PERHITUNGAN NERACA ENERGI

Perhitungan panas untuk bahan dalam fasa cair/gas

Qi/o = ∫
T=298
n.Cp.dT (Van Ness, 1975)

Data perhitungan Cp untuk air:

Cpl = 18,2964 + 0,472118 T + (-0,00133878) T2 + 0,000001314 T3
(Reklaitis,1983)
Cpv = 7,9857 + 0,00046332 T + 1,402810-6 T2 + (-6,5784),10-10 T3
(Reklaitis,1983)

Data perhitungan Cp untuk abu:

Cp = 0,1800 + 0,000078 T (Reklaitis,1983)

Data perhitungan Cp untuk HCl:

Cpl = 17,7227 + 0,904261 T + (-0,00564496) T2 + (1,13383),10-5 T3
(Reklaitis, 1983)
6,969 + (-2,236),10-4 T + (7,333),10-7 T2 + (-1,776),10-5 T3
(Reklaitis,1983)

Tabel L.B.1 Kontribusi Gugus untuk Perhitungan ΔHf298 (kJ/mol)

Gugus Nilai
-CH2- -26,80
Saya
--CH-- 8,67

-OH -208,04
-O- -138,16
(Sumber: Perry, 1999)

Nilai ΔHf 298 untuk senyawa bio-polimer

1. Glukosa (C6H12O6)
‫׀‬
ΔHf298 = 5(-OH) + 5(-CH-) + 1 (- O -) + 1 (- CH2 -)
= -1049,55 kJ/mol
-5830,8333 kJ/kg
 2. Pati (C6H10O5)
‫׀‬
ΔHf298 = 3(-OH) + 5(-CH-) + 2 (- O -) + 1 (- CH2 -)
= -772,71 kJ/mol
= -4769,8148 kJ/kg

Nilai kapasitas panas (Cp) untuk senyawa bio-polimer:

C6H10O5
‫׀‬ ‫׀‬ ‫׀‬
Cp = 1(-CH2OH) + 1(-O-) + 1(-CH2-) + 2(-CHOH) + 1(-CHOH) + 1(-CHOH)
= 1(17,5) + 1(8,4) + 1(6,2) + 2(18,2) + 1(4,4) + 1(26,6)
= 99,5 kal/mol,K = 2,5698 kJ/kg.K

Serat ((C6H10O5)n
‫׀‬ ‫׀‬ ‫׀‬
Cp = 1(-CH2OH) + 1(-O-) + 1(-CH2-) + 2(-CHOH) + 1(-CHOH) + 1(-CHOH)
= 1(17,5) + 1(8,4) + 1(6,2) + 2(18,2) + 1(4,4) + 1(26,6)
= 99,5 kal/mol,K = 2,5698 kJ/kg.K

3. Protein (CH3(CHNH2) COOH)

H O
‫׀‬ ‫׀‬ ‫׀׀‬
Cp = 1(H-N-) + 16(-CH2-) + 1(-C-OH) + 1(-CH-)
= 1(8,8) + 16(7,26) + 1(19,1)
= 144,06 kal/mol,K = 2,5719 kJ/kg.K

4. Fat
a. Asam palmitat (10%) (CH3(CH2)14(COOH)
O
‫׀׀‬
Cp = 1(-CH3) + 14(-CH2-) + 1(-CH2-) + 1(-C-OH)
= 1(8,8) + 14(7,26) + 1(19,1)
= 129,54 kal/mol,K = 2,1136 kJ/kg.K
b. Asam stearat (3%) (CH3(CH2)16 COOH
O
‫׀׀‬
Cp = 1(-CH3) + 16(-CH2-) + 1(-C-OH)
= 1(8,8) + 16(7,26) + 1(19,1)
= 144,06 kal/mol,K = 2,1188 kJ/kg.K

c. Asam oleat (30%) (CH3(CH2)7CHCH(CH2)7COOH

O
‫׀׀‬ ‫׀׀‬
Cp = 1(-CH3) + 14(-CH2-) + 1(-CH2-) + 1(-C-OH) + 2(-C-H)
= 1(8,8) + 14(7,26) + 1(19,1) + 2(5,10)
= 139,74 kal/mol,K = 2,0699 kJ/kg.K

