Because it operates at a higher temperature difference between the hot and the cold source. This implies in higher entropy generation.

Dear mustapa,

carnot cycle is a ideal cycle. It uses some assumptions like 

cylinder piston is frictionless

only isentropic, adiabatic processes take place.

Cylinder head act as heat souce as well as sink but inside temperature remain constant even during heat rejection and absorption. 

But rankine cycle use stream as working fluid.  Further modifed rankine cycle represents real steam power plant process. It has polytropic process. It has entropy, heat losses.

Thats why you get lesser efficiency for rankine cycle.

Simply carnot is ideal cycle and rankine is real cycle.good luck, 

with regards

P.Thirumal

If the same hot and cold reservoirs are used, the maximum and minimum posible temperatures are defined, Th and Tc respectively. In the Carnot cycle, these are the temperatures for the evaporation and condensation processes. In the Rankine cycle you have irreversibilities associated to temperature differences between these reservoirs and the working fluid, required for heat transfer between them.

Moreover, the average temperature of the working fluid in the boiler is lower because the fluid is in subcooled state at the boiler inlet. The subcooled or saturated state is required at the outlet of the condenser because compression of a multiphase mixture up to the boiler pressure is not used due to its low efficiency.

I hope this can help you.

The Carnot process contains the simplification of isentropic transformations which require perfect heat insulation (sub-processes without losses), which isn´t possible due to the 2nd law of thermodynamics: Every thermodynamic transformation causes entropy production, or in other words energetic processes in reality always have losses. The rankine cycle takes this into account. 

the Carnot cycle is a hypothetical cycle with no consideration to entropy generation. it assumes perfect heat transfer in the heat source and heat sink. the working fluid also remains in the same phase which is unlike the Rankine cycle where the fluid undergoes phase change. under similar heat source and heat sink temperature a practical thermodynamic cycle such as Rankine cycle would achieve at best about 50% of the Carnot cycle efficiency.

All my colleagues above are right. To sum it up (or put it simple), while Carnot cycle is dealing with ideal, reversible processes, the Rankine cycle deals with real processes affected by irreversibilities (internal or external). Irreversible processes and losses of heat (for example, due to friction) prevent the ideal from taking place at every cycle transformation.

All opinion were correct and correlated in oneway.

Carnot cycle is ideal with best combination of process, but that cycle is not possible to execute in real. 

In real irreversibility and unavailability are involved.

Irrespective of working medium, for same temperature ends (source and sink) and different cycle no one will give higher efficiency than Carnot cycle because of it's unrealitic processes.

The efficiency of a Carnot cycles is the highest possible between Th (high temperature) and Tc (low temperature) because all the heat is given to the fluid at the highest temperature and the fluid rejects the waste heat at the lowest temperature. The advantage of the Carnot cycle is true regardless of the internal irreverisbility of the working fluid. The efficiency of a Carnot cycle can be matched (with regenerative cycles) but never exceeded. Another cycle performing isothermal heat-addition and heat-rejection processes is the Stirling cycle.

Assuming that both cycles are ideal, in the Carnot cycle, all of the heat addition occurs at the maximum temperature, and all of the heat rejection occurs at the minimum temperature.  In the Rankine  cycle, heat is added when the liquid working fluid exits the high pressure pump.  Hence, heating liquid, boiling liquid, and even the superheat occurs at  temperatures lower than the maximum cycle temperature when the superheat gas enters the expansion turbine.  The thermodynamic efficiencies for both the Stirling and Ericsson cycle are the same as the Carnot cycle, since again, heat is added at a constant maximum temperature and heat is rejected at a constant minimum temperature.  If you plot the cycle processes on a T-s diagram, then the ratio of heat rejection to heat addition is minimum.  Since cycle efficiency is the net work (energy sought, i.e., heat addition minus heat rejection) divided by the heat addition (energy that costs), then the Carnot cycle has a higher ratio than the Rankine cycle.  Sorry for the amount of detail.

It is a simple fundamental fact that no practical engine works with 100% efficiency. Rankine is the functional and practical model of carnot. Carnot is theoretical and ideal and its efficiency is hardly achieved by any working cycles. hence, rankine must be less compared to carnot due to losses.

