Question

A hot water stream at 80 oC enters a mixing chamber with mass flow rate of 3.6 kg/s and mixed with cold water at 20 oC. If the mixture temperature is 42 oC, determine the mass flow rate of cold stream in kg/min. Assume all the streams being at 250 kPa

Answer 1

Explanation:

The mixing chamber will be well insulated when steady operating conditions exist such that there will be negligible heat loss to the surroundings. Therefore, changes in the kinetic and potential energies of the fluid streams will be negligible and there are constant fluid properties with no work interactions.

   $T < T_{sat}$ at 250 kPa = $127.41^{o}C$

   $h_{1}$ approx equal to $h_{f}$ at $80^{o}C$

              = 335.02 kJ/kg

    $h_{2}$ ≈ $h_{f}$ at $20^{o}C$

                      = 83.915 kJ/kg

and,    $h_{3}$ ≈ $h_{f}$ at $42^{o}C$ = 175.90 kJ/kg

Therefore, mass balance will be calculated as follows.

   $m^{o}_{in} - m^{o}_{out} = \Delta m^{o}_{system} \rightarrow m^{o}_{1} + m^{o}_{2} = m^{o}_{3}$

And, energy balance will be given as follows.

      $E^{o}_{in} - E^{o}_{out} = \Delta E^{o}_{system}$

As we are stating steady conditions,

     $\Delta m^{o}_{system}$ and $\Delta E^{o}_{system}$ cancel out to zero.

So,    $E^{o}_{in} = E^{o}_{out}$

     $m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = m^{o}_{3}(h_{3})$

On combining the relations, we solve for $m^{o}_{2}$ as follows.

   $m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = (m^{o}_{1} + m^{o}_{2})(h_{3})$

   $m^{o}_{2} = (\frac{(h_{1} - h_{3})}{(h_{3} - h_{2})}) \times m^{o}_{1}$

              = $\frac{(335.02 - 175.90)}{(175.90 - 83.915)} \times 0.5$  

       $m^{o}_{2}$ = 0.865 kg/s

                       = 51.9 kg/min      (as 1 min = 60 sec)

Thus, we can conclude that the mass flow rate of cold stream is 51.9 kg/min.