The document discusses food process engineering which includes converting raw materials into ready or processed foods through various unit operations like heat transfer, drying, evaporation, and mechanical separations. It explains that food processes can be broken down into a small number of basic unit operations and outlines the aims of the food industry as extending shelf life, increasing variety, providing nutrients, and generating income. Food processes are usually represented through flow charts that show the flow of materials and energy.

1. Sudarminto Setyo Yuwono
Dasar Keteknikan Pengolahan

2. Apple Cooling
qfrig
Driver

3. INTRODUCTION
• Food process engineering: includes the part of human activity to convert raw material to be
ready or processed foods
• The main objective: to study the principles and laws governing the physical, chemical, or
biochemical stages of different processes, and the apparatus or equipment by which such
stages are industrially carried out
• The study of process engineering is an attempt to combine all forms of physical processing
into a small number of basic operations, which are called unit operations
• Food processes may seem bewildering in their diversity, but careful analysis will show that
these complicated and differing processes can be broken down into a small number of unit
operations
• Important unit operations in the food industry are fluid flow, heat transfer, drying,
evaporation, contact equilibrium processes (which include distillation, extraction, gas
absorption, crystallization, and membrane processes), mechanical separations (which
include filtration, centrifugation, sedimentation and sieving), size reduction and mixing.

4. The aims of the food industry
• 1. To extend the period during which a food remains wholesome (the
shelf life) by preservation techniques which inhibit microbiological or
biochemical changes and thus allow time for distribution, sales and
home storage.
• 2. To increase variety in the diet by providing a range of attractive
flavours, colours, aromas and textures in food (collectively known as
eating quality, sensory characteristics or organoleptic quality); a
related aim is to change the form of the food to allow further
processing (for example the milling of grains to flour).
• 3. To provide the nutrients required for health (termed nutritional
quality of a food).
• 4. To generate income for the manufacturing company

5. Food processes are usually
schematized by means of
flow charts
diagrams indicate different
manufacturing steps, as well
as the flow of materials and
energy in the process.
There are different types of
flow charts; the most
common use “blocks” or
“rectangles.” connected by
arrows to indicate the way in
which the
materials flow.

6. DIMENSIONS AND UNITS
• All engineering deals with definite and measured
quantities, and so depends on the making of
measurements
• To make a measurement is to compare the unknown with
the known
• record of a measurement consists of three parts: the
dimension of the quantity, the unit which represents a
known or standard quantity and a number which is the ratio
of the measured quantity to the standard quantity

7. 1.2 BESARAN DAN SATUAN
 Besaran :
Sesuatu yang dapat diukur  dinyatakan dengan angka (kuantitatif) Contoh :
panjang, massa, waktu, suhu, dll.
 Mengukur :
Membandingkan sesuatu dengan sesuatu yang lain yang sejenis yang ditetapkan
sebagai satuan.
contoh : panjang jalan 10 km
Besaran Fisika baru terdefenisi jika :  ada nilainya (besarnya)
 ada satuannya
nilai
satuan
1.4

8.  Satuan :
Ukuran dari suatu besaran ditetapkan sebagai satuan.
Contoh :
 Sistem satuan : ada 2 macam
1. Sistem Metrik : a. mks (meter, kilogram, sekon)
b. cgs (centimeter, gram, sekon)
2. Sistem Non metrik (sistem British)
 Sistem Internasional (SI)
Sistem satuan mks yang telah disempurnakan  yang paling banyak
dipakai sekarang ini.
Dalam SI :
Ada 7 besaran pokok berdimensi dan 2 besaran pokok tak berdimensi
 meter, kilometer  satuan panjang
 detik, menit, jam  satuan waktu
 gram, kilogram  satuan massa
 dll.
1.5

9. NO Besaran Pokok Satuan Singkatan Dimensi
1 Panjang Meter m L
2 Massa Kilogram kg M
3 Waktu Sekon s T
4 Arus Listrik Ampere A I
5 Suhu Kelvin K θ
6 Intensitas Cahaya Candela cd j
7 Jumlah Zat Mole mol N
7 Besaran Pokok dalam Sistem internasional (SI)
NO Besaran Pokok Satuan Singkatan Dimensi
1 Sudut Datar Radian rad -
2 Sudut Ruang Steradian sr -
Besaran Pokok Tak Berdimensi
1.6

