SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 188 of 200

Marks: +1, -0

An adiabatic heat exchanger is used to heat cold water at 15°C entering at the rate of 10kg/sec by hot air at 90°C entering at the rate of 10 kg/sec. if the exit temperature of cold water IS 29°C then what will be the temperature of hot air at the exit.

C_p

for air = 1.008 KJ/Kg.

C_p

for water = 4.18 KJ/Kg:

26.54°C
31.88°C
58.65°C
89.12°C

Validate

Solution:

Given:
For air:

T_i = 90°C, T_o = 20°C, m = 10

kg/s,

C_p = 1.008

kJ/kgK
For water:

T_i = 15°C, T_o = ?, m = 10

kg/s,

C_p = 4.18

kJ/kgK

⇒ 10 × 1.008 × (90 – T_f) = 10 × 4.18

× (29 -15)

⇒ T_f = 31.6° C

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