Problem 5.6.1: A single effect evaporator operates at a pressure of 1.89 psi. What will be the heating surface necessary to concentrate 9900 pound per hour of 10 percent caustic soda to 41 percent, using steam at 240 degrees Fahrenheit and 90 percent quality. The heating surface is 4 feet below the liquid level. Steam condensate leaves at 232 degrees Fahrenheit. Heat transfer coefficient of the internal heat exchanger is given as 220 Btu per hour per square foot per degree Fahrenheit. Height of liquid above heating surface, H is 4 feet. Heat capacity of liquor, c_Pf is 0.96 Btu per pound per degrees Fahrenheit. Temperature of liquor, T_F is 64 degrees Fahrenheit. Boiling point elevation of the solution, BPE is 54 degrees Fahrenheit. Density of liquor, rho is 83.66 pounds per cubic foot. Heat capacity of concentrated mix, c_pm is 0.78 Btu per pound per degree Fahrenheit.

Solution:

Mass flow rate of liquor, F_o is 9900 pound per hour. Weight fraction of solids in the feed, x_o is 0.1. Weight fraction of solids in the final exiting stream, x₁ is 0.41. Temperature of steam, T_o is 240 degrees Fahrenheit.

Quality of steam, x is 90 percent. From the steam tables, properties of steam, enthalpy of fluid, h_f is 208.45 Btus per pound. Latent heat of vaporization, h_fg or lambda is 952.1 BTUs per pound. Temperature of condensate, T_c is 232 degrees Fahrenheit; enthalpy of condensate h_c is 200.35 BTU per pound. Operating pressure of evaporator is 1.89 psi.

Solute balance around the evaporator can be applied to give weight of concentrated solution leaving the evaporator. Note that the caustic soda does not leave with vapors from the top of the evaporator. 

F₁ is equal to x_o time F_o over x₁ or 2415 pound per hour.

Water balance around the evaporator gives the amount of vapors leaving the evaprator.

Water entering the evaporator is 0.9 time 9900 or 8910 pounds.

Water leaving with concentrated solution is 0.59 time 2415 or 1425 pounds.

Water leaving as vapors from the evaporator (D₁) is 8910 minus 1425 or 7485 pound per hour.

Pressure at the surface of the liquor is the same as evaporator pressure, i.e, 1.89 psi.

Pressure at the boiling location (near the heat exchanger) can be found by adding the static head to liquor 

P is equal to P₁ plus H time rho over 144 or 4.214 psi. Note that factor of 144 has been used to get the answer in psi.

From steam stables, we can see this vapor pressure corresponds to a saturation temperature, T_s of 156 degrees Fahrenheit.

Concentrated liquir has a boiling point elevation or rise is 54 degrees Fahrenheit, so liquor will boil at T₁' 156 plus 54 or 210 degrees Fahrenheit.

At evaporator pressure of 1.89 psi, latent heat of vaporization is 1023.3 BTUs per pound

Enthalpy of the entering feed is

Heat received by feed, Q_c is F₀(c_Pm.T₁' minus c_PfT_F) plus D₁λ₁ or 8.673 x 10⁶ Btus per hour.

Enthalpy of steam entering the heat exchanger, x.λ_o plus h_f or 1065.34 Btus per pound. Enthalpy of condensate, h_c is 200.35 Btus per pound. Enthalpy provided by steam is 1065.34 minus 200.35 or 864.99 Btus per pound.

Amount of steam required, D_o is Q_c over q_h or 1.089 x 10⁴ pounds per hour. Temparture difference for the heat exchanger, ΔT is equal to T_o minus T₁' or 240 minus 210 or 30 degrees Fahrenheit. Heat exchanger area, A can be found from the relation Q_c is equal to UA delta T. So, area is U over ΔT or 1.089 x 10⁴ over 220 over 30 or 1,427 square feet.