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Power cycles basics
The device which produces net power output is called an ENGINE. The thermodynamic cycles on which the engines work are called POWER CYCLES.
“Thermodynamic cycle consists of a linked sequence of thermodynamic processes that involve transfer of heat and work into and out of the system, while varying pressure, temperature, and other state variables within the system, and that eventually returns the system to its initial state.”
In the process of passing through a cycle, the working fluid (system) may convert heat from a warm source into useful work, and dispose of the remaining heat to a cold sink, thereby acting as a heat engine. POWER CYCLES work continuously to generate the output power.
An important application of thermodynamics is the analysis of power cycles through which the energy absorbed as heat can be continuously converted into mechanical work . A thermodynamic analysis of the heat engine cycles provides valuable information regarding the design of new cycles or for improving the existing cycles.
These four processes make a cycle..
1.Compression
2. Heat Addition
3.Expansion
4.Heat rejection
Compression and expansion are opposite to each other, Whereas heat addition and heat rejection are opposite to each other. Compression and Expansion for all processes are isentropic , because it is the ideal one for both and actual process is somewhat close to it.
In this chapter of Power cycles we will study in details following five cycles
1) Carnot cycle ( Isothermal cycle)
2) Otto cycle (Constant volume cycle)
3) Brayton cycle (Constant Pressure cycle)
4) Diesel Cycle
5) Dual Cycle ( Limited pressure cycle)
Power cycle 1: CARNOT CYCLE
Ans :
not cycle is an ideal cycle in which heat addition and heat rejection is carried out at constant temperature(isothermal process). Actually it is not practically possible to carry out the heat addition and rejection at constant temperature. Carnot cycle is used for comparing the actual cycles. It is obvious that any actual cycle while operating between same temperatures can not have the efficiency more than that oc Carnot cycle.
Process 12 Isentropic compression
Process 23 Isothermal heat addition
Process 34 isentropic expansion
Process 41 isothermal heat rejection
Carnot heat engine is an ideal heat engine and is not possible in practice due to following reasons.
i) Alternate adiabatic and isothermal process is not possible.
ii) Heat addition and heat rejection at constant temperature is not possible.
iii) All processes are reversible which is not possible in practice.
iv) Isothermal process needs very slow movement whereas adiabatic process needs very fast movements, Both of these cannot be achieved in one cycle .
A Carnot engine working between 650K and 310 K produced work of 150 KJ, Find the thermal efficiency and heat added during cycle.
An ideal engine working on Carnot cycle receives heat at 590 K and rejects at 295 K. If it absorbs heat at the rate of 35 KJ/sec Calculate the work done per Sec.
In a Carnot cycle engine, the temperature of the source and sink are and .If heat supplied is 84 KJ/s. Find the power developed by the engine.
PROBLEMS B
A Carnot cycle receives heat at and heat is supplied at the rate of 84 KJ/sec .Find the temperature at which the heat is rejected if the work produced is 45 KJ/sec.
A Carnot engine operates between two reservoirs at temperature T1 and T2. The work output of engine is 0.6 time heat rejected. The difference in temperature between source and sink is .Calculate thermal efficiency, sink temp and source temp.
A Carnot engine is operating between two temperatures such that the difference between temperatures . It the efficiency of the engine is 52.3 %. Calculate the Sink and source temperature. If the heat supplied is 300 KJ. Find the Work Done.
An Engineer claims that his engine develops 3.75 KW. On testing it is found that engine consumes 0.44 kg of fuel per hour having calorific value 42000 KJ/kg. The maximum and minimum temp in cycle were and . Find whether the engineer is justified in his claim or not…
An inventor created petrol engine operating between and which consumes 0.12 kg/hr of petrol having 46000 KJ/kg calorific value and produces 0.735 kW. Check the validity of his claim..
An inventor created engine which operates between temperatures 600K and 300 K. According to his claim it develops the 17.5 KW of work by taking the heat of 30 KW. Check the validity of his claim.
Power cycle 2:OTTO CYCLE
Ans :
The airstandardOtto cycles is the idealized cycle for the sparkignition internal combustion engines (SI engines or Petrol Engines).This cycles is shown above on PV and TS diagrams.
This cycle is also named as 'Constant Volume Cycle' because heat addition and rejection taees place at constant volume. .The Otto cycle 1234 consist of following four process :
Process 12 ; Reversible adiabatic compression of air
Process 23: Heat addition at constant volume
Process 34 : Reversible adiabatic expansion of air
Process 41 : Heat rejection at constant volume
Air standard efficiency of the Otto cycle is given by ,
where r is the compression ratio and gamma is the index of adiabatic compression for air.
From the equation it is clear that efficiency is the function of Compression ration only. With increase in r, the efficiency also increases.
Ans: Air standard efficiency of the Otto cycle is given by ,
where r is the compression ratio and gamma is the index of adiabatic compression for air.
From the equation it is clear that efficiency is the function of Compression ration only. With increase in r, the efficiency also increases.
The graph above demonstrates the variation of efficiency with the compression ratio. Due to practical limitations of 'Knocking possibility', the compression ratio is limited between the range of 710 for SI engines.
In an otto cycle the temperature at the beginning and end of compression are and . Determine air standard efficiency of the cycle. Take for air.
{Ans =46%}
An Engine working on Otto cycle has efficiency of 50% and index of adiabatic compression is 1.4 ,find compression ratio. If the initial pressure and temperatures are 1 bar and . Find the pressure,temperature and volume at the end of compression. Take ,, , for air.
{Ans = r=5.65}
In an Otto cycle, air at 1 bar and 290 k is compressed isentropically up to 40 bar .
Calculate: 1) comp. ratio 2) Air std. eff. 3) Temp at the end of comp.
Take ,, , for air.
{Ans=r=13.94, T2=831.96K, eff=65%}
An engine working on Otto cycle has cylinder diameter of 150 mm and stroke of 225 mm. The clearance volume is .Find the air standard efficiency of the engine. Take index of compression as 1.4. Take ,, , for air. { Eff =43.6 %}
Calculate the ideal air standard cycle efficiency of a petrol engine operating on Otto cycle. The cylinder bore is 50 mm, a stroke is of75 mm and the clearance volume is of 21.3 cm3. Take ,, , for air.
{r= 7.913, Eff =0.5628}

