Hi everyone!

Today I wanna share my knowledge regarding the basics of mechanical engineering wherein one of those basic concepts is the Otto cycle.

*Image Source*

### WHAT IS OTTO CYCLE?

Otto Cycle is the ideal cycle for the spark-ignition engines which is named in honour of the German engineer named Nikolaus Otto who made a very big contribution to the evolution of the internal combustion engines (ICE).

Otto Cycle is a 4-process cycle wherein it is composed of 2 isometric processes (constant volume process) and 2 isentropic processes (constant entropy process). These 4 processes are as follow

- Process 1 – 2: Isentropic Compression (
)*S*_{1}= S_{2} - Process 2 – 3: Isometric Addition of Heat (
)*V*_{2}= V_{3} - Process 3 – 4: Isentropic Expansion (
)*S*_{3}= S_{4} - Process 4 – 1: Isometric Rejection of Heat (
)*V*_{4}= V_{1}

Wherein ** S** stands for entropy and

**stands for volume.**

*V*In order for us to easily understand the concept of those 4 processes we need to make ourselves be familiar of the P-V (Pressure vs. Volume) and T-S (Temperature vs. Entropy) diagrams for this type of cycle, which is shown in the picture below.

*Image Source*

In order to obtain the performance of the Otto Cycle we are bombarded with lots of formula that our professors or instructors instructed as to familiarize or even memorized, which are as follows for each process of the Otto Cycle.

- Process 1 – 2: Isentropic Compression
- This process is also known as the
, wherein the piston of the cylinder compresses the mixture of air and the fuel.*Compression Stroke* - This process is mathematically proven with these following relations:
- Compression Ratio (
)*r*_{k}

- Pressure (
), Volume (*P*) and Temperature (*V*) Relations*T*

- Compression Ratio (

- This process is also known as the

- Process 2 – 3: Isometric Addition of Heat
- This process is also known as the
, wherein the spark plug of the engine ignites the compressed mixture of air and fuel.*Power Stroke* - This process is mathematically proven with these following relations:
- Pressure (
*P*), Volume () and Temperature (*V*) Relations*T*

- Heat Addition, (
)*Q*_{A}

- Pressure (

- This process is also known as the

Process 3 – 4: Isentropic Expansion

- This process is also known as the
wherein the piston of the engine pushes the cylinder up thereby releasing the burned gases.*Exhaust Stroke* - This process is mathematically proven with these following relations:
- Pressure (
), Volume (*P*) and Temperature (*V*) Relations*T*

- Pressure (

- This process is also known as the
Process 4 – 1: Isometric Rejection of Heat

- This process is also known as the
wherein the piston of the cylinder moves down the cylinder thereby resulting to an intake of air.*Intake Stroke* - This process is mathematically proven with these following relations:
- Pressure (
), Volume (*P*) and Temperature (*V*) Relations*T*

- Heat Rejection, (
**Q**)_{R}

- Pressure (

- This process is also known as the

Now here are the other parameters necessary in validating the performance of the Otto Cycle.

Net Work Done (

)*W*_{n}

Cycle Efficiency (

), can be obtained in two ways if both the W*e*_{NET}and Q_{A}are given or if you want it real quick, we can obtain ideal cycle efficiency by using the compression ratio, r_{k}.

Mean Effective Pressure (

)*P*_{ME}

So let us now solve a problem relating to the ideal Otto cycle.

An ideal Otto engine, operating on the hot-air standard with k = 1.34, has a compression ratio of 5. At the beginning of compression the volume is 6 ft

^{3}, the pressure is 13.75 psia and the temperature is 100 degrees Fahrenheit. During the constant-volume heating, 340 BTU are added per cycle. Find the following:

- Clearance (
),c- Temperature at the end of isometric addition of heat or the temperature at the beginning of isentropic expansion (
),T_{3}- Pressure at the end of isometric addition of heat or the pressure at the beginning of isentropic expansion (
),P_{3}- Cycle efficiency (
) and,e- the Mean Effective Pressure (
).P_{ME}

Note: This problem is an exercise obtain from page 114 of the book entitled "Thermodynamics" which was authored by Hipolito Sta. Maria. All solutions and screenshots are made by me in making this article.

In solving for this review problem, we students/engineers need to know and master the 3C’s of Mechanical Engineering namely the: ** Concept, Constants and Conversions.** So we will now start computing for what is asked for this sample problem.

Clearance (

),*c*

Based on the formula in obtaining the compression ratio (), we can use it to get the clearance. Using the formula, deriving the formula so that it now appears to be looking for the clearance and then substituting the data we had:*r*_{k}

We arrived with a clearance of 0.25 or 25%.Temperature at the end of isometric addition of heat or the temperature at the beginning of isentropic expansion (

),*T*_{3}

In obtaining the temperature at the beginning of the isentropic expansion or at the end of isometric addition of heat process (), the first thing to do is to obtain the temperature at the beginning of the isometric addition of heat process which is*T*_{3}. Using the temperature relations of*T*_{2}and*T*_{1}, deriving the formula so that it now appears to be looking for*T*_{2}and then substituting the data we had:*T*_{2}

We arrived with a temperature of 967.92 degrees Rankine, say 968 degrees Rankine.