10.  Dimensi
Cara besaran itu tersusun oleh besaran pokok.
 Besaran Turunan
Besaran yang diturunkan dari besaran pokok.
1. Untuk menurunkan satuan dari suatu besaran
2. Untuk meneliti kebenaran suatu rumus atau persamaan
- Metode penjabaran dimensi :
1. Dimensi ruas kanan = dimensi ruas kiri
2. Setiap suku berdimensi sama
- Guna Dimensi :
1.7

11. Contoh :
a. Tidak menggunakan nama khusus
NO Besaran Satuan
1 Kecepatan meter/detik
2 Luas meter 2
b. Mempunyai nama khusus
NO Besaran Satuan Lambang
1 Gaya Newton N
2 Energi Joule J
3 Daya Watt W
4 Frekuensi Hertz Hz
1.8

12. Besaran Turunan dan Dimensi
NO Besaran Pokok Rumus Dimensi
1 Luas panjang x lebar [L]2
2 Volume panjang x lebar x tinggi [L]3
3 Massa Jenis [m] [L]-3
4 Kecepatan [L] [T]-1
5 Percepatan
[L] [T]-2
6 Gaya massa x percepatan [M] [L] [T]-2
7 Usaha dan Energi gaya x perpindahan [M] [L]2 [T]-2
8 Impuls dan Momentum gaya x waktu [M] [L] [T]-1
massa
volume
perpindahan
waktu
kecepatan
waktu
1.9

13. Faktor Penggali dalam SI
NO Faktor Nama Simbol
1 10 -18 atto a
2 10 -15 femto f
3 10 -12
piko p
4 10 -9
nano n
5 10 -6
mikro μ
6 10 -3 mili m
7 10 3 kilo K
8 10 6 mega M
9 10 9 giga G
10 10 12 tera T
1.10

14. 1. Tentukan dimensi dan satuannya dalam SI untuk besaran turunan berikut :
a. Gaya
b. Berat Jenis
c. Tekanan
d. Usaha
e. Daya
Jawab :
b. Berat Jenis = = =
= MLT-2 (L-3)
= ML-2T-2 satuan kgm-2
berat
volume
Gaya
Volume
MLT -2
L3
a. Gaya = massa x percepatan
= M x LT -2
= MLT -2 satuan kgms-2
c. Tekanan = = = MLT -2 satuan kgm-1s-1
gaya
luas
MLT -2
L2
d. Usaha = gaya x jarak = MLT -2 x L = ML 2 T -2 satuan kgm-2s-2
e. Daya = = = ML 2 T -1 satuan kgm-2s-1
usaha
waktu
ML 2 T -2
T
Contoh Soal
1.11

15. 2. Buktikan besaran-besaran berikut adalah identik :
a. Energi Potensial dan Energi Kinetik
b. Usaha/Energi dan Kalor
Jawab :
a. Energi Potensial : Ep = mgh
Energi potensial = massa x gravitasi x tinggi
= M x LT-2 x L = ML2T-2
Energi Kinetik : Ek = ½ mv2
Energi Kinetik = ½ x massa x kecepatan2
= M x (LT-1) 2
= ML2T-2
Keduanya (Ep dan Ek) mempunyai dimensi yang sama  keduanya identik
b. Usaha = ML2T-2
Energi = ML2T-2
Kalor = 0.24 x energi = ML2T-2
Ketiganya memiliki dimensi yang sama  identik
1.12

16. Dimensionless Ratios
• It is often easier to visualize quantities if they are expressed in ratio form and ratios have the
great advantage of being dimensionless
• For example, specific gravity is a simple way to express the relative masses or weights of
equal volumes of various materials. The specific gravity is defined as the ratio of the weight
of a volume of the substance to the weight of an equal volume of water
• SG = weight of a volume of the substance/ weight of an equal volume of water .
Dimensionally, SG=[F]/ [L]-3 divided by[F]/ [L]-3 = 1
• it gives an immediate sense of proportion
• This sense of proportion is very important to food technologists as they are constantly
making approximate mental calculations for which they must be able to maintain correct
proportions
• Another advantage of a dimensionless ratio is that it does not depend upon the units of
measurement used, provided the units are consistent for each dimension
• Dimensionless ratios are employed frequently in the study of fluid flow and heat flow. These
dimensionless ratios are then called dimensionless numbers and are often called after a
prominent person who was associated with them, for example Reynolds number, Prandtl
number, and Nusselt number