5. Glukosa (C6H12O6)
‫׀‬ ‫׀‬
Cp = 1(-CH2OH) + 1(-O-) + 1(-OH) + 2(-CH) + 3(-CHOH)
‫׀‬ ‫׀‬
= 1(8,4) + 1(8,4) + 1(10,7) + 2(4,4) + 3(18,2)
= 100 kal/mol K = 2,3444 kJ/kg.K

Hidroksi metil furfural (C6H6O3)

--C=O ‫׀‬
Cp = 1(-O-) + 1(-CH2OH) + 1 (‫׀‬ ) + 2(-CH=) + 2(-C=)
H
= 1(8,4) + 1(17,5) + 1(12,66) + 2(5,3) + 2(2,9)
= 54,96 kal/mol K = 1,8250 kJ/kg.K

7. Fruktosa (C6H12O6)
‫׀‬ ‫׀‬ ‫׀‬
Cp = 1(-O-) + 2(-CH2OH) + 1(-CH) + 2(-CHOH) + 1(-C-OH)
‫׀‬ ‫׀‬
= 1(8,4) + 2(17,5) + 1(4,4) + 2(18,2) + 1(26,6)
= 110,8 kal/mol K = 2,5755 kJ/k.K
Panas penguapan (Hvl)

H2O = 40656,2 J/mol = 2258,6778 kJ/kg (Reklaitis,1983)

HCl = 16150,3 J/mol = 442,4658 kJ/kg (Reklaitis,1983

Tangki Perebusan

Temperatur dasar = 25°C = 298 K

303 303 303

F1Pati ∫298 Cp.dT + F1air ∫298 Cp.dT + F1protein ∫298 Cp.dT +

303 303 303
F1abu ∫ Cp.dT + F1lemak ∫ Cp.dT + F1serat ∫ Cp.dT +
298 298 298

303

∫ Cp.dT
F2udara
298

= (887,6263 kg/jam) (12,849 kJ/kg) + (132,5758 kg/jam) (21 kJ/kg)

(130,0505 kg/jam) (12,8595 kJ/kg) + (21,4646 kg/jam) (7,0217
kJ/kg) + (63,1313 kg/jam) (13,2345 kJ/kg) + (27,7778 kg/jam)
(12,849 kJ/kg) + (631,3131 kg/jam) (21 kJ/kg)
= 30.462,3078 kJ/jam

323 323 323

F3Pati ∫298 Cp.dT + F3udara∫298 Cp.dT + F3protein ∫298 Cp.dT +

323 323 323

F3abu ∫ Cp.dT + F3lemak ∫ Cp.dT + F3serat ∫ Cp.dT

298 298 298

(887,6263 kg/jam) (64,2450 kJ/kg) + (763,8889 kg/jam) (105

kJ/kg) + (130,0505 kg/jam) (64,2975 kJ/kg) + (21,4646 kg/jam)
(35,3715 kJ/kg) + (63,1313 kg/jam) (66,1725 kJ/kg) + (27,7778
kg/jam) (64,2450 kJ/kg)
= 152.317,184 kJ/jam

Qo – Qi
 Keluar-Masuk
(152.317,184-30.462,3078) kJ/Jam
= 121854,8762kJ/Jam
Uap superpanas pada 1 atm, 150̊C, 2776,3 kJ/Kg
(Reklaitis, 1983)
Uap jenuh pada 1 atm, 100̊C 2676 kJ/Kg pada 100̊C
(Reklaitis, 1983)
HL (100̊C) = 419,064 kJ/Kg
(Reklaitis, 1983)
Lamda = [H (150̊C) – Hv (100̊C)] + [Hv (100̊C) – Hl (100̊C)]
(2776,3-2676) + (2676-419,064)
= 2357,236 kJ/kg
Jumlah steam yang diperlukan (m) = Q/lamda
121854,8762kJ/Jam
=
2357,236kJ/kg
51.694 kg/jam

