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.

Third stream’s temperature is  and its relative humidity is 50%.

Explanation:

According to the Psychrometric chart.

For the first stream, we have the following.

  ,       R.H = 40%

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

    = 89 kj/kg ,    

For the second stream, we have the following.

       R.H = 80%

        V = 1 L/s

     W = 8.5 gm of moisture/kg of dry air

  = 36.5 kj/kg

Now,

              =

              = 3.286 kg/s

              =

              = 1.2077 kg/s

We will calculate the value of as follows.

            = 3.286 kg/s  + 1.2077 kg/s

             = 4.5 kg/s

Now, heat balance will be as follows.

        = 74.7855 kj/kg

Hence, the moisture balance will be calculated as follows.

        = 16.155

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

Values of and against and are:

           R.H = 50%

Thus, we can conclude that third stream’s temperature is  and its relative humidity is 50%.

може ли да зареждате повече информация, моля