Question

In: Mechanical Engineering

ericsson cycle

Consider an ideal Ericsson cycle with air as the working fluid executed in a steady-flow system. Air is at a temperature of 270 degrees and a pressure of  120 kPa at the beginning of the isothermal compression process, during which 150 kJ/kg of heat is rejected. Heat transfer to air occurs at a temperature of 1200K. Determine

(a) the maximum pressure in the cycle

(b) the network output per unit mass of air

(c) the thermal efficiency of the cycle

Solutions

Expert Solution

An Ericsson cycle involves an isothermal heat-addition process and an isothermal heat-rejection process. the Ericsson cycle differ from the Carnot cycle in that the two isentropic processes in the Carnot cycle are replaced by two constant-pressure regeneration processes. The cycle utilizes regeneration, a process during which heat is transferred to a thermal energy storage device called a regenerator during one part of the cycle and is transferred back to the working fluid during another part of the cycle. 

the p-v and T-s diagrams are shown in the diagram above and the processes are as follows:

 1 → 2: isothermal compression, 

2 → 3: isobaric heating in the combustion chamber,

3 → 4: isothermal expansion in the turbine,

 4 → 1: isobaric heat transfer in surroundings

            (A). The maximum pressure

the change in entropy can be given by:

S4-S3=qout/TL: (-150/300K

     =-0.5KJ/Kg/K

The change of entropy can also be given by:

cp In(T4/T3) - R In(p4/p3)

from the T-s diagram, T3=T4=TL

the change in entropy can therefore be given by:

-0.5KJ/KG/K = Cp In1-(0.287kJ/KG/K) In( P4/P3)

-0.5KJ/KG/K = 0 - (0.287KJ/KG/K) In (P4/120Kpa)

p4 = 685.12Kpa which is our maximum pressure

 

     (B)The net work output

Wnet = qin - qout

for an ideal reversible cycle, (qin/qout)=(TH/TL)

qin = qoutX(TH/TL)  

       = 150X(1200/300) = 600KJ/Kg

The net work is given by:

qin-qout

600KJ/kg - 150KJ/kg

Wnet = 450KJ/Kg

   

       (c) The thermal efficiency of the cycle

 

  thermal efficiency = 1 - (TL/TH)

                                 = 1 - (300/1200)

                         0.75 = 75%

 

 

 


(a). maximum pressure = 685.12 Kpa

(b.). the net work output = 450kJ/Kg

(c). the thermal efficiency = 75% 

 

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