2. Reaktor Hidrolisis
323 323 323

Panas Masuk = F4Pati ∫

298
Cp.dT + F4Udara∫ Cp.dT + F4Protein
298
∫
298
Cp.dT+ F4Abu

323 323 323 303

∫
298
∫ F4SeratCp.dT+F∫ 5UdaraCp.dT
Cp.dT+ F 4LemakCp.dT+
298 298
∫298

303
+ F5HCl ∫298 Cp.dT
= (887,6263 kg/jam) (64,2450 kJ/kg) + (763,8889 kg/jam) (105
kJ/kg) + (130,0505 kg/jam) (64,2975 kJ/kg) + (21,4646 kg/jam)
(35,41 kJ/kg) + (63,1313 kg/jam) (66,1725 kJ/kg) + (27,7778
(kg/jam) (64,2450 kJ/kg) + (238,6364 kg/jam) (21 kJ/kg) +
(140,1515 kg/jam) (11,9831 kJ/kg)
= 159.008,8242 kJ/jam
 353 353 353
Panas Keluar = F6HCL ∫298 Cp.dT + F6Pati ∫298 Cp.dT + F6Udar ∫ Cp.dT+ F6Protein
298

353 353 353 353

∫298 Cp.dT + F 6Abu∫298Cp.dT + F 6LemakCp.dT+

∫298 F6SeratCp.dT ∫298 +

353
FGlukosa ∫298 Cp.dT
= (140,1515 kg/jam) (131,8141 kJ/kg) + (17,75253 kg/jam)
(141,3390 kJ/kg) + (905,8726 kg/jam) (231 kJ/kg) + (130,0505
(kg/jam) (141,4545 kJ/kg) + (21,4646 kg/jam) (77,902 kJ/kg) +
(63,1313 kg/jam) (145,5795 kJ/kg) + (27,7778 kg/jam)
(141,3390 kJ/kg) + (966,5264 kg/jam) (128,942 kJ/kg)
= 388.050,5596 kJ/jam

Reaksi: C6H10O5 + H2O → C6H12O6

Panas reaksi pada suhu 25̊C (298 K)

r = 1043,8485 kg/jam
r. ΔHr25 = [(-5830,8333) - (-4769,8148 + (-13564,6272))] kJ/kg x 1043,8485
kg/jam
= 13.051.873,19 kJ/jam

Panas yang dilepas steam (Q) = r. ΔHr 25 + (Qo - Qi)

= 13.051.873,19 + (388.050,5596-159.008,8242)
= 13.280.914,92 kJ/jam
Uap super panas pada 1 atm, 150̊C, 2776,3 KJ/Kg
(Reklaitis, 1983)
Uap jenuh pada 1 atm, 100̊C, 2676 KJ/Kg pada 100̊C
(Reklaitis, 1983)
HL (100̊C) = 419,064 KJ/Kg
(Reklaitis, 1983)
Lamda = [H (150̊C) - Hv (100̊C)] + [Hv (100̊C) - Hl (100̊C)]
[2776,3-2676,1] + [2676,1-419,04]
= 2357,236 kJ/kg
13.280.914,92 kJ/ jam
Jumlah steam yang diperlukan (m) = Q/lamda =
2357,236kJ/kg

5634,1049 kg/jam

3. Pendingin-01
Temperatur dasar = 25°C = 298 K
Panas masuk ke Cooler = panas keluar Reaktor = 388.050,5596 kJ/jam

333 333 333

F7HCL ∫298 Cp.dT + F7Pati ∫298 Cp.dT + F7Air ∫298 Cp.dT+ F7Protein

333 333 333 333

∫
298
Cp.dT + F 7Abu∫Cp.dT + F 7LemakCp.dT+
298
∫ F7SeratCp.dT
298
∫298 +

333 333
F7Glukosa ∫ Cp.dT + F7HMF ∫ Cp.dT
298 298

=(140,1515 kg/jam) (83,8817 kJ/kg) + (17,75253 kg/jam)

(89,9430 kJ/kg) + (905,8726 kg/jam) (147 kJ/kg) + (130,0505
(kg/jam) (90,0165 kJ/kg) + (21,4646 kg/jam) (49,574 kJ/kg) +
(63,1313 kg/jam) (92,6415 kJ/kg) + (27,7778 kg/jam) (89,9430
kJ/kg) + (956,8611 kg/jam) (82,054 kJ/kg) + (9,6653 kg/jam)
(63,8750 kJ/kg)
= 246.765,5597 kJ/jam
Panas yang diserap air pendingin (Q) = Qo–Qi
 (246.765,5597-388.050,5596) kJ/jam
- 141.284,9999 kJ/jam
H(25°C) = 2528,9566 kJ/kg (Reklaitis, 1983)
H(50°C) = 2576,4251 kJ/kg (Reklaitis, 1983)
Lamda = H[25°C] – H[50°C]
(2528,9566 – 2578,4251) kJ/kg
= -49,4685 kJ/kg
Jumlah air pendingin yang diperlukan (m) = Q/lamda
−141.284,9999 kJ/ jam
=
−49,4685 kJ/kg
2.856,0599 kg/jam
4. Pemanas-01
333 333 333

