Thermodynamics: Otto Cycle

In summary: For homework, use the homework help forums. Otherwise, use the engineering forum (or thermal engineering, etc).In summary, the conversation discusses an air-standard Otto cycle with a cold cycle assumption and properties evaluated at 300 K. The air intake and exhaust are also at 300 K and 1 bar, with a compression ratio of 9 and a volume of 1 liter at bottom dead center. The conversation includes questions about the mass of air in the cycle, work produced during expansion, work consumed during compression, energy transferred during heating, and the thermal efficiency of the cycle. The equations and attempts at solving the problem involve using isentropic processes, constant volume processes, and specific volume to determine temperature and internal energy. The
  • #1

Homework Statement

Analyze an air-standard Otto cycle assuming it is a cold cycle with properties evaluated at 300 K. Air intake and exhaust are at 300 K, 1 bar. After the combustion (heat in) process, the temperature is 2500K. The compression ratio is 9. The volume at bottom dead center is 1 liter.

a. What is the mass of air in the cycle?

b. How much work is produced during expansion?

c. How much work is consumed during compression?

d. How much energy is transferred into the air during heating?

e. What is the thermal efficiency of the cycle?

Homework Equations

The biggest problem I am having is identifying which temperature and pressure corresponds to what process. Does air intake and exhaust mean that T1 = 300K?

1 to 2: Isentropic Compression (BDC to TDC)
2 to 3: Constant Vol. heat transfvr to air while piston at TDC.
3 to 4: Isentropic expansion
4 to 1: Constant Vol. where heat rejected from air at BDC

Also, I can do c,d,e once a and b are known, but I need help with those.

The Attempt at a Solution

b) efficiency of turbine n_t = (h1-h2)/(h1-h2s)
h2 = h1 - n_t(h1-h2s)
= 2804.2 -0.75(2804.2 - 2084.63) = 2264.5225
Power of pump Wp_dot = m_dot*(h4-h3)
h4 = h3 + Wp_dot/m_dot = h3 + v3(p4-p3)/n_p

Where n_ p = pump efficiency
h = enthalpy
p = pressure
m_dot = mass flow rate
v = specific volume
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  • #2
can anyone please help me get started?
  • #3
Just a few questions for you to think about (well and for me to know where you are)

Have you drawn the cycle out on paper?
Remember this is air standard - so what equations are you using?
Why are you using a turbine in your attempt at a solution?
Why are you using a mass flow rate - what type of system is an otto cycle?
Remember cycle efficiency is not the same as component efficiency.

Think about these and let me know what you think.
  • #4
Yeah I have it drawn out on paper.

For air standard: W_12 = u2 -u1
W_34 = u3-u4
Q_23 = u3-u2
Q_41 = u4-u1
Thermal efficiency n = 1 - [(u4-u1)/(u3-u2)]
For isentropic processes: v_r2 = (V2/V1)*v_r1
v_r4 = (V4/V3)*v_r3
where v_r is the reduced specific volume, which i can use to get the temperature and specific internal energy
For constant Volume: p3 = p2*(T3/T2)
p4 = p1*(T4/T1)

In my book, the cycle consists of a pump, boiler, turbine, and condenser. I am using n_t to get h2.

Isnt the mass flow rate constant throughout the process?
  • #5
A pump,boiler, turbine, and condenser is a Rankine/Carnot cycle that typically tuns on steam.

The Otto cycles are the 4 stroke spark ignition cycle that can have an air standard equivilant.

There is a rather major fundamental mistake in something here because you are trying to solve a completely different cycle than you have the information for. Some of your equations are based on the ideal gas law and some are based on steam cycle principals.

If this doesn't solve your problems I would suspect that there is something majorly wrong with the question (info) could you give scan of it?
  • #6
Oh sorry, my mistake. You were right. I typed in something i didnt meant to. The question is correct. I am still having trouble determining which is which temperature, pressure, etc...
Here are my attempts:

Assuming T1 = 300K, u1 = 214.07 kJ/kg, v_r1 = 621.2

Then v_r2 = v_r1/r = 69.02

Interpolating: T2 = 702.75 K, u2 = 514.5 kJ/kg

mass m = (p1*V1)/[(R/MW)T1] = 1.16g, R = 8.3145

W_cycle = m[(u3-u4) - (u2-u1)], need u3 and u4

T2/T1 = (V1/V2)^(k-1), k= 1.4

V2 = 0.00499 L

p2 = p1[(T2*V1)/(T1*V2)] = 1670.676 bar
p3 = p2(T3/T2)
  • #7
The internal energy can be found from CvT (specific heat at const vol)

and Cp/Cv = k

This would allow you to find Q in by m*cv*delta_T

and this in turn would allow you to find u3 as you know u2.

EDIT: This is all from memory and I've not done this in quite a while so I may be wrong.

I believe this stuff is in Thermodynamics - cengel and boles. I don't have my books with me so I can't check.

Also I think your V2 value is a bit off which is giving you the incorrect pressures for p2 and p3. [Vmax/Vmin is Rc]
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  • #8
What would be by T1,T2,T3, and T4?
  • #9
please tell me how to interpolate in otto cycle to find temperature
  • #10
Please start a new thread to ask your question. From the appropriate forum, just click the "New Topic" [PLAIN] [Broken] button.

Also, which forum is the appropriate one depends on whether this is for a homework assignment or not.
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1. What is the Otto Cycle and how does it work?

The Otto Cycle is a thermodynamic cycle used in internal combustion engines to convert fuel into energy. It consists of four processes: intake, compression, power, and exhaust. During the intake process, the fuel-air mixture is drawn into the engine. In the compression process, the mixture is compressed by the piston. The power process is where the fuel is ignited, causing an explosion and pushing the piston down. Finally, the exhaust process releases the waste products from the engine.

2. What is the purpose of the Otto Cycle?

The purpose of the Otto Cycle is to convert chemical energy from fuel into mechanical energy that can be used to power an engine. This cycle is commonly used in automobiles and other internal combustion engines to generate power and propel the vehicle.

3. What are the key factors that affect the efficiency of the Otto Cycle?

The key factors that affect the efficiency of the Otto Cycle include the compression ratio, the temperature and pressure of the intake air, and the fuel-air mixture ratio. A higher compression ratio and intake air temperature can increase efficiency, while a leaner fuel-air mixture can improve fuel efficiency but may decrease power output.

4. How does the Otto Cycle differ from the Diesel Cycle?

The main difference between the Otto Cycle and the Diesel Cycle is the type of fuel used. The Otto Cycle uses a spark ignition, where the fuel is ignited by a spark plug, while the Diesel Cycle uses a compression ignition, where the fuel is ignited by the heat generated from compressing the air in the cylinder. This results in different operating principles and efficiency levels for each cycle.

5. What are the practical applications of the Otto Cycle?

The Otto Cycle has a wide range of practical applications, with the most common being in internal combustion engines for vehicles such as cars, motorcycles, and small aircraft. It is also used in gas turbines and other power generation systems. Additionally, the principles of the Otto Cycle are used in the design and optimization of various heating and cooling systems.

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