17. Suhu dan komposisi
• C, F, K
• Fraksi mol, konsentrasi
• Suatu wadah berisi 50g air dan 50 g NaOH, berapa
fraksimol masing-masing
• Albumin 2% berat memiliki densitas 1,028g/cm3.
Berat molekul albumin 67000 g/g mol. Berapa fraksi
mol masing-masing komponen

18. Neraca Massa
• Sangat penting dalam menentukan efisiensi proses
dan memprediksi hasil akhir proses
• Rumus umum => massa in = massa out +
akumulasi
• Neraca massa:
– Proses-proses yang tidak terjadi reaksi kimia
– Proses-proses yang terjadi reaksi kimia

19. Proses yang tidak terjadi reaksi kimia
• Proses yang tidak mengalami reaksi kimia:
– Pengeringan,
– pembekuan,
– pemekatan,
– kristalisasi,
– Pencampuran, dsb
• Reaksi kimia mungkin terjadi pada proses tersebut
namun tidak terlalu mempengaruhi massa total

20. Tahapan perhitungan
• Gambar diagram
• Tulis reaksi kimia jika ada
• Tulis dasar-dasar perhitungan
• Hitung neraca massanya

21. Contoh neraca massa
• Larutan soda api (NaOH), sebanyak 1000 kg/jam
mengandung 10% NaOH di pekatkan pada
evaporator sehingga kadarnya menjadi 60%. Hitung
larutan NaOH pekat yang dihasilkan.
• Cabe 100 kg berkadar air 80% dikeringkan hingga
kadar air 10%. Berapa kilogram cabe kering yang
dihasilkan

22. Dikerjakan dan dikumpulkan
• Proses produksi selai buah dilakukan dengan cara
memekatkan bubur buah dari kadar padatan 10%
menjadi 30%. Pemekatan dilakukan dalam 2 tahap
evaporator. Pada evaporator yang pertama kadar
padatan meningkat menjadi 22%. Hitung selai buah
yang dihasilkan untuk tiap 100 kg/jam bubur buah
yang dipakai.

23. Tugas dikerjakan dan dikumpulkan
• Adonan biskuit diperoleh dengan mencampurkan Terigu
sebanyak 60% berat, gula 10%, telur 10%, garam 3%, mentega
12% dan air 5%. Jika diketahui kadar protein terigu dan telur
sebesar 10% dan 15% berapa kadar protein adonan.
• Proses pembuatan daging burger dilakukan dengan
mencampurkan daging sapi dengan lemak sapi. Daging sapi
memiliki kadar protein 15%, lemak 20% dan air 63%,
sedangkan lemak sapi berkadar protein 3%, lemak 80%, air
15%. Berapa daging sapi dan lemak sapi yang ditambahkan
untuk memperoleh adonan daging burger sebanyak 100 kg
dengan kadar lemak 25%?

24. Batas proses (boundary)
• Batas proses dapat digunakan untuk menyederhanakan suatu
proses
• Dapat diperluas atau diperkecil

25. Dikerjakan
• Nira tebu 1000 kg/jam berkadar gula 20%
dipekatkan hingga kadar gula 60%. Nira pekat
selanjutnya dikristalisasi pada suhu 20oC.
Konsentrasi kejenuhan gula pada suhu 20oC
sebesar 40%. Berapa kg/jam kristal gula yang
dihasilkan? Diasumsikan kristal gula tidak
mengandung air

26. Proses pencampuran
• Draw a diagram and set up equations representing total mass
balance and component mass balance for a system involving the
mixing of pork (15% protein, 20% fat, and 63% water) and backfat
(15% water, 80% fat, and 3% protein) to make 100 kg of a mixture
containing 25% fat.

