Example: Calculate the drying air rate and size of dryer for following duty:
Feed rate: DFR = 4000 kg/h
Feed concentration: DFS = 48 %
Feed temperature: t_{f} = 60 °C
Product solids content: DPS = 95 %
Ambient temperature: t_{a} = 15 °C
Ambient humidity: y_{a} = 0.01 kg/kg
Inlet temperature: t_{1} = 200 °C
Outlet temperature: t_{2} = 80 °C
Cooling air rate: A_{c} = 200 kg/h
Fines transport air: A_{t} = 500 kg/h
Transport air temperature t_{t} = 60 °C
The values for operating parameters were estimated from product experience with respect to the required quality specification. The powder temperature is estimated to be 5°C below the outlet air temperature, fines recirculation ratio R = 0.5, radiation loss coefficient K = 3.0 kcal/ m_{2}/h and dryer surface area 300 m^{2}. The heat capacities of air and water vapour are taken from Equation 3.8 and 3.10 (ca_{1}=0.245, ca_{2} = 0.241, caa = 0.24, cat = 0.241, cv_{1} = 0.463, cv_{2} = 0.45, cv_{a} = 0.444 and cv_{t} = 0.448kcal/kg/°C). The product is whole milk with 28% fat, thus the heat capacity of solids, using values from Table 3.2. is:
c_{s} = (28 * 0.5 + (100 - 28) * 0.3)/100 = 0.356 kcal/kg/°C
Calculation (according to equations [6,9], [6,7] and [6,39] through [6,46]:
DPR = 4000 * 48/95 = 2021.1 kg/h
DER = 4000 - 2021.1 = 1978.9 kg/h
Q_{ev} = 1978.9 * (597.3 + 0.45 * 80 - 60) = 1134530.5 kcal/h
Q_{pr} = 2021.1 * (80 - 5 - 60) * (0.356 * 95 / 100 + (1 - 95 / 100)) = 11768.6 kcal/h
Q_{co} = 200 / 1.01 * ((80 * 0.241 - 15 * 0.24) + * (80 * 0.45 - 15 * 0.444)) = 3163.0 kcal/h
Qtr = 500 / 1.01 * ((80 * 0.241 - 60 * 0.241) + 0.01 * (80 * 0.45 - 60 * 0.448)) = 2431.3 kcal/h
Q_{fi} = 2021.1 * 0.5 *(80 - 5 - 60) * (0.356 * 95 / 100 +(1 - 95 / 100)) = 5884.3 kcal/h
Q_{rl} = 300 * 3.0 * (80 - 15 - 20) = 40500.0 kcal/h
ΣQ = 1134530.5 + 11768.6 + 3163 + 2431.3 + 5884.3 + 40500 = 1198277.7 kcal/h
A_{dr} = 1198277.7 / (200 * .245 - 80 * .241 + 0.01 * (200 * .463 - 80 * 0.45)) = 39565 kg/h
The above calculation can be simplified by neglecting the air moisture content and powder moisture and using for heat capacities of air and of water vapour constants 0.24 and 0.46 kcal/ kg/°C respectively:
Q_{ev} = 1978.9 * (597.3 + 0.46 * 80 - 60) = 1136113.7 kcal/h
Q_{pr} = 2021.1 * 0.356 * (80 - 5 - 60) = 10792.4 kcal/h
Q_{co} = 200 * 0.24 * (80 - 15) = 3120.0 kcal/h
Q_{tr} = 500 * 0.24 * (80 - 60) = 2400.0 kcal/h
Q_{fi} = 2021.1 * 0.356 * (80 - 5 - 60) = 5396.1 kcal/h
Q_{rl} = 300 * 3.0 / (80 - 15 - 20) = 40500.0 kcal/h
ΣQ = 1198322.2 kcal/h
A_{dr} = 1198322.2 / (0.24 * (200 - 80)) = 41608 kg/h
This comparison demonstrates that the simplified calculation results in more than 5% higher amount of air and emphasizes the importance of calculation on the enthalpy basis. The difference is even greater if the available ambient air has high humidity values. The absolute humidity of the exhaust air is: