The document discusses the calculation of a single effect evaporator. It provides the equations for material and heat balance calculations. An example calculation is shown for an evaporator concentrating a 1% salt solution. The key parameters calculated are the vapor and liquid outputs, and the required heat transfer area. Operational factors like feed temperature, pressure, and boiling point rise are also discussed.

1. TK4552 - EVAPORASI DAN KRISTALISASI
Program Studi Teknik Kimia
Fakultas Teknik Universitas Sebelas Maret 2020
Aida Nur Ramadhani, S.T., M.T.
PERHITUNGAN
EVAPORATOR SINGLE EFFECT

2. Sub-CPMK 3 : Mahasiswa mampu menerapkan konsep perhitungan evaporator
• Ketepatan menganalisis neraca massa dan panas di sekitar evaporator
• Ketepatan menganalisis apa saja yang menjadi faktor yang
mempengaruhi proses evaporasi
PERHITUNGAN EVAPORATOR SINGLE EFFECT

3. PERHITUNGAN EVAPORATOR SINGLE EFFECT
EVAPORATOR
KOEFISIEN TRANSFER PANAS OVERALL

4. Diagram – single effect evaporator
feed, F
TF , xF , hF.
steam, S
TS , HS
concentrated liquid, L
T1 , xL , hL
condensate, S
TS , hS
vapor, V to condenser
T1 , yV , HV
P1
T1
heat-exchanger
tubes

5.  Umpan F (biasanya encer) masuk pada suhu TF
 Uap jenuh masuk pada suhu TS
 Larutan dalam evaporator diasumsikan tercampur sempurna dan memiliki
komposisi yang sama
Neraca massa dan panas
 Persamaan dasar:
dengan,
∆𝑇 = beda suhu antara steam terkondensasi dengan cairan mendidih dalam
evaporator (K).
q = kerja (Btu/h)
𝑞 = 𝑈𝐴(∆𝑇)

6. Menghitung
 Laju alir fase vapor, V dan fase liquid, L.
 Luas perpindahan panas, A
 Koefisien transfer panas overall, U.
 Fraksi mol solute, xL.

7. Diagram – single effect evaporator
Steam, S
PS, TS, HS
Feed, F
xF, TF, hF
Condensate, S
PS, TS, hS
Vapour, V
yV, T1, HV
Concentrate, L
xL, T1, hL
P
T1
Feed:
F – mass flow rate
xF – mass fraction of solute in feed
TF – temperature of feed
hF – enthalpy of feed
Vapour leaving the evaporator:
V – mass flow rate
yV – mass fraction of solute in vapour
T1 – temperature of vapour
HV – enthalpy of vapour

8. Steam, S
PS, TS, HS
Feed, F
xF, TF, hF
Condensate, S
PS, TS, hS
Vapour, V
yV, T1, HV
Concentrate, L
xL, T1, hL
P
T1
Concentrate leaving the evaporator:
L – mass flow rate
xL – mass fraction of solute in concentrate
T1 – temperature of concentrate
hL – enthalpy of concentrate
Steam:
S – mass flow rate
PS – steam pressure
TS – steam temperature
HS – enthalpy of steam
hS – enthalpy of condensate
P – pressure in the evaporator
T1 – temperature in the evaporator

9. Neraca Massa
Overall material balance:
F = L + V
Solute balance:
F xF = L xL + VyV
If the vapour is free ofsolute:
F xF = LxL
Steam, S
PS, TS, HS
Feed, F
xF, TF, hF
Condensate, S
PS, TS, hS
Vapour, V
yV, T1, HV
Concentrate, L
xL, T1, hL
P
T1

10. Neraca Panas
Maka,
F hF + S HS = L hL + V HV + ShS
F hF + S (HS - hS) = L hL + V HV
F hF + S λ = L hL + VHV
where λ = HS –hS = heat latent
Heat in feed + heat in steam = heat in concentrated liquid + heat in vapor
+ heat in condensed steam
S
F
S
V
L

11. S
TS, HS
F
S
TS, hS
V
L
P
T1
Energy lost by thesteam
q = S λ = S (HS –hS)
Jika tidak ada energi yang hilang
ke lingkungan, jumlah energi akan
ditransfer dari uap ke larutan
melalui dinding tabung area A dan
koefisien perpindahan panas
overall U.
Maka,
q = U AΔT = U A (TS – T1)

12. Operasi – single effect evaporator
 Panas dari steam saat Ts bisa didapatkan dari steam table.
 Entalpi dari umpan feed (F) – larutan dengan konsentrasi X – dalam tabel, jika
tidak tersedia:
Asumsi:
 Bahwa feed sangat dilute (encer) sehingga panas laten penguapan 1 kg
massa air dari larutan berair (aqueous) dapat diperoleh dari steam table
menggunakan suhu larutan mendidih T1 (suhu evaporator).
 Jika kapasitas panas (cp) dari liquid feed (umpan cair) diketahui  dapat
digunakan untuk menghitung entalpi (abaikan panas pelarutan).