F13HCL ∫
298
Cp.dT + F13Pati ∫
298
Cp.dT + F13Udara∫ Cp.dT+ F13Protein
298

333 333 333

∫
298
Cp.dT+ F 13Lemak ∫
298
Cp.dT+ F 13Glukosa ∫
298
Cp.dT + F13HMF

333

∫
298
Cp.dT

(140,1515 kg/jam) (83,8817 kJ/kg) + (17,75253 kg/jam)

(89,9430 kJ/kg) + (901,3432 kg/jam) (147 kJ/kg) + (0,0033
(kg/jam) (90,0165 kJ/kg) + (62,8156 kg/jam) (92,6415 kJ/kg) +
(952,0768 kg/jam) (82,054 kJ/kg) + (9,6653 kg/jam) (63,8750
kJ/kg)
= 230.409,0215 kJ/jam
 353 353 353
F14HCl ∫
298
Cp.dT + F14Pati ∫
298
Cp.dT + F14Udara∫ Cp.dT+ FProtein
298

353 353 353

∫298 Cp.dT + F 14Lemak 298

∫ Cp.dT+ F 14Glukosa 298
∫ Cp.dT+ F 14HMF

353

∫298 Cp.dT
= (140,1515 kg/jam) (131,8141 kJ/kg) + (17,75253 kg/jam)
(141.339 kJ/kg) + (901.3432 kg/jam) (231 kJ/kg) + (0,0033
(kg/jam) (141,4545 kJ/kg) + (62,8156 kg/jam) (145,5795 kJ/kg)
+ (952,0768 kg/jam) (128,942 kJ/kg) + (9,6653 kg/jam)
(100,3750 kJ/kg)
= 362.071,3195 kJ/jam
Panas yang dilepas steam (Q) = Qo – Qi
(362.071,3195-230.409,0215) kJ/jam
= 131.662,298 kJ/jam
Uap superpanas pada 1 atm, 150°C, 2776,3 kJ/Kg pada 150°C
(Reklaitis, 1983)
Uap jenuh pada 1 atm, 100°C 2676 kJ/Kg
(Reklaitis, 1983)
HL (100°C) = 419,064 kJ/Kg
(Reklaitis,
1983)
Lamda = [H(150°C) – Hv (100°C)] + [Hv (100°C) – Hl(100°C)]
= [2776,3-2676] + [2676-419,064]
= 2357,236 kJ/kg
131.662,298 kJ/ jam
Jumlah steam yang diperlukan (m) = Q/lamda =
2357,236kJ/kg

= 55,8545 kg/jam
5. Evaporator-1
Panas masuk ke Evaporator-01 = Panas keluar Heater-01
= 362.071,3195 kJ/jam
 376 376 376

