Answer:
(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp
(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp
(c) The volume flow rate at exit is V₃ =V₁ + V₂
Explanation:
Solution
Now
The system comprises of two inlets and on exit.
Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁
Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂
Mass flow rate enthalpy of fluid from exit be m₃ and h₃
Mixing chambers do not include any kind of work (w = 0)
So, both the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)
(a) Applying the mass balance of mixing chamber, min = mout
Applying the energy balance of mixing chamber,
Ein = Eout
min hin =mout hout
miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃
T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +
T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp
The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp
(b) From the ideal gas equation,
v =RT/PT
v₃ = RT₃/P₃
The volume flow rate at the exit, V₃ =m₃v₃
V₃ = m₃ RT₃/P₃
Substituting the value of T₃, we have
V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)
V₃ = R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)
Now
The mixing process occurs at constant pressure P₃=P₂=P₁.
Hence V₃ becomes:
V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp
V₃ =V₁ + V₂ + RQin/P₃Cp
Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp
(c) Now for an adiabatic mixing, Qin =0
Hence V₃ becomes:
V₃ =V₁ + V₂ + r * 0/P₃Cp
V₃ =V₁ + V₂ + 0
V₃ =V₁ + V₂
Therefore the volume flow rate at exit is V₃ =V₁ + V₂
