Material & Energy balance example: Spray Drier

In the previous blog, material and energy balance in process design, I revised common principles of mass and energy balance. Balances on spray driers, kilns, furnace and filters has been described briefly. In this blog, I will describe the details of spray drier mass and energy balance. 

The figure below shows the components of a spray drier system configuration. Drying air is first heated in the heater to the desired temperature as determined partly by the product property.The material to be dried is sprayed by spray system( nozzles) into the drying chamber. The dried product is collected by product
recovery system (cyclones). The air is recycled by blower back to the air heater.

Suppose 5.02kg/min of protein isolate, 60% solids at temperature of 15degreeC is to be dried to final specification of 93% protein and 5% moisture at 35degreeC. The amount of dry air should be determined from air heater energy balance but the final temperature can be assumed to be equal to the product temperature. 

Mass balance

Mass balance

                      5.02=Mp+Mw

Solid balance

         0.6*5.02=0.93*Mp

        Mp=3.24Kg/min

Thus, mass flow rate of water in the product

        Mw=0.05*3.24

             =0.162Kg/min 

 

Energy balance

Assume 10% heat gained on the heater is lost

         

            m_a            Air heater                

        T=25^oC   ---------------------------->   100^oC

      H=.013kg/kg dry air                               H=0.013kg/kgdry air  

      h=70.01kj/kg                                          h=135.5kj/kg

 where H is the relative humidity and h is the enthalpy

          Heat gained by air on heating from 25-100^oC is

          Q = m_a (135.5-70.01) 

              = 65.5*m_a

          E_in= E_out + E_loss 

_{Energy in, consists of the enthalpy of the feed(Ef) and the drying air (Eia) while energy out is the sum of the air out(Eoa), the enthalpy of the dried product(Ep) and latent heat of vaporization of water.(Revap)}

          E_f +Eⁱ_a = E^o_a+E_p+R_{evap   ..................................EQ. 1}

                  E_f=5.02*C_f*(288-273)   feed at 15C

                  Eⁱ_a=Cⁱa*mⁱ_a*(373-273)  air at 100C

                  E_p=C_p*m_p (323-273)     product at 35C

                  E^o_a=Co_a*mⁱa (323-273) air at 35C

                  R_evap=lambda*m^o_w

Now lets determine the properties

          C_p protein=2.0082+1.2089*10^-3*T - 1.3129*10^-6*T²

          C_p at 15^oC=2.026 kJ/kg.^oK     

                    35^oC=2.0486 kJ/kg.^oC

          Specific heat capacity of air

                        C_a=1.005+1.88*H 

                         H_i=0.0013 kg/kg dry air at 25^oC

Assume the relative humidity of the out let air to be 0.032kg/kgdry air. It can be checked back using water balance

Therefore C_pⁱ_a=1.02944kj/kg °C   C_p^o_a=1.06516KJ/kg.oc

Substituting in the above energy balance equations,

        

            E_f=5.02*2.026*(288-273) =152.55 kJ/min

            Eⁱ_a=Mⁱ_a*1.0294*(373-273) =102.94*mⁱ_a

            E_p=2.04*3.24*(323-273) =330.48 kJ/min

            E^o_a=1.06516*m^o_a*(323-273) =53.25*m^o_a

            R_evap=2360*0.162=382.32 kJ/min

             E_loss=0.1*65.5*m^oa=6.55*m^o_a

The amount of air required for the drying process is not known initially. The air leaving the dryer is also not known. So as initial iteration we will assume they are equal. Since the relative humidity of the out let air has been assumed initially, it will be recalculated back and the iteration continues until the relative humidity calculated is equal to assumed.

                               mⁱ_a=m^o_a

 Using equation 1, after substituting the calculated values

     152.55+102.94*mⁱ_a=330.48+53.25*m^o_a+382.32+6.55*m^o_a

                  (102.94-53.25-6.55) moa=712.8 

                           m^o_a=16.5 kg/min

  Total amount of air (dry +moisture)

                   mia=16.5+0.013*16.5=16.71 kg/min

 Total amount of air from drier, 

(previously we calculated the amount water in the product is Mw=0.162 kg/min)

                =16.71+0.162=16.87 kg/min

 Amount of water in the air at the out let = Amount of water in the inlet air + amount of water evaporated

                 16.5*0.013+0.162= 0.377 kg/min

  Relative humidity of exit air

                 3.0995/101.5 = 0.023 kg/kg dry air 
This value is lower than the assumed 0.032, thus, more iterations are required.

From psychometric chart specific volume of air=1.075m³/Kg dry air

Therefore volumetric flow rate=1.075*16.5=17.74 m³/min it is recycled to the fresh air to reduce fresh air requirement. The final answer depends on the the value of m^o_a at the end of the iteration.

     The overall thermal efficiency

                        h=T₁-T₂/ (T₁-T_o)*100

       From psychometric chart, the corresponding to saturated condition of out let temperature of air  at 25^oC

                         h=100-35/ (100-25) =87%

       Therefore we can say that the design is good

I hope it is clear enough. If you have questions please leave them in the comment box or contact me. Please subscribe if you would like to get email updates of more balance and design examples.