Question

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

Answer 1

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%