# Energy Balance on Rv1 Essay

Component Molecular mass

(kg kmol-1) Component Molecular mass

(kg kmol-1)

Na+ 23 NaHCO3 84

Ca2+ 40 Na2CO3 106

Cl- 35.5 CO2 44

NaCl 58.5 H2O 18

CaCl2 111 CaCO3 100

Table 1: Relative molecular mass values for all components in system

Data was taken from Tuesday 11:28-11:40. Flow rates and temperatures throughout this period were fairly constant and so the mean values were found using the density of water at the relevant temperature (it was assumed that all streams in the process have the density of water). Data for the density of water was taken from Perry (1999). An example calculation for the mean product flow rate is shown below:

m ̇=ρV ̇

m ̇=(4.63*60*991.77)/1000=275.7 kg h^(-1)

A mass

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Rearranging for the feed rate of Sodical, S:

S=7 kg h^(-1)

The amount of CaCl2 in the Sodical was found, using the contamination, to be 0.35 kg h-1 (0.01 kmol h-1). Using a stoichiometric balance, the amount of dosing agent required and water, sodium chloride, carbon dioxide and calcium carbonate produced were found. The results are shown below:

Sodical feed rate NaCl feed rate CaCl2 feed rate NaHCO3 feed rate Solid feed rate H20 produced

(kg h-1) 7.00 6.64 0.35 1.65 8.65 0.18

(kmol h-1) 0.11 0.01 0.02 0.01 H20 required Total feed rate CO2 produced CaCO3 produced NaCl produced Na2CO3 produced

(kg h-1) 268.51 277.16 0.43 0.10 1.15 1.04

(kmol h-1) 14.92 0.01 0.01 0.02 0.01

Table 2: Feed and production rates in RV1

In order to find the flow rates on every line of the system four equations were derived:

b=r+Y (a)

0.2b=e+r (b)

x=r/R (c)

x=e/(e+P) (d)

Equation (a) was substituted into equation (b) and rearranged to make r the subject:

r=(0.2Y-e)/0.8 (e)

Equations (c) and (d) were set equal to one another and equation (e) substituted in:

r/R=e/(e+P)=(0.2Y-e)/0.8R (f)

The equation for (f) was solved for e and produced the following equation:

0.05e^2-e(0.05Y-0.05P-0.95R)-0.05YP=0 (g)

Substituting in all known values, R=280.2 kg h-1, P=275.7 kg h-1, Y=0.1 kg h-1, and