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

Answer:

The exit temperature of cold water is 37.6°C

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:

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