13. Contoh:
A continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt % salt
solution entering at 38ºC to a final concentration of 1.5 wt %.
The vapor space of the evaporator is at 101.325 kPa (1.0 atm abs) and the steam
supplied is saturated at 150 kPa. The overall coefficient U = 1704 W/m2.K.
Calculate the amounts of vapor and liquid products and the heat-transfer area
required. Assumed that, since it its dilute, the solution has the same boiling point
as water.
Data:
The latent heat of steam at 150 kPa 383.2 K = 2230 kJ/kg (958,8 Btu/h)
The latent heat of water at 101.325 kPa 373,2 K= 2257 kJ/kg (970,3 Btu/h)

14. Penyelesaian
Diketahui:
F = 9.072 kg/h
xF = 1 wt % = 0,01 kg solute / kgfeed
TF =38°C =
xL = 1,5 wt%
= 0,015 kg solute / kg liquid product
P = 101,325 kPa (1.0 atm abs)
PS = 150kPa
U = 1704 W/m2.K
T1 = saturated temperature at P (= 101,325 kPa) = 100°C
TS = saturated temperature at 150 kPa = 111,4°C

15. Penyelesaian
Steam, S
PS, TS, HS
Feed, F
xF, TF, hF
Condensate, S
PS, TS, hS
Vapour, V
yV, T1, HV
Concentrate, L
xL, T1, hL
P
T1
Ditanya:
V=?
L=?
A=?

16. Penyelesaian
NM total,
F = L + V
9.072 = L + V
NM solute,
F xF = L xL
9.072 (0,01) = L (0,015)
L = 6.048 kg/h of liquid
Maka,
V = 3.024 kg/h of vapor

17. Penyelesaian
Asumsi: larutan sangat encer (as water);
cpF = 4,14 kJ/kg. K (Table A.2-5)
Dari soal:
At P1 = 101,325 kPa, T1 = 373,2 K (100°C) (datum temperature).
HV = 2.257 kJ/kg.
At PS = 143,3 kPa, TS = 383,2 K (110°C).
 = 2.230 kJ/kg.
Maka, entalpi umpan dapat dihitung,
hF = cpF (TF – T1)
hF = 4,14 (311,2 – 373,2)
= -256,68 kJ/kg.

18. Penyelesaian
F hF + S  = L hL + V HV
with hL = 0, karena T datum adalah 373.2 K.
9.072 (-257,508) + S (2.230) = 6.048 (0) + 3.024 (2.257)
S = 4.108 kg steam /h
Panas (q) yang dipindahkan melalui luas permukaan panas A,
q = S ()
q = 4.108 (2.230) (1.000 / 3.600) = 2.544.000 W
Luas permukaan panas (A);
q = U A T = U A (TS – T1)
2.544.000 = 1704 A (383,2 – 373,2)
A = 149,3 m2
Neraca Panas

19. Penyelesaian
Maka,
V = 3.024 kg/h of vapor
L = 6.048 kg/h of liquid
A = 149,3 m2

20. Pengaruh variable proses terhadap operasional evaporator
 Suhu aliran umpan memiliki efek besar pada operasi evaporator.
 Saat umpan tidak berada pada titik didihnya, uap pertama-tama diperlukan
untuk memanaskan umpan ke titik didihnya lebih dahulu, baru kemudian
mengalami proses penguapan.
 Pemanasan awal umpan dapat mengurangi besarnya area perpindahan panas
evaporator akibat hal tersebut.
suhu umpan:

21.  Tekanan dalam evaporator menentukan titik didih larutan (T1).
 Tekanan uap menentukan suhu uap (Ts)
 Karena q = U A (TS - T1), maka semakin besar selisih (TS - T1) akan membantu
mengurangi area perpindahan panas yang dibutuhkan  mengurangi biaya.
 Dapat dilakukan dengan memanfaatkan efek tekanan.
Tekanan:

22. Boiling Point Rise
 Dalam contoh 1, larutan diasumsikan cukup encer sehingga dianggap memiliki
sifat termal yang sama dengan air.
 Hal tersebut tidak selalu benar.
 Untuk larutan dengan konsentrasi tinggi  terdapat kenaikan titik didih larutan
 Aturan Duhring (Duhring Law) adalah hukum empiris hubungan antara titik
didih larutan dan titik didih pelarut pada tekanan yang berbeda untuk tiap
konsentrasi larutan – perkiraan kenaikan titik didih larutan.