27. • Draw a diagram and set up a total mass and component balance equation for a crystallizer
where 100 kg of a concentrated sugar solution containing 85% sucrose and 1% inert, water-
soluble impurities (balance, water) enters. Upon cooling, the sugar crystallizes from solution.
A centrifuge then separates the crystals from a liquid fraction, called the mother liquor. The
crystal slurry fraction has, for 20% of its weight, a liquid having the same composition as the
mother liquor. The mother liquor contains 60% sucrose by weight.

28. Neraca massa jika terjadi reaksi kimia
• Beberapa proses pengolahan kemungkinan terjadi
reaksi kimia
– Fermentasi
– Pembakaran
– Netralisasi
• Dasar perhitungan bukan dari massa tetapi dari
perubahan mol
• Setelah itu baru dikonversikan ke massa

29. contoh
• Pembakaran C
• Pembuatan sodium sitrat C6H5Na3O7 dari asam
sitrat C6H8O7 dengan NaOH
• Gas LPG : Propana (C3H8) dan Butana (C4H10),
serta sejumlah kecil Etana (C2H6,) dan Pentana
(C5H12).

30. Tahapan
• Konversikan semua massa menjadi mol
• Dari reaksi kimia hitung jumlah mol yang
dibutuhkan serta mol produk
• Neraca massa diperoleh dengan mengkonversi mol
bahan dan mol produk menjadi massa

31. Harap dikerjakan
• Larutan NaOH diproduksi dengan cara
menambahkan larutan Na2CO3 berkadar 10% ke
dalam aliran bubur Ca(OH)2 yang berkadar 25%.
Bagaimana komposisi bubur akhir (komponen dan
kadarnya) jika reaksi 90% sempurna. Gunakan
dasar 100 kg/jam aliran bubur Ca(OH)2
• Ca(OH) 2 + Na2CO3 => 2NaOH + CaCO3
• MR Ca(OH) 2= 74,1; MR Na2CO3 = 106

32. contoh
• Bahan bakar mengandung 5 %mol H2, 30 %mol CO, 5
%mol CO2, 1 %mol O2, dan 59 %mol N2. Dibakar dengan
media udara. Untuk 100 kg mol bahan bakar hitung mol
gas buang dan komponennya, jika :
• A. Pembakaran sempurna, udara pas
• B. Pembakaran 90% sempurna, udara pas
• C. Udara berlebih 20%, pembakaran sempurna 80%

33. • Bubur susu berkadar air 80%. Pada proses
fermentasi bubur susu, Laktosa C12H22O11
dioksidasi
• Untuk 100g bubur susu, jika sebanyak 1 g laktosa
yang dioksidasi, berapa kadar air bubur susu
setelah fermentasi

34. • Selai dibuat dengan formulasi 45 bagian adalah buah dan 55 bagian
adalah gula.. Untuk menghasilkan gel yang baik, maka kandungan
padatan terlarut selai minimal 65%. Proses pembuatan meliputi
pencampuran bubur buah, gula, dan pektin lalu dievaporasi
sehingga diperoleh selai. Pektin yang ditambahkan pada pembuatan
selai adalah pectin 100 grade (untuk tiap 1 kg pektin memerlukan
gula 100kg). Jumlah pektin yang ditambahkan bergantung pada
jumlah gula yang digunakan. Jika bubur buah mengandung padatan
terlarut 10% hitung kebutuhan bubur buah, gula dan pektin yang
ditambahkan untuk menghasilkan 100 kg selai. Pektin tidak
mengandung padatan terlarut.

35. • Sodium sitrat (Na2C6H6O7) dibentuk dengan
mereaksikan larutan asam sitrat (C6H8O7)
10%(berat) dengan bubur NaOH 50% (berat).
Untuk tiap 100 kg larutan asam sitrat, buat neraca
massanya (reaksi berlangsung sempurna)
• Berat atom O : 16; C : 12; H : 1; Na : 23

36. C12H22O11 + 12O2 =>12CO2 + 11H2O
• Suatu larutan asam sitrat (C6H8O7) 12%(berat) direaksikan dengan
NaOH sehingga terbentuk Sodium Sitrat (Na2C6H6O7). Sodium sitrat
yang terbentuk dipekatkan sehingga diperoleh larutan dengan
konsentrasi 35% berat. Larutan lalu didinginkan pada suhu 15oC
untuk mengkristalkan sodium sitrat. Jika kelarutan sodium sitrat pada
suhu 15oC sebesar 20% berat, hitung kristal sodium sitrat yang
diperoleh, untuk setiap 100 kg asam sitrat
• Asumsi : - Sodium sitrat dalam bentuk anhydrous
• Berat atom O : 16; C : 12; H : 1; Na : 23
• Reaksi berlangsung secara sempurna