F16Pati ∫298 Cp.dT + F16Udara∫ Cp.dT + F16Protein

298
∫
298
Cp.dT + F16Lemak

376 376 376

∫298 Cp.dT+ F16Glukosa ∫298 Cp.dT+ F 16HMF ∫

298
Cp.dT+ F15Udara

376 376

∫298 Cp.dT+ HVL+ F15HClCp.dT+

∫298 HVL
= (17,7525 kg/jam) (200,4444 kJ/kg) + (405,6044 kg/jam) (327,6
kJ/kg) + (0,0033 kg/jam) (200,6082 kJ/kg) + (62,8156 kg/jam)
(206,4582 kJ/kg) + (952,0768 kg/jam) (182,8632 kJ/kg) +
(9,6653 kg/jam) (142,3500 kJ/kg) + (495,73876 kg/jam) (327,6
(kJ/kg) + (2258,6778 kJ/kg) + (140,1515 kg/jam) (186,9364
kJ/kg) + (442,4658 kJ/kg)
1.695.209,326 kJ/jam
Qo– Qi
(1.695.209,326-362.071,3195) kJ/jam
1.333.138,007 kJ/jam
Uap superpanas pada 1 atm, 150°C, H (150°C) = 2776,3 kJ/Kg
(Reklaitis, 1983)
Uap jenuh pada 1 atm, 100°C, HV (100°C) = 2676 kJ/Kg
(Reklaitis, 1983)
HL (100°C) = 419,064 kJ/Kg
(Reklaitis, 1983)
Lamda = [H(150°C) – Hv (100°C)] + [Hv (100°C) – Hl(100°C)]
= [2776,3-2676] + [2676-419,064]
= 2357,236 kJ/kg
Jumlah steam yang diperlukan (m) = Q/lamda
1.333.138,007 kJ/ jam
=
2357,236kJ/kg
= 565,5514 kg/jam
6. Pendingin-2
376 376 376

F16Pati ∫298 Cp.dT + F16Udara∫ Cp.dT + F16Protein

298
∫298 Cp.dT + F16Lemak

376 376 376

∫298 Cp.dT+ F16GlukosaCp.dT+

∫298 F16HMFCp.dT ∫298
=(17,7525 kg/jam) (200,4444 kJ/kg) + (405,6044 kg/jam) (327,6

kJ/kg) + (0,0033 kg/jam) (200,6082 kJ/kg) + (62,8156 kg/jam)

(206,4582 kJ/kg) + (952,0768 kg/jam) (182,8632 kJ/kg) +

(9,6653 kg/jam) (142,3500 kJ/kg)

= 324.879,5141 kJ/jam
333 333 353
F17Pati ∫298 Cp.dT + F17Udara∫ Cp.dT + F17Protein
298
∫298 Cp.dT+ F17Lemak

333 333 353

∫298 Cp.dT+ F 17Glukosa∫298

Cp.dT+F17HMFCp.dT∫
298

= (17,7525 kg/jam) (89,9430 kJ/kg) + (405,6044 kg/jam) (147

kJ/kg) + (0,0033 kg/jam) (90,0165 kJ/kg) + (62,8156 kg/jam)

(92,6415 kJ/kg) + (952,0768 kg/jam) (82,054 kJ/kg) + (9,6653

kg/jam) (63,8750 kJ/kg)

= 145.779,2692 kJ/jam

Panas yang diserap air pendingin (Q) = Qo– Qi

(145.779,2692-324.879,5141) kJ/jam
- 179.100,2449 kJ/jam
2528,9566 kJ/kg pada 25°C
(Reklaitis, 1983)
2576,4251 kJ/kg pada 50°C
 (Reklaitis, 1983)
Lamda = H[25°C] – H[50°C]
= 2528,9566 – 2576,4251
= -47,4685 kJ/kg
Jumlah air pendingin yang diperlukan (m) = Q/lamda
−179.100,2449kJ/ jam
=
−47,4685kJ/kg
3.773,0336 kg/jam
7. Evaporator-2
333 333 333

F20Pati ∫298 Cp.dT + F20Kilang∫ Udara

Cp.dT + F20Protein
298
∫298 Cp.dT + F20Lemak

333 333

∫298 Cp.dT + F 20Glukosa∫Cp.dT

298

= (17,7525 kg/jam) (89,9430 kJ/kg) + (405,6044 kg/jam) (147

kJ/kg) + (0,0033 kg/jam) (90,0165 kJ/kg) + (62,8156 kg/jam)

(92,6415 kJ/kg) + (952,0768 kg/jam) (82,054 kJ/kg)

= 145.161,8981 kJ/jam
383 383 383
F23Pati ∫
298
Cp.dT + F23Udara∫ Cp .dT + F23Protein
298
∫
298
Cp.dT + F23Lemak

383 383 383

∫
298
∫ .dT+ F22AirCp .dT+∫F22Udara.HVL
Cp.dT+ F23GlukosaCp
298 298

(17,7525 kg/jam) (218,433 kJ/kg) + (89,2330 kg/jam) (357

(kJ/kg) + (0,0033 kg/jam) (218,6115 kJ/kg) + (62,8156 kg/jam)

