Ortho-phosphoric acid (H3PO4) is produced as a dilute aqueous solution that must be concentrated before further use. In one facility, 100 tons/day of a 28 wt% P2O5 solution [see part (a) of this problem] at 125°F is to be concentrated in a single evaporator to 42 wt% P2O5. Heat is supplied to the evaporator by condensing saturated steam at 27.5psia. The evaporator is to operate at 3.7psia, and there is a boiling point elevation of 37°F for the 42 wt% P2O5 solution in the evaporator (see Section 6.5c). The heat of solution of H3PO4 at 77°F may be taken to be —5040 Btu/lb-mole H3 PO4 relative to H3PO4 (1) and H2O(l). The heat capacity of the 28% solution is 0.705 Btu/ (lbm ∙ °F) and that of the 42% solution is 0.583 Btu/ (lbm ∙°F).

(a) It is conventional for the compositions of phosphoric acid solutions to be expressed in terms of wt% P205. Write the stoichiometric equation for the formation of ortho-phosphoric acid (MW = 98.00) from phosphorus pent oxide (MW = 141.96), and use it to derive the expression wt % H3PO4 = l.381 (wt% P2O3)

(b) Calculate the ratio (lbm water evaporated/lbm feed solution).

(c) Suppose the water evaporated is subsequently condensed at a constant pressure of 3.7psia. Determine the condensate flow rate in gal/min. How much heat (Btu/min) can be recovered through condensation of this water? At what temperature is this heat available? (To put it another way. if this heat were to be transferred to another stream, what is an upper bound on the temperature of that stream?)

(d) How much steam (lbm/h) must be supplied to the system to evaporate the required amount of water? Recast your answer in terms of lbm steam per lbm water evaporated.