37. Recycle
• Proses pengulangan ke tahap sebelumnya dengan
tujuan memperbaiki sifat produk sesuai kebutuhan
• Banyak digunakan pada proses
– Evaporasi
– Kristalisasi
– Fermentasi
• Tahap :
– perluas batasan proses
– Hitung yang direcycle

38. Contoh Recycle dikerjakan
• Pada suatu proses produksi sodium sitrat,
1000kg/jam larutan sodium sitrat berkadar 10%
dipekatkan di suatu evaporator bersuhu 353K
sehingga diperoleh kadar 40%. Larutan lalu
dimasukkan ke kristalizer yang bersuhu 303K
sehingga diperoleh kristal Na sitrat berkadar air 5%.
Larutan jenuh yang mengandung 30% Na sitrat lalu
direcycle ke evaporator. Hitung berapa laju aliran
recycle dan produk yang dihasilkan.

39. Harap dikerjakan
• Pada industri gula, larutan gula 1000 kg/jam
berkadar 25% dipekatkan hingga berkadar 55%.
Larutan tersebut lalu dimasukkan ke kristalizer
sehingga diperoleh kristal gula berkadar air 15%.
Larutan jenuh berkadar gula 40% selanjutnya
direcycle ke evaporator lagi. Kristal gula yang
dihasilkan lalu dikeringkan hingga berkadar air 5%.
Hitung jumlah larutan yang direcycle dan gula yang
dihasilkan.

40. • Evaporator berkapasitas menguapkan air sebanyak
10 kg/jam sehingga kadar padatan berubah dari
5,5% menjadi 25%. Untuk meningkatkan kualitas
produk, sebagian konsentrat di-recycle dan
dicampurkan dengan bahan masuk dengan
menggunakan pompa berkapasitas 20 kg
campuran/jam. Hitung berapa banyak aliran
konsentrat yang dihasilkan serta aliran re-cycle nya.

41. FLUID FLOW THEORY
• Many raw materials for foods and many finished
foods are in the form of fluids.
• Thin liquids - milk, water, fruit juices,
Thick liquids - syrups, honey, oil, jam,
Gases - air, nitrogen, carbon dioxide,
Fluidized solids - grains, flour, peas.
• The study of fluids can be divided into:
– the study of fluids at rest - fluid statics, and
– the study of fluids in motion - fluid dynamics.

42. FLUID STATICS
• very important property : the fluid pressure
• Pressure is force exerted on an area
• force is equal to the mass of the material multiplied by
the acceleration due to gravity.
• mass of a fluid can be calculated by multiplying its volume
by its density
• F = mg = Vρg
• F is force (Newton) or kg m s-2, m is the mass, g the
acceleration due to gravity, V the volume and ρ the
density.

43. The force per unit area in a fluid is called the fluid pressure. It is exerted
equally in all directions.
• F = APs + ZρAg
• Ps is the pressure above the surface of the
fluid (e.g. it might be atmospheric pressure
• total pressure P = F/A = Ps + Zρg
• the atmospheric pressure represents a
datum P = Zρg

44. EXAMPLE . Total pressure in a tank of
peanut oil
• Calculate the greatest pressure in a spherical tank,
of 2 m diameter, filled with peanut oil of specific
gravity 0.92, if the pressure measured at the
highest point in the tank is 70 kPa.

