Question

Air at 20°C is to be heated to 50°C by mixing it in steady flow with a quantity of air at 100°C. Assuming that the mixing process is adiabatic and neglecting changes in kinetic and potential energy, calculate : 

(i) The ratio of mass flow of air initially at 100°C to that initially at 20°C. 

(ii) The effectiveness of heating process, if the atmospheric temperature is 20°C.

Answer 1

(i) Let, x = ratio of mass flows. 

Stream 1 = air at 20°C (T₁ = 20 + 273 = 293 K) 

Stream 2 = air at 100°C (T₂ = 100 + 273 = 373 K) 

Stream 3 = air at 50°C (T₃ = 50 + 273 = 323 K) 

If, c_p = Specific heat of air constant pressure

(ii) Let the system considered be a stream of air of unit mass, heated from 20°C to 50°C. Increase of availability of system

= b₃ – b₁ = (h₃ – h₁) – T₀(s₃ – s₁) = c_p(T₃ – T₁) – T₀(s₃ – s₁) 

= 1.005(323 – 293) – 293(s₃ – s₁) [\(\because\) T₀ = 20 + 273 = 293 K]

Also, s₃ – s₁ = c_p log_e \(\cfrac{T_3}{T_1}\) = 1.005 log_e \(\cfrac{323}{293}\) = 0.0979 kJ/kg K

∴ Increase of availability of system 

= 1.005 × 30 – 293 × 0.0979 = 1.465 kJ/kg. 

The system, which is the air being heated, is ‘surrounded’ by the air stream being cooled. Therefore, the loss of availability of the surroundings is given by, x(b₂ – b3). 

i.e., Loss of availability of surroundings

The low figure for the effectiveness is an indication of the highly irreversible nature of the mixing process.