A certain quantity of air pressure of 1 bar and temperature is compressed reversibly and adiabatically until the pressure is 7 bar in an Otto cycle engine. 460 KJ of heat per kg of air is now added at constant volume determine.
1) Compression ratio of the engine
2) Temperature at the end of compression.
3) Temperature at end of head addition. Take ,, , for air.
{Ans=r=3.97, T2=604 K,T2=1255 K}
The pressure and temperature of air at the beginning of compression in an Otto cycle is 103 kPa and respectively. The heat added per kg of air is 1850 kJ. The compression ratio is 8. Determine thermal efficiency, Maximum temperature in the cycle and Maximum pressure in the cycle. Take ,, , for air.
{ Ans : eff=0.564, T3=3265.8 K,P3=8970.3 kPa }
In an Otto cycle the temperature at the beginning of compression is and at the end of compression is . Calculate
a) Compression ratio b) Efficiency c) Work done if QA=210 KJ/kg Take ,, , for air.
{Ans= r=5.24, air=48%, T3= 1070.29K}
Petrol engine working an Otto cycle has compression ratio 8 It consumes 1kg of air per minute if maximum and minimum temperature during the cycle is 2000k and 300K. Find power developed [ work /sec] Take ,, , for air.
{Ans= air=0.564, T2=689.219K, work done= 8.749 KJ/ sec}
Petrol engine working on Otto cycle has compression ratio 7.5. It consumes 0.8 kg of air per min. If maximum and minimum temp. in the cycle are and . Determine the Power developed by the engine. Take ,, , for air.
{Ans eff=0.55, T2=667.176K, QA=7.577KJ/Kg, work done=4.16735KJ/Kg}