45. • Density of water = 1000 kg m-3
Density of oil = 0.92 x 1000 kg m-3 = 920 kg m-3
Z =greatest depth = 2 m
and g = 9.81 m s-2
Now P = Zρg
= 2 x 920 x 9.81 kg m-1 s-2
= 18,050 Pa = 18.1 kPa.
• To this must be added the pressure at the surface of 70 kPa.
• Total pressure = 70 + 18.1 = 88.1 kPa.
• the pressure depends upon the pressure at the top of the tank, the
depth of the liquid

46. Expressing the pressure
• absolute
pressures
• gauge
pressures
• head

47. EXAMPLE. Head of Water
• Calculate the head of water equivalent and mercury to standard
atmospheric pressure of 100 kPa.
• Density of water = 1000 kg m-3, Density of mercury = 13,600 kg m-3
g = 9.81 m s-2
and pressure = 100 kPa
= 100 x 103 Pa = 100 x 103 kg m-1s-2.
Water Z = P/ ρ g
= (100 x 103)/ (1000 x 9.81)
= 10.2 m
Mercury Z = (100 x 103)/ (13,600 x 9.81)
= 0.75m

48. FLUID DYNAMICS
• In most processes fluids have to be moved
• Problems on the flow of fluids are solved by
applying the principles of conservation of mass and
energy
• The motion of fluids can be described by writing
appropriate mass and energy balances and these
are the bases for the design of fluid handling
equipment.

49. Mass Balance
• ρ1A1v1 = ρ2A2v2
• incompressible
ρ1 = ρ2
so in this case
• A1v1 = A2v2
(continuity equation)
• area of the pipe at
section 1 is A1 , the
velocity at this section,
v1 and the fluid density
ρ1 , and if the
corresponding values
at section 2 are A2, v2,
ρ2

50. EXAMPLE. Velocities of flow
• Whole milk is flowing into a centrifuge through a full
5 cm diameter pipe at a velocity of 0.22 m s-1, and
in the centrifuge it is separated into cream of
specific gravity 1.01 and skim milk of specific
gravity 1.04. Calculate the velocities of flow of milk
and of the cream if they are discharged through 2
cm diameter pipes. The specific gravity of whole
milk of 1.035.

51. Solving
• ρ1A1v1 = ρ2A2v2 + ρ3A3v3
• where suffixes 1, 2, 3 denote respectively raw milk, skim milk and
cream.
• since the total leaving volumes equal the total entering volume
• A1v1 = A2v2 + A3v3
• v2 = (A1v1 - A3v3 )/A2
• ρ1A1v1 = ρ2A2(A1v1 – A3v3)/A2 + ρ3A3v3
• ρ1 A1v1 = ρ2 A1v1 - ρ2 A3v3 + ρ3 A3v3
• A1v1(ρ1 - ρ2 ) = A3v3(ρ3 - ρ2 )

52. • A1 = (π/4) x (0.05)2 = 1.96 x 10-3 m2
• A2 = A3 = (π/4) x (0.02)2 = 3.14 x 10-4 m2
v1 = 0.22 m s-1
ρ1 = 1.035, ρ2 = 1.04, ρ3 = 1.01
• A1v1(ρ1 - ρ2 ) = A3v3(ρ3 - ρ2 )
• -1.96 x 10-3 x 0.22 (0.005) = -3.14 x 10-4 x v3 x (0.03)
v3 = 0.23 m s-1
• v2 = (A1v1 - A3v3 )/A2
• v2 = [(1.96 x 10-3 x 0.22) - (3.14 x 10-4 x 0.23)] / 3.14 x 10-4
= 1.1m s-1

53. Energy Balance
• Referring Fig. before we shall consider the changes in the total energy of unit
mass of fluid, one kilogram, between Section 1 and Section 2.
• Firstly, there are the changes in the intrinsic energy of the fluid itself which include
changes in:
(1) Potential energy = Ep = Zg (J)
(2) Kinetic energy = Ek = v2/2 (J)
(3) Pressure energy = Er = P/ρ (J)
• Secondly, there may be energy interchange with the surroundings including:
(4) Energy lost to the surroundings due to friction = Eƒ (J).
(5) Mechanical energy added by pumps = Ec (J).
(6) Heat energy in heating or cooling the fluid
• In the analysis of the energy balance, it must be remembered that energies are
normally measured from a datum or reference level.
• Ep1 + Ek1 + Er1 = Ep2 + Ek2 + Er2 + Ef - Ec.
• Z1g + v1
2/2 + P1/ρ1 = Z2g + v2
2/2 + P2/ρ2 + Ef - Ec.
• Zg + v2/2 + P/ρ = k Persamaan Bernouilli