(224,9865 kJ/kg) + (952,0768 kg/jam) (199,274 kJ/kg) +

(316,3714 kg/jam) (357 kJ/kg)+(316,3714 kg/jam) (2258,6778

kJ/kg)

1.067.117,096 kJ/jam
Panas yang dilepas steam (Q) = Qo–Qi
(1.067.117,096-145.161,8981) kJ/jam
= 921.955,1979 kJ/jam
Uap super panas pada 1 atm, 150°C, H (150°C) = 2776,3 kJ/kg
(Reklaitis, 1983)
Uap jenuh pada 1 atm, 100°C 2676 kJ/kg
(Reklaitis, 1983)
HL (100°C) = 419,064 kJ/kg
(Reklaitis, 1983)

Lamda = [H(150°C) – Hv (100°C)] + [Hv (100°C) – Hl (100°C)]

= [2776,3-2676] + [2676-419,064]
= 2357,236 kJ/kg
Jumlah steam yang diperlukan (m) = Q/lamda
921.955,1979kJ/ jam
=
2357,236kJ/kg
= 391,1171 kg/jam

8. Cooler-3
383 383 383

F21Pati ∫
298
Cp.dT + F21Air ∫
298
Cp.dT + F21Protein ∫
298
Cp.dT + F21Lemak

383 383

∫
298
Cp .dT+ F21GlukosaCp∫.dT
298

(17,7525 kg/jam) (218,4330 kJ/kg) + (405,6044 kg/jam) (357

kJ/kg) + (0,0033 kg/jam) (218,6115 kJ/kg) + (62,8156 kg/jam)

(224,9865 kJ/kg) + (952,0768 kg/jam) (199,274 kJ/kg)

= 339.972,9183 kJ/kg
 333 333 333
F23Pati ∫298 Cp.dT + F23Air ∫298 Cp.dT + F23Protein 298
∫ Cp.dT+ F23Lemak

333 333

∫298 Cp.dT+ F 23Glukosa∫298

Cp.dT

(17,7525 kg/jam) (89,9430 kJ/kg) + (405,6044 kg/jam) (147

kJ/kg) + (0,0033 kg/jam) (90,0165 kJ/kg) + (62,8156 kg/jam)

(92,6415 kJ/kg) + (952,0768 kg/jam) (82,054 kJ/kg)

= 145.161,8981 kJ/jam

Panas yang diserap air pendingin (Q) = Qo– Qi

(145.161,8981-339.972,9183) kJ/jam
- 194.811,0202 kJ/jam
H(25°C) = 2528,9566 kJ/kg (Reklaitis, 1983)
2576,4251 kJ/kg (Reklaitis, 1983)

Lamda = H[25°C] – H[50°C]

= 2528,9566 – 2576,4251
-47,4685 kJ/kg
Jumlah air pendingin yang diperlukan (m) = Q/lamda
−194.811,0202 kJ/ jam
=
−47,4685kJ/kg
4.104,00624 kg/jam

Kristalizer
Panas masuk ke Kristalizer = panas keluar Cooler-03
= 145.161,8981 kJ/jam
 288 288 288
F25Pati ∫298 Cp.dT + F25Udara∫ Cp.dT + F25Protein
298
∫298 Cp.dT+ F25Lemak

288 288 288

∫298 Cp.dT + F 25Glukosa ∫298 Cp.dT + F25Molase ∫298 Cp.dT+ F 24Molase

288 288

∫298 Cp.dT+ F ∫298 Cp.dT

24K udara

=(17,7525 kg/jam) (-25,698 kJ/kg) + (53,5398 kg/jam) (-42 kJ/kg)

+ (0,0033 kg/jam) (-25,719 kJ/kg) + (62,8156 kg/jam) (-26,469

kJ/kg) + (727,7006 kg/jam) (-23,444 kJ/kg) + (22,4376 kg/jam)

(-25,755 kJ/kg) + (201,9386 kg/jam) (-25,755 kJ/kg) + (35,6932

kg/jam) (-42 kJ/kg)

- 28.705,76263 kJ/jam

Panas yang diserap air pendingin (Q) = Qo– Qi

= (- 28.705,76263-145.161,8981) kJ/jam
- 173.867,6607 kJ/jam
2528,9566 kJ/kg (Reklaitis, 1983)
2499,2755 kJ/kg (Reklaitis, 1983)

