Question

Two humid streams are adiabatically mixed at 1 atm pressure to form a third stream. The first stream has a temperature of 40C, a relative humidity of 40%, and a volumetric flow rate of 3 L/s, while the second stream has a temperature of 15C, a relative humidity of 80%, and a volumetric flow rate of 1 L/s. Calculate the third stream’s temperature and relative humidity.

Answer 1

Answer: Third stream’s temperature is $33.2^{o}C$  and its relative humidity is 50%.

Explanation:

According to the Psychrometric chart.

For the first stream, we have the following.

  $T_{1} = 40^{o}C$,       R.H = 40%

      V = 3L/s ,         W = 19 gm of moisture/kg of dry air

    $h_{1}$ = 89 kj/kg ,     $v_{1} = 0.913 m^{3}/kg$

For the second stream, we have the following.

   $T_{2} = 15^{o}C$

       R.H = 80%

        V = 1 L/s

     W = 8.5 gm of moisture/kg of dry air

  $h_{2}$ = 36.5 kj/kg

  $v_{2} = 0.828 m^{3}/kg$

Now,

     $ma_{1} = \frac{V}{v_{1}}$

              = $\frac{3}{0.913}$

              = 3.286 kg/s

     $ma_{2} = \frac{V}{v_{2}}$

              = $\frac{1}{0.828}$

              = 1.2077 kg/s

We will calculate the value of $ma_{3}$ as follows.

  $ma_{3} = ma_{1} + ma_{2}$

            = 3.286 kg/s  + 1.2077 kg/s

             = 4.5 kg/s

Now, heat balance will be as follows.

     $ma_{1}h_{1} + ma_{2}h_{2} = ma_{3}h_{3}$

        $h_{3}$ = 74.7855 kj/kg

Hence, the moisture balance will be calculated as follows.

        $ma_{1}W_{1} + ma_{2}W_{2} = ma_{3}W_{3}$

       $W_{3}$ = 16.155

Now, again using the psychrometric chart we will find the value of  temperature and relative humidity as follows.

Values of $T_{3}$ and $R.H_{3}$ against $W_{3}$ and $h_{3}$ are:

     $T_{3} = 33.2^{o}C$

           R.H = 50%

Thus, we can conclude that third stream’s temperature is $33.2^{o}C$  and its relative humidity is 50%.