54. Water flows at the rate of 0.4 m3 min-1 in a 7.5 cm diameter pipe at a pressure of 70
kPa. If the pipe reduces to 5 cm diameter calculate the new pressure in the pipe.
Density of water is 1000 kg m-3.
• Flow rate of water = 0.4 m3 min-1 = 0.4/60 m3 s-1.
• Area of 7.5 cm diameter pipe = (π/4)D2
= (π /4)(0.075)2
= 4.42 x 10-3 m2.
So velocity of flow in 7.5 cm diameter pipe,
v1 = (0.4/60)/(4.42 x 10-3) = 1.51 m s-1
• Area of 5 cm diameter pipe = (π/4)(0.05)2
= 1.96 x 10-3 m2
and so velocity of flow in 5 cm diameter pipe,
v2 = (0.4/60)/(1.96 x 10-3) = 3.4 m s-1
• Now
• Z1g + v1
2/2 + P1 /ρ1 = Z2g + v2
2/2 + P2 / ρ2
and so
0 + (1.51)2/2 + 70 x 103/1000 = 0 + (3.4)2/2 + P2/1000
0 + 1.1 + 70 = 0 + 5.8 + P2/1000
P2/1000 = (71.1 - 5.8) = 65.3
P2 = 65.3k Pa.

55. Water is raised from a reservoir up 35 m to a storage tank through a 7.5 cm diameter pipe. If it
is required to raise 1.6 cubic metres of water per minute, calculate the horsepower input to a
pump assuming that the pump is 100% efficient and that there is no friction loss in the pipe.
1 Horsepower = 0.746 kW.
• Volume of flow, V = 1.6 m3 min-1 = 1.6/60 m3 s-1 = 2.7 x 10-2 m3 s-1
• Area of pipe, A = (π/4) x (0.075)2 = 4.42 x 10-3 m2,
• Velocity in pipe, v = 2.7 x 10-2/(4.42 x 10-3) = 6 m s-1,
• And so applying eqn
Z1g + v1
2/2 + P1/ρ1 = Z2g + v2
2/2 + P2/ρ2 + Ef - Ec.
• Ec = Zg + v2/2
• Ec = 35 x 9.81 + 62/2
= 343.4 + 18
= 361.4 J
• Therefore total power required
• = Ec x mass rate of flow
= EcVρ
= 361.4 x 2.7 x 10-2 x 1000 J s-1
= 9758 J s-1
• and, since 1 h.p. = 7.46 x 102 J s-1,
• required power = 13 h.p.

56. VISCOSITY
• Viscosity is that property of a fluid that gives rise to forces
that resist the relative movement of adjacent layers in the
fluid.
• Viscous forces are of the same character as shear
forces in solids and they arise from forces that exist
between the molecules.
• If two parallel plane elements in a fluid are moving relative
to one another, it is found that a steady force must be
applied to maintain a constant relative speed. This force is
called the viscous drag because it arises from the action of
viscous forces.

57. If the plane elements are at a distance Z apart, and if their relative velocity is v, then
the force F required to maintain the motion has been found, experimentally, to be
proportional to v and inversely proportional to Z for many fluids. The coefficient of
proportionality is called the viscosity of the fluid, and it is denoted by the symbol µ
(mu).
From the definition of viscosity we can write
F/A = µ v/Z

58. Unit of Viscosity
• N s m-2 = Pascal second, Pa s,
• The older units, the poise and its sub-unit the centipoise,
• 1000 centipoises = 1 N s m-2, or 1 Pa s.
• the viscosity of water at room temperature 1 x 10-3 N s m-2
• acetone, 0.3 x 10-3 N s m-2;
• tomato pulp, 3 x 10-3;
• olive oil, 100 x 10-3;
• molasses 7000 N s m-3.
• Viscosity is very dependent on temperature decreasing sharply as
the temperature rises. For example, the viscosity of golden syrup is
about 100 N s m-3 at 16°C, 40 at 22°C and 20 at 25°C.

