ME2121 - Example on Entropy

Entropy Generation in a Mixing Chamber

EXAMPLE

Water at 200 kPa and 10°C enters a mixing chamber at a rate of 150 kg/min where it is mixed steadily with steam entering at 200 kPa and 150°C. The mixture leaves the chamber at 200 kPa and 70°C, and heat is lost to the surrounding air at 20 °C at a rate of 190 kJ/min. Neglecting the changes in kinetic and potential energies, determine the rate of entropy generation during this process. Tb=20°C 190 kJ/min Water P1=200kPa, T1=10°C, 150 kg/min

Steam P2=200kPa, T2=150°C

Mixture P3=200kPa, T3=70°C

State 1:

P 1=200 kPa; T 1 =10°C



h 1 =h f@10 C =42.022kJ/kg °

s1 =s f@10 C =0.1511kJ/kg.K  °

State 2:

P 2=200 kPa; T 2 =150°C



h 2 =2769.1kJ/kg s2 =7.2810kJ/kg.K 

State 3:

P 3=200 kPa; T 3 =70°C



h 3 =h f@7 0 C =293.07kJ/kg °

S 3=s f@7 0 C =0.95511kJ/kg.K  °

Substituting,

190 kJ/min= [150 x 42.022 +

which gives

m & 2 x 2769.1

& 2 ) x 293.07] kJ/min  (150 + m

 –

m & 2 =15.29 kg/min

The rate of entropy generation during this process can be determined by applying the rate form of the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature of the extended system is 20°C = 293K:

+

S& in

&  S out  

& S gen



Rate of  net entropy tr  ansfer  by heat and mass

Rate of  entropy generation

& 1s1  m & 2s 2 m

Sgen







& 3s 3 m





dSsystem / dt

(kW/K)



Rate of  change in entropy

& 3s 3 m



& Q out

T b

& 1s1  m & 2s 2 m





S& gen



0

& Q out

T b

= (165.29 x 0.9551  – 150 x 0.1511  – 15.29 x 7.2810) +

190 293

= 24.53 kg/min.K

D i s c u s s i o n  Note

that entropy is generated during this process at a rate of 24.53 kJ/min. This entropy generation is caused by the mixing of two fluid streams (an irreversible process) and the heat transfer between the mixing chamber and the surroundings through a finite temperature difference (another irreversible process).