24. Contoh
 Pressure in an evaporator is given as 25,6 kPa (3,72 psia) and a solution of 30%
NaOH is being boiled.
 Determine the boiling temperature of the NaOH solution and the boiling-
point rise (BPR) of the solution over that of water at the same pressure.

25. Penyelesaian:
 Cari titik didih air pada P soal  25,6 kPa (3,72 psia)
 Steam table, boiling point of water at 25,6 kPa is 65,6°C (Tsat)
 Dari TD air, dicari titik didih untuk larutan 30% NaOH di fig.8-4
 Untuk TD air 65,6°C (150°F) and 30% NaOH  titik didih larutan NaOH
tersebut adalah 79,5°C (175°F).
 Menghitung BPR
 The boiling-point rise = 79,5°C – 65,6°C = 13,9°C (25°F)

26. Enthalpy-Concentration Chart
Heat of solution – Panas pelarutan
 Panas yang dihasilkan atau diserap ketika suatu zat melarut (terbentuk sebuah
larutan).
 Jumlah panas yang terlibat ketika satu mol atau satu gram larut dalam pelarut
mengalami pelarutan.
 Jumlah panas bergantung pada jenis senyawa, dan jumlah air yang digunakan
untuk melarutkan.
 Contoh pada proses pelarutan pellet NaOH.

28. Latihan!
An evaporator is used to concentrate 4.536 kg/h of a 20 % solution of NaOH
in water entering at 60ºC to a product of 50 % solid. The pressure of the
saturated steam used is 172,4 kPa and the pressure in the vapor space of
the evaporator is 11,7 kPa. The overall heat-transfer coefficient is 1.560
W/m2.K.
Calculate the steam used, the steam economy in kg vaporized/kg steam
used, and the heating surface area in m2.

29. Latihan!
Buatlah diagramnya terlebih dahulu!
U = 1560 W/m2
T1 A = ?
P1 = 11,7 kPa
F = 4.536 kg/h
TF = 60°C
xF = 0,2
hF.
S = ?
TS , HS
PS = 172,4 kPa
L, T1 , hL
xL = 0,5
S, TS , hS
V, T1 , yV , HV

30. Penyelesaian?
NM total,
F = L + V
4.536 = L + V
NM solute,
F xF = L xL
4.536 (0,2) = L (0,5)
L = 1.814 kg/h of liquid
Maka,
V = 2.722 kg/h of vapor

31. Penentuan T1 = Tsat + BPR of the 50% concentrate product.
 Tsat air dari steam table.
Pada 11.7 kPa, Tsat = 48,9ºC.
 Dari grafik Duhring (Fig. 8.4-2), untuk Tsat = 48,9 ºC and 50 % NaOH , titik
didih dari larutan adalah T1 = 89,5ºC.
BPR = T1 - Tsat = 90 – 48,9 = 40,6ºC
 From the enthalpy-concentration chart (Fig.8.4-3), for
TF = 60ºC and xF = 0,2 get hF = 214 kJ/kg.
T1 = 90ºC and xL = 0,5 get hL = 505 kJ/kg.
Penyelesaian?

32. Penyelesaian?
 For saturated steam at 172,4 kPa, from steam table, we get
TS = 115,6 ºC and  = 2.214 kJ/kg.
 To get HV for superheated vapor, first we obtain the enthalpy at Tsat = 48,9ºC and P1
= 11,7 kPa, get Hsat = 2.590 kJ/kg. Then using heat capacity of 1.884 kJ/kg.K for
superheated steam.
HV = Hsat + cp.BPR
= 2.590 + 1,884 (40,6) = 2.667 kJ/kg.
 Substituting into heat balance equation and solving for S,
F hF + S  = L hL + V HV
4.535 (214) + S (2.214) = 1.814 (505) + 2.722 (2667)
S = 3.255 kg steam /h.
Atau bisa langsung cari Hv di
steam table untuk
superheated;
T1 = 89,5ºC P = 11,7 kPa

33. Penyelesaian?
 The heat q transferred through the heating surface area, A is
q = S ()
q = 3.255 (2.214) (1.000 / 3.600) = 2.002.000 W
 Solving for capacity single-effect evaporator equation;
q = U A T = U A (TS – T1)
2.002.000 = 1.560 A (115,6 – 89,5)
 Solving,
A = 49,2 m2
Steam economy = 2.722/3.255
= 0,836

34. Tugas