Question

In an industrial bottling plant, the glass bottles must be washed, dried, and cooled before going to the filling department. The plant operates at atmospheric pressure at a capacity of 1000 bottles/hr. During washing process the bottles retain 0.1kg H2O/kg glass and enters the drier at 25C. Determine the heat load on the dryer if the dry bottles leave at 100C. Each bottle weighs 0.5 kg. Assume the heat capacity over the temperature of operation is 0.18 kcal/kg.K for glass and 1.00 kcal/kg.K for water.

Answer 1

N = 1000 bottles/h

Ratio = 0.1 kg water / 1 kg glass

Ti = 25°C

Cpglass = 0.18 kcal / gC = 0.753 J/gC

Cpw = 1  kcal / gC = 4.184 J/gC

Assume it is evaporating at 100°C:

Vaporization = = 2264.76 J/g

bottle mass = 0.5 kg of glass...

Assume a basis of 1 hour

so

Ntota = 1000 botles

mass of bottle = 0.5*1000 = 500 kg of glass

mass of water = 0.1 kg of water / kg of glass* 500 kg glass = 50 kg of water

Qglass = m*C*(Tf-Ti) = 500*0.753 (100-25) = 28237.5 kJ (per hour)

Qwater = m*C*(Tf-Ti) + m*Hvap = 50*4.184 * (100-25) + 50*2264.76 = 128928 kJ ( per hour )

total = 28237.5 +128928 = 157165.5 kJ (per hour)

1 h = 3600 s

Q duty = 157165.5 / 3600 = 43.6570 kW