Lamda = H[10°C] – H[25°C] = (2499,2755 – 2528,9566) kJ/kg

- 29,6811 kJ/kg

Jumlah air pendingin yang diperlukan (m) = Q/lamda

−173.867,6607 kJ/ jam

=
−29,6811kJ/kg

5.857,8577 kg/jam

10. Pengering Rotary

 288 288 288
F25Udara ∫298 Cp.dT + F25Glukosa ∫298 Cp.dT + F25Protein ∫298 Cp.dT+ F25Lemak

288 288 288

∫298 Cp.dT+ F 25Pati∫298

Cp.dT+F25MolaseCp.dT ∫
298

=(53,5398 kg/jam) (-42 kJ/kg) + (727,7006 kg/jam) (-23,444

kJ/kg) + (0,0033 kg/jam) (-25,719 kJ/kg) + (62,8156 kg/jam) (-

26,4690 kJ/kg) + (17,7525 kg/jam) (-25,6980 kJ/kg) + (22,4376

kg/jam) (-25,755 kJ/kg)

- 22.005,71959 kJ/jam
348 348 348

F26Udara ∫298 Cp.dT + F26Molase ∫298 Cp.dT + F27Udara∫ Cp.dT+¿ F27Glukosa

298

348 348 348

∫298 Cp.dT + F27Pati ∫298 Cp.dT + F27Lemak ∫298 Cp.dT + F27Protein

348

∫298 Cp.dT
(48,988917 kg/jam) (210 kJ/kg) + (22,4376 kg/jam) (128,7750

kJ/kg) + (4,5509 kg/jam) (210 kJ/kg) + (727,7006 kg/jam)

(172,2 kJ/kg) + (17,7525 kg/jam) (128,4900 kJ/kg) + (62,8156

(kg/jam) (132,3450 kJ/kg) + (0,0033 kg/jam) (128,5945 kJ/kg)

= 150.037,5805 kJ/jam

Panas yang dilepas steam (Q) = Qo–Qi

(150.037,5805 - (-22.005,71959)) kJ/jam
= 172.043,3001 kJ/jam
Uap super panas pada 1 atm, 150°C, H (150°C) = 2776,3 KJ/Kg
(Reklaitis, 1983)
Uap jenuh pada 1 atm, 100°C 2676 KJ/Kg pada 100°C
(Reklaitis, 1983)
419,064 KJ/Kg pada 100°C
(Reklaitis, 1983)

Lamda = [H(150°C) – Hv (100°C)] + [Hv (100°C) – Hl(100°C)]

= [2776,3-2676] + [2676-419,064]
2357,236 kJ/kg
Jumlah steam yang diperlukan (m) = Q/lamda

172.043,3001kJ/ jam
=
2357,236kJ/kg

72,9852 kg/jam

11. Kondensor

Panas yang keluar pada alur 15 = panas masuk ke kondensor

376 376

= F15Udara∫ Cp.dT + HVL+ F15HCL ∫ Cp.dT + HVL

298 298

(495,73876 kg/jam) (327,6 kJ/kg) + (2258,6778 kJ/kg) +

(140,1515 kg/jam) (186,9364 kJ/kg) + (442,4658 kJ/kg)

1.370.329,812 kJ/jam

303 303

F15Udara ∫
298
Cp.dT + F15Udara.HVL+ F15HCL ∫
298
Cp.dT+¿ F15HCL.HVL

= (495,73876 kg/jam) (21 kJ/kg) +(495,73876 kg/jam) (2258,6778

(kJ/kg) + (140,1515 kg/jam) (11,9831 kJ/kg) + (140,1515

kg/jam) (442,4658 kJ/kg)

1.193.816,341 kJ/jam
Panas yang diserap air pendingin (Q) = Qo–Qi
(1.193.816,341-1.370.329,812) kJ/jam
- 176.513,4712 kJ/jam
2528,9566 kJ/kg (Reklaitis, 1983)
2576,4251 kJ/kg (Reklaitis, 1983)
Lamda = H[25°C] – H[50°C]
= 2528,9566 – 2576,4251
= -47,4685 kJ/kg
Jumlah air pendingin yang diperlukan (m) = Q/lamda
−176.513,4712 kJ/ jam
=
−47,4685kJ/kg
3.718,5391 kg/jam