Question

A heat exchanger is used to heat cold water at 8 C entering at a rate of 1.2 kg/s by hot air at 90 C entering at rate of 2.5 kg/s. The heat exchanger is not insulated, and is loosing heat at a rate of 28 kJ/s. If the exit temperature of hot air is 20 C, the exit temperature of cold water is:___________ (use constant specific heats at room temperature)

Answer 1

The exit temperature of cold water is 37.6°C

Step-by-step explanation:

Mass flow rate of water,
`\dot{m}_(water) = 1.2 kg/s`

Mass flow rate of air,
`\dot{m}_(air) = 2.5 kg/s`

`T_(in) = 90^(0)C\\ T_(out) = 20^0C`

`c_(air) = 1010\\c_(water) = 4186`

`\dot{Q}_(lost) = 28000 J/s`

Applying the conservation of heat law between air, water and the surrounding:

`[\dot{m}c(T_(in) - T_(out))]_(air) = [\dot{m}c(T_(in) - T_(out))]_(water) + \dot{Q}_(lost)`

We get the exit temperature of cold water by making
`T_(out)` the subject of the formula above:

`(T_(out))_(water) =\frac{ [\dot{m}c(T_(in) - T_(out))]_(air) + [\dot{m} c T_(in)]_(water) - \dot{Q}_(lost)}{(\dot{m}c)_(water)} \\\\(T_(out))_(water) = ((2.5*1010*(90-20) + [1.2*4186*(8+273)] - 28000)/(1.2*4186) \\\\(T_(out))_(water) = 310.6 K\\\\(T_(out))_(water) = 37.6^0 C`