Problem - With the aid of figure below, write a solver for Problem 8.5.6 from the Textbook using Excel.
8.5-6. Evaporation of Sugar Solution in Double Effect Evaporator. A double-effect evaporator with reverse feed is used to concentrate 4536 kg/h of a 10 wt %sugar solution to 50%. The feed enters the second effect at 37.8°C. Saturated steam at 115.6°C enters the first effect and the vapor from this effect is used to heat the second effect. The absolute pressure in the second effect is 13.65 kPa abs. The overall coefficients are U1 = 2270 and U2 = 1705 W/m2 K. The heating areas for both effects are equal. Use boiling-point-rise and heat-capacity data from Example 8.5-1. Calculate the area and steam consumption.
Saturation temperature of pure water at ??1 is ????,2 , ??1 = ????,2 + ??????(??1)
Saturation temperature of pure water at ??2 is ????,3 , ??2 = ????,3 + ??????(??2)
Steam is fed at pressure ???? and ???? ≠ P1 ≠ ??2
You can modify the triple effect forward feed solver provided or you can write your own solver. You can use the functions written for triple effect forward feed solver. You can use the following functions:
Hsat,liq (T) = 4.1882 T,
Where T in oC and saturated liquid enthalpy is in kJ/kg
Hsat,liq (T) = 107715 T + 2501.4
Where T in oC and saturated vapor enthalpy is in kJ/kg
BPR(x) = 1.78x + 6.22x2
Where x is the weight fraction of the solution and boiling point rise, BPR is in oC
cp(x) = 4.19 - 2.35x
Where cp is the heat capacity of liquids in kJ/kg ·K with solutes and x is the weight fraction of the solutes.
If you are planning to modify the triple effect solver, please be very careful with the cells to be modified and the cells that are standing for known values, this is a reverse feed system. Also, you don't have a third effect and the equations associated with that. For the correct mass and energy balances, Classwork 1 solutions may aid you, but you may need to modify the solutions slightly to account for the lack of the third evaporator.
Correct equations for the mass, energy balances:
Correct inputs& constraints for the solver:
Convergence of the solver solutions:
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Effect 1
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Effect 2
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xF
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0.1
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L2
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2807.13
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kg/h
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TF
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37.8
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C
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V2
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1728.87
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kg/h
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S
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2129.508001
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kg/h
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x2
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0.161589
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Ts1
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115.6
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C
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Ts2
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85.87607
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C
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T1
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88.32106885
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T2
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52.35822
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C
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x1
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0.5
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Ts3
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51.90818
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C
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L1
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907.2
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kg/h
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Ps2
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13.65
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kPa
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F
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4536
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kg/h
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V1
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1899.930046
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kg/h
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cpsuperheat
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1.884
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kJ/kgK
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Textbook : https://4lfonsina.files.wordpress.com/2012/11/christiegeankoplis-transportprocessandunitoperations.pdf