Next thing to obtain is the mass of the air (), which can be obtained by using the Universal Gas Constant formula and by using the data provided for the start of isentropic compression process.*m*_{a}

wherein R is the univeral gas constant, for air we have 53.34 ft – lb_{f}per unit lb_{m}per unit degree Rankine.

Using the formula and substituting all the data;

we obtained a mass of 0.398 lb_{m}.

Next thing to do is to obtain the value of the specific heat capacity of the air at constant volume and at a k of 1.34. We can obtain theby using the formula for*C*_{}v, here is the solution:*C*_{}v

And thus, we obtain aof 156.88 ft – lb*C*_{v}_{f}per unit lb_{m}per unit degree Rankine.

Finally since we now have the mass of the air (), the specific heat capacity of the air at constant volume and at a k of 1.34 (*m*_{a}), and the temperature at the beginning of isometric addition of heat process, (*C*_{}v) and since we are provided with 340 BTU of energy at the heat addition process. We can now obtain the temperature at the end of isometric addition of heat process by using the formula for calculating*T*_{2}and deriving it so that it appears that it is looking for*Q*_{A}. The solution goes like this:*T*_{3}

And thus, we obtain a temperature of 5210 degrees Rankine which is the same as being 4750 degrees Fahrenheit.Pressure at the end of isometric addition of heat or the pressure at the beginning of isentropic expansion (

),*P*_{3}

In order to obtain the pressure at the beginning of isentropic expansion, the first thing to do is to obtain the pressure at the start of isometric addition of heat process which isand this can be obtain by using the pressure and volume relations during the isentropic compression process. The solution goes like this:*P*_{2}

And we arrived with a pressure of 118.83 psi for P_{2}.

Finally we can now obtain the pressure at the beginning of isentropic expansion or at the end of isometric addition of heat process, P_{3}by applying the relations of temperature and pressure relation during the isometric addition of heat process. The solution goes like this:

Thus, we had obtained a pressure of 639.6 psia for P_{3}.Cycle efficiency (

) and,*e*

Since we are provided with the compression ratio of the ideal Otto engine, we can directly obtained the efficiency by using the formula wherein compression ratio (r_{}k) is mainly used. The solution goes like this:

And thus, we obtained a cycle efficiency of 0.4214 or 42.14%.the Mean Effective Pressure (

).*P*_{ME}

In solving for the mean effective pressure of the ideal Otto engine, we need to obtain the net work done by the engine and since we are provided with the heat added during the isometric addition of heat process and at the same time we have computed the cycle efficiency previously, we directly obtain the net work done by the ideal Otto engine by using the formula and equating it so that it appears that it looks for the net work done. The solution goes like this:

And thus, we arrived an energy of 143.28 BTU for the net work done by the system.

So by now we are a step closer in obtaining the mean effective which can be obtained by utilizing the formula as found below. And since we hadn’t obtained the value of the volume at the end of isentropic compression,. We can obtain it by using the formula for the compression ratio (*V*_{2}) so that we can obtain the ***V*r*_{k}_{2}. And the solution goes like this:

And we are able to obtain a volume of 1.2 ft^{3}. And substituting the data we have, we can now obtain the mean effective pressure and the solution goes like this:

Thus, the mean effective pressure of the ideal Otto engine is 161.3 psia. Notice that I multiplied 778 to the value of the work done to the system, the very reason for that is to convert BTU into ft-lb_{f}since 1 BTU is equal to 778 ft-lb_{f}. Additionally, I multiplied 144 to the difference of V_{1}and V_{2}for the very reason that it is still in the units of ft^{3}and taking into account the ft found in the numerator, it becomes ft^{2}and since 1 ft^{2}is equal to 144 inch^{2}, and since pressure is always expressed in the units of pound per square inch or psi in English system of units.

Before I end this article regarding Otto Cycle, I would like to advice mechanical engineering students to not memorize the formula, the key to mastering the Otto cycle is through mastering the concept by way of familiarizing the P-V and T-S Diagram of the Otto Cycle and mastering the fundamentals of Thermodynamics.

I guess that would be all.

Much love and respect.

Joseph Ace Tigas | @josephace135

Registered Mechanical Engineer (RME)

matkodurko (59)· 4 months agoWow...here, take my like :D I honestly have to say I haven't read it cuz Im saving my brain energy for my upcoming machine learning master thesis session :) But glad that engineering has its place here on Steemit :)

josephace135 (62)· 4 months agoThank you for passin' by bro. I hope you are successful in your upcoming machine learning masteral thesis. all hail to engineering people @matkodurko.

reewritesthings (53)· 4 months agoWow I can see the effort you've poured into this post. Too bad all the information here can't be registered to my tiny little brain.

josephace135 (62)· 4 months agoYeah, i got exhausted after I finished composing this article. HAHAHAHAHA writing is very hard as compared to solving. Ahh @reewritesthings we are all unique, you too have your own forte. Cheer up.

reewritesthings (53)· 4 months agoTrue! I look forward to learning more from you. :)

josephace135 (62)· 4 months agoohh I felt pressured with that statement @reewritethings

reewritesthings (53)· 4 months agohhahaha don't be

josephace135 (62)· 4 months agookay lemme handle this and give me enormous amount of time. i lost my speed in solving problems like this one hahaha

bssman (61)· 3 months agoPlease how did you write your formulas in markup?

josephace135 (62)· 3 months ago@bssman, those are screenshots, I used the equation features in MS Word Application, i don't know how to make mathematical equations using markdown.

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