Question

Suagr syrup is concentrated from 10% solids to 40% solids at the rate of 100 kg/h using steam in a jacketed tube. The syrup enters the tube at 15 C and is evaporated under reduced pressure at 80 C. The steam is supplied at 115 C. Assuming boiling point remains constant. Other pertinent properties are

Specific Heat of syrup= 3.96 kJ/kg

Latent Heat of vaporization= 2309 kJ/kg

Latent heat of steam condensation= 2217 kJ/kg

Specific heat of water= 4.2 kJ/kg

Calculate the amount of steam needed per hour assuming 1) condensate exits at boiling point temperature ,and 2) condensate exits at steam temperature.

Answer 1

Initially sugar syrup contains 10% solids i.e. 1 Kg sugar syrub contains 0.1 Kg solids and 0.9 Kg water.
This is concentrated to 40% solids i.e. 1 Kg sugar syrup contains 0.4 Kg solids.

​(0.1 Kg solids)/(0.1 Kg solids + mass of water) = 0.4
​Mass of water = 0.15 Kg

Water to be removed = 0.9-0.15 = 0.75 Kg
(a)
​Consider 1 Kg of sugar syrup containing 10% solids:
​The entire sugar solution has to be heated to it's specific heat and then 0.75 Kg water is to be supplied with latent heat.
​Heat required for 1 Kg of sugar syrup at 10% solids = (0.1)(3.96)(80-15) + (0.9)(4.2)(80-15) + (0.75)(2309)
= 25.74 + 245.7 + 1732.75 = 2003 kJ

​Steam required = 2003/2217 = 0.903 kg

Therefore, for 100 Kg/h, steam required will be 90.3 Kg/h

(b) If the vapour is heated to 115 deg C,
​Additional heat required will be to superheat the steam.
C_p,steam = 2 kJ/Kg
​Additional heat required = 0.75(2)(115-80) = 52.5 kJ

Steam required for 1 Kg of 10% solids sugar syrup = 2003 + 52.5 = 2056 kJ

Steam required = 2056/2217 = 0.927 kg

Therefore, for 100 Kg/h, steam required will be 92.7 Kg/h