Simplement parce que le cycle de Rankine comprend 2 isobares au lieu de 2 isothermes. Pour obtenir le rendement maximum il faut échanger la chaleur à température constante, ce qui n'est pas le cas dans un cycle de Rankine.

there is  energy loss for the non- isothermal heat transfer

Carnot cycle is composed of all internally and externally reversible processes. On the contrary, (Ideal) Rankine cycle is composed of all internally reversible processes but externally irreversible processes related with the Boiler and Condenser heat transfer processes. If we approach the Rankine cycle efficiency to the Carnot cycle one, we need to make the Rankine cycle to have two reversible isothermal process. However, it is impossible to fabricate the pump to compress the wet steam state at the present technology. Instead of making the pump, we adopt nowadays the extracting regenerative cycle with reheat process to approach the Rankine cycle to the Carnot cycle. I hope this answer will help your understanding for the reason of adopting Rankine cycle instead of Carnot cycle. 

In the Rankine cycle, the mean temperature at which heat is supplied is less than the maximum temperature, so that the efficiency is less than that of a Carnot cycle working between the same maximum and minimum temperatures.

Moreover, The reason lies in the nature of the Rankine cycle and that of the fluid. There are modifications introduced to the Rankine cycle to improve the efficiency but the nature of the fluid and the intrinsic nature of Rankine cycle still constrain the maximum efficiency available. if you look at the T-s diagram of a simple Rankine cycle, you will observe that much of the heat gained in the boiler is lost in the condenser (the area under the condenser process is a large fraction of the area under the boiler process). One could improve the efficiency of Rankine cycle by having as much difference between boiler and condenser temperatures but beyond certain limits, extreme values of boiler and condenser temperature become practically unfeasible. Please note that given a high and low temperature, even the idealised Carnot cycle has some limited, less than 100 % efficiency. If you compare the Rankine cycle efficiency to this efficiency, it will be found to be significantly close.

@Mustapa, you probably do not have access to basic books in Thermodynamics in USA (Univ. of South Australia), but you do have access to the net. Why did you not try the net before posting this question? 

Thank's a lot for all helping answer my question. 

@Vijay Raghavan, I have look my course material on this topic. I try to find another addition knowledge and more understanding about this. That's why I post here. :-)

The Carnot cycle is an ideal cycle. As such the processes are considered reversible. It therefore represents the most efficient cycle that can operate between two constant temperature reservoirs. Its thermal efficiency is therefore theoretical.

Because the Carnot cycle is an ideal theroritcal one assume reversible process without any losses it used  just to give indication about the highst level of effeciency. while the efficiency of rankine, cycle is the actual value so it must be lower than the carnot efficiency

In layman language, the Rankine cycle is the practical version of the Carnot cycle. 

Ideal is not practical and when it is practical, the performance is much lesser, so as the case of Rankine cycle. 

However, from scientific point of view, there is different explanation as given above, by number of eminent researchers. 

Actually the Carnot cycle consist of two isothermal and two isentropic processes which are reversible and theoreticaly to get maximum efficiency. But it actual practice we can achieve an isothermal process by very slow and prolonged one. Similarly in an isentropic process we need infinity temperature source or nil heat transfer between the system and surrounding both are not possible. All the process are only an ideal.

In the case of Rankine cycle the heat addition and heat removal takes place in constant pressure (Isobaric) processes which is practically  possible, but, the work done during the constant pressure process is less than that of the work done during the isothermal process. That's why the efficiency of the Carnot cycle is always greater than the efficiency of any cycle.

@Mustapa, I have given a rather easy explanation with figures in the attached file. I hope it brings clarity. You might also find this answer easier to comprehend than many others given. If I recommended text books, it is because, it takes too much effort to put together such an answer, with figures etc. I hope you can develop the ability to read on your own and learn. Good luck in your studies. 

Thank you @Vijay for your help. :-)

Dear Mustapa,

Carnot cycle is a standard cycle which efficiency is 100%. In Carnot cycle lot of assumption is considered. For maintaining the standards, other cycle is compared with Carnot cycle. On the other hand, the Rankine cycle is a practical cycle which has different types of losses like heat loss, entropy loss etc. that's why the Rankine cycle efficiency is lower than the Carnot cycle.

No Mr. Washim Akram.

Carnot cycle does not gives 100% efficiency it efficiency depends upon the temperature limits but never goes 100%.

Practically the Rankine cycle cannot compete with a Carnot cycle due to the necessity to condense the low temperature steam at the bottom of the Rankine cycle to phase change the steam to water for pumping with a latent energy loss of 2.2 Mega-Joules per liter of water.