# 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.

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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.

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(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, cp = 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) = cp(T3 – T1) – T0(s3 – s1

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

Also, s3 – s1 = cp loge $\cfrac{T_3}{T_1}$ = 1.005 loge $\cfrac{323}{293}$ = 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.