Question:
An evaporator is a device in which a feed stream is concentrated by boiling a solution. A combination of heating and reduced pressure are used to induce the boiling. Consider a single- stage evaporator as shown schematically below. The feed flow rate is 40,000 lb/hr of a sugar solution, and the feed is at a temperature of 200°F and a concentration of 5% solids. It is to be concentrated to a "thick liquor" of 45% solids by condensing saturated steam at 250°F inside heating tubes that are in the boiling liquid. The water vapor leaving the boiling solution becomes saturated at the condenser, where it has a temperature of 125°F.
The temperature of the boiling liquid is 150°F. The overall heat transfer coefficient U between the heating tubes and the boiling liquid is 600BTU/hrft 2 °F.
The specific heats of all solutions (water and sugar solution) can be assumed to be 1 BTU/lb°F. There is no precipitation of solids in the evaporator, heat losses are negligible, and all condensates leave at the condensing temperature.
You are to calculate the following:
a. The area (in ft2) of heating surface needed for the heating tubes;
b. The steam consumption per hour;
c. The pounds of water evaporated per pound of steam.
Hint: Do a material balance to find the amount of vapor that must be removed from the feed the increase its concentration as specified. Then, do a heat balance, accounting for the latent heat of the condensing steam in the heating tubes and the vapor being removed from the feed, in addition to the fact that the feed temperature is different than the temperature in the evaporator.