An industrial dryer for reducing the amount of water in a candy product is to be

Question

An industrial dryer for reducing the amount of water in a candy product is to be designed. A schematic of the dry is shown in the figure. The dryer will operate steadily at atmospheric pressure with air entering at 25°C, 40% relative humidity Thermal losses from the jacket of the dryer are negligible. The air is heated by heat exchange with saturated steam that enters at 10 bar with a quality of 1.0. The heated air is then blown over the product (candy) and exits the dryer at 70°C and 18% relative humidity at atmospheric pressure. The product enters the dryer at a rate of 5,000 kg/hr at 25°C with a moisture mass fraction of 38%. The dry candy has a specific heat capacity of 3.5 kJ/kg- K. Due to water loss, the product exits the dryer at a mass flow rate of 4325 kg/hr at 44°C. Estimate the necessary air and steam flow rates for this dryer in units of kg/hr. State any assumptions that you employ. Air at 70°C 18% rh, 101.3 kPa Steam 10 bar quality Steam 10 bar quality-0 Candy, 25 C, 5000 kghr 38% water Candy, 44 C 4325 kg/h Air at 25 C, 40% rh, 101.3 kPa

Explanation / Answer

The mean rate of heat transfer for such applications can be expressed as:

q = m cp dT / t

q = mean heat transfer rate (kW (kJ/s))

m = mass of the product (kg)

cp = specific heat of the product (kJ/kg.oC)

dT = Change in temperature of the fluid (oC)

t = total time over which the heating process occurs (seconds)

he = evaporation energy of the steam (kJ/kg)

Mass=5000. Kg/hr

Cp=4.19 Kj/Kgdeg C

Temp(44-25)

q=5000*4.19*(44-25)/3600(sec)

= 110 Kw

Amount of Steam Rate Kg/Hr

Heat Transfer Rate/evaporation energy of the steam (kJ/kg)

he=2013 @ 10 Bar

so, steam flow Rate=110/2013====0.0546 kg/sec

=196.72 kg/hr.

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