# Time needed for Evaporation

An apparent overall heat transfer coefficient is 735 Btu/(hr.sqft.^{o}F) for a forced circulation evaporator concentrating sulfite liquor under certain special conditions. How long will it take to concentrate 20000 lb of a feed liquor containing 5% solute by weight to a final concentration of 15% if the steam temperature is 230.8^{o}F and the temperature corresponding to the pressure in the vapor space is 210.7^{o}F. The heating surface is 100 sqft. Indicate all the assumptions made in obtaining the result. How will the result be affected if each of the assumptions is not made.

Calculations:

Assumptions:

(i) Steady state operation

(ii) No elevation in boiling point due to solute concentration.

(iii) No elevation in boiling point due to hydrostatic head.

(iv) No sub-cooling of condensate.

(v) Feed is at the temperature of 210.7^{o}F

Rate of heat transfer, Q = U A DT

= 735 x 100 x (230.8 - 210.7) / 3600 = 410.375 Btu/sec

Mass flow rate of steam, (m_{S}):

m_{S }l_{S} = 410.375

l_{S} = 958 Btu/lb (from Steam Tables - for the temperature 230.8^{o}F )

Therefore,

m_{S} = 410.375/958 = 0.42837 lb/sec

Making heat balance for the solution: (taking 210.7^{o}F as the datum temperature)

By denoting mass flow rate of vapor as, m_{V},

m_{V}l_{V} = m_{S}l_{S}

l_{V} = 971 Btu/lb (from Steam Tables - for the temperature 210.7^{o}F )

Therefore,

m_{V} = 410.375/971 = 0.42263 lb/sec

i.e., Mass of vapor evaporated = 0.42263 lb/sec

Time needed for evaporation:

Solute balance:

If the product is denoted as P,

20000 x 0.05 = P x 0.2

P = 5000 kg

Water in feed = water evaporated + water in concentrated solution

20000 x 0.95 = 0.42263 x Dt + P x 0.8

19000 = 0.42263 x Dt + 4000

Dt = 15000/0.42263 = 35492 sec = 9 hr 52 min

Time needed for evaporation = 9 hour and 52 minutes.

If the assumptions were not made:

If the assumption (i) is not made, then we have to make a unsteady balance (For Batch Evaporation). The results will be different for these two operations

If the assumption (ii) is not made, then the DT will be equal to, (T_{steam} - T_{boiling point of water corresponding to the pressure inside the evaporator} - boiling point elevation). This will lead to the increase of time needed for evaporation.

Assumption (iii) also has the same effect of assumption (ii).

Assumption (iv): If sub-cooling of steam is allowed, it will give more heat to the solution, which in turn will reduce the time needed for evaporation.

Assumption (v): If feed temperature is less than this, it leads to more steam requirement. So with the available heat transfer rate, it leads to increased evaporation time.