Compression ratio of otto cycle is 8 at the beginning of compression air is at 1 bar 300k. The heat added is 1900 KJ/Kg. Calculate air std efficiency and mean effective pressure.
Take ,, , for air.
{Ans: eff=56.42%,work done=1072.93KJ/kg, MEP=1422.98KN/m3}
In an Otto cycle compression ration is 7. Air is taken at 1 bar and 313 K. heat added is 2510 kJ/kg. Find 1) Cycle efficiency 2) Workdone 3) Maximum temp in cycle 4) Maximum pressure in cycle 5) MEP
Take ,, , for air.
{ Ans: eff = 0.541, Workdone 1358 kJ/kg ,T3=4202 K,P3=93.9 bar,MEP=17.6 bar}
A fourstroke,four cylinder petrol engine is having 250 mm bore and 375 mm stroke. The clearance volume is 0.01052 m3. The engine works on otto cycle. The initial pressure and temperatures are 1 bar and . If the maximum pressure is limited to 25 bar. Find the following
1) Air standard efficiency
2) The mean Effective pressure.
Take ,, , for air.
{ eff= 0.565, MEP = 1.346 bar }
In an Otto cycle air at 1 bar and 290k is compressed until pressure becomes 15 bar the heat is added at constant volume until the pressure rises to 40 bar. Calculate air standard efficiency and mean effective pressure.Take ,, , for air.
{eff=0.5385, MEP=5.68 bar }
SOLUTION TO THESE PROBLEMS IS GIVEN IN THE PRINTED NOTES
Power cycle 3:DIESEL CYCLE
Air standard diesel cycle is a idealized cycle for diesel engines(compression ignited engines).
Process 12 : Reversible adiabatic compression
Process 23 ; Constant Pressure heat addition
Process 35 : Reversible adiabatic compression
Process 41; Constant volume heat rejection
Air standard efficiency of Diesel engine is given by,
where
Following graph shows the variation of efficiency of diesel engine with the compression ratio (r) and the cutoff ratio (rho).
From the graph it is clear that,as the compression ratio increases the efficiency also increases. And as the cutoff ratio increases the efficiency decreases.
Ans : The formula for efficiency of Otto cycle and diesel cycle is given by,
….................................Otto cycle
…......... Diesel cycle
Observing both formulas carefully reveals that there is one additional term K in the diesel cycle, whose value is
This constant K is a function of i.e. cutoff ratio, as value of
The cutoff ratio can take values as below and corresponding value of K
3 
2.5 
2 
1.5 

K 
1.31 
1.24 
1.17 
1.09 
Thus from table it is clear that the value of K is always greater than Unity(1) and hence the efficiency of diesel engine will be always less than Otto cycle for same compression ratio.
Ans: Otto cycle intakes airfuel mixture and compresses it, since gasoline has low flash point, if we increase the compression ratio of Otto cycle there is possibility of SELF IGNITION. Means the fuel will get ignited by itself before the ignition by spark plug. This creates undesirable effect of Knocking and power loss.
Whereas in diesel cycle only air is taken and compressed, and fuel is injected through injector. So there is no such possibility. So the Compression ratio of diesel engine is higher than Otto cycle
For Otto cycle : 6 to 10
For Diesel engine : 16 to 20
Point 
Otto Cycle 
Diesel Cycle 
Heat Addition 
Constant Volume 
Constant Pressure 
Heat rejection 
Constant Volume 
Constant Volume 
Compression ratio 
Less {6:1 to 10:1} 
High { 15:1 to 22:1} 
Efficiency More 
Less 
More 
Application of Power cycles 
Petrol engines 
Diesel engines 
Power cycle 4:DUAL CYCLE
Ans : This cycle has heat addition split into two processes, part is at constant volume and part is at constant pressure. Hence the process of heat addition 23 is split into two processes, as 23’ & 3’3.
In actual practice in case of diesel engine heat addition is partly at constant volume and partly at constant pressure.This cycle is very close to the actual cycle because the heat addition in actual engines does not occur exactly at constant volume or at constant pressure but it is a combination of both.
Process 12: Reversible adiabatic(Isentropic) compression
Process 23’: Constant Volume heat addition
Process 3’3: Constant Pressure heat addition
Process 34: Reversible adiabatic(Isentropic) expansion
Process 41: Constant volume heat rejection
Power cycle 5:BRAYTON CYCLE
Process 12 Isentropic compression
Process 23 Constant pressure heat addition
Process 34 isentropic expansion
Process 41 Constant pressure heat rejection
The Brayton cycle is a theoretical cycle for simple gas turbine. This cycle consists of two isentropic and two constant pressure processes. The cycle is similar to the diesel cycle in compression and heat addition. The isentropic expansion of the diesel cycle is further extended followed by constant pressure heat rejection.The efficiency of the ideal Brayton cycle is given by
Application : This cycle is used in Gas turbines
Cycle 
Pv Diagram 
TS Diagram 

Power cycles 1 :Carnot Cycle



Power cycles 2 :OttoPower cycle 4: Cycle
