Vsw for the turbo engine.

One thing that wasn't stated in the original problem was the volumetric efficiency for the non-turbo engine of .9. Was that given and you left it out?

Part 2 looks good as well. At 4000 RPM the crank spins 14.4° in 600 μsec, which puts the crank at 2.4° after TDC.

I'll quit for now and look at what you have for the 3rd part.

• 1 person
Mark44
Mentor
For part 3, you need to convert the fuel density from Kg/m3 to g/cm3. That's not actually that hard.

## 850 \frac{kg}{m^3} = 850 \frac{kg}{m^3} * 10^3 \frac g {kg} * \frac 1 {10^6} \frac{m^3}{cm^3}##

On the right side, everything after the first * is a unit, meaning that is equal to 1 and has no units such as grams, centimeters, etc. Multiplying by 1 is always permitted, and doesn't change the value of what is multiplied by it.

Simplifying the right side above, the kg units cancel, the m3 units cancel, and we are left with units of g/cm3. The number simplifies to 850 * 103 * 10-6 = 850 * 10-3 = .850 in units of g/cm3. With that fuel density, I think you have what you need to finish your calculation.

I did the same sort of thing to convert the crank speed from RPM to °/sec.

$$4000 \frac{rev}{min} = 4000 \frac{rev}{min} * 360 \frac{°}{rev} * \frac{1}{60}\frac{min}{sec}$$
The revolution units cancel, the min units cancel, and we end up with 24,000 °/sec, same as you got.

• 1 person
Thanks for all your help and I appreciate you working through the problem with me instead of just giving me the answer. I actually learned something!

Mark44
Mentor
You're welcome!