59. Newtonian and Non-Newtonian Fluids
• F/A = µ v /Z = µ(dv/dz) = t
 t = k(dv/dz)n power-law equation
• Newtonian fluids (n = 1, k = µ )
• Non-Newtonian fluids (n ≠ 1)
• (1) Those in which n < 1. The viscosity is apparently
high under low shear forces decreasing as the shear
force increases. Pseudoplastic (tomato puree)
• (2) Those in which n > 1. With a low apparent viscosity
under low shear stresses, they become more viscous
as the shear rate rises. Dilatancy (gritty slurries such as
crystallized sugar solutions).
• Bingham fluids have to exceed a particular shear stress
level (a yield stress) before they start to move.
• Food : Non-Newtonian

60. STREAMLINE AND TURBULENT FLOW
• STREAMLINE, flow is calm, in slow the pattern and
smooth
• TURBULENT, the flow is more rapid, eddies develop and
swirl in all directions and at all angles to the general line of
flow.
rv2D/mv = Dvr/m =Reynolds number (Re), dimensionless
• D is the diameter of the pipe
• For (Re) < 2100 streamline flow,
For 2100 < (Re) < 4000 transition,
For (Re) > 4000 turbulent flow.

61. EXAMPLE . Flow of milk in a pipe
Milk is flowing at 0.12 m3 min-1 in a 2.5-cm diameter pipe. If the
temperature of the milk is 21°C, is the flow turbulent or streamline?
• Viscosity of milk at 21°C = 2.1 cP = 2.10 x 10-3 N s m-2
Density of milk at 21°C = 1029 kg m-3.
Diameter of pipe = 0.025 m.
Cross-sectional area of pipe = (p/4)D2
= p/4 x (0.025)2
= 4.9 x 10-4 m2
Rate of flow = 0.12 m3 min-1 = (0.12/60) m3 s-1 = 2 x 10 m3 s-1
• So velocity of flow = (2 x 10-3)/(4.9 x 10-4)
= 4.1 m s-1,
and so (Re) = (Dvr/m)
= 0.025 x 4.1 x 1029/(2.1 x 10-3)
= 50,230
and this is greater than 4000 so that the flow is turbulent.

62. ENERGY LOSSES IN FLOW
• Friction in Pipes
• Energy Losses in Bends and Fittings
• Pressure Drop through Equipment
• Equivalent Lengths of Pipe

63. Friction in Pipes
• Eƒ : the energy loss due to friction in the pipe.
• Eƒ : proportional to the velocity pressure of the fluid and to a factor related
to the smoothness of the surface over which the fluid is flowing.
• F/A = f rv2/2
• F is the friction force, A is the area over which the friction force acts, r is
the density of the fluid, v is the velocity of the fluid, and f is a coefficient
called the friction factor (depends upon the Reynolds number for the flow, and
upon the roughness of the pipe).
• P1 - P2 = (4frv2/2)(L1 - L2)/D
 DPf = (4frv2/2) x (L/D) (Fanning-D'Arcy equation)
• Eƒ = DPf/r = (2fv2)(L/D)
• L = L1 - L2 = length of pipe in which the pressure drop, DPf = P1 - P2 is the
frictional pressure drop, and Eƒ is the frictional loss of energy.

64. Friction factors in pipe (Moody graph)

65. predicted f
• f = 16/(Re) streamline flow, Hagen-Poiseuille
equation 0 < (Re) < 2100
• ƒ = 0.316 ( Re)-0.25/4 ( Blasius equation for smooth
pipes in the range 3000 < (Re) < 100,000)
• roughness ratio = Roughness factor (e)/pipe
diameter (turbulent region)
•

66. ROUGHNESS FACTORS FOR PIPES
Material
Roughness factor
(e)
Material
Roughness factor
(e)
Riveted steel 0.001- 0.01
Galvanized
iron
0.0002
Concrete 0.0003 - 0.003
Asphalted
cast iron
0.001
Wood staves 0.0002 - 0.003
Commercial
steel
0.00005
Cast iron 0.0003 Drawn tubing Smooth

67. EXAMPLE Pressure drop in a pipe
Calculate the pressure drop along 170 m of 5 cm diameter horizontal steel pipe
through which olive oil at 20°C is flowing at the rate of 0.1 m3 min-1
• Diameter of pipe = 0.05 m,
Area of cross-section A
= (π/4)D2
= π /4 x (0.05)2
= 1.96 x 10-3 m2
• From Appendix 4,
• Viscosity of olive oil at 20°C = 84 x 10-3 Ns m-2 and density = 910 kg m-3,
and velocity = (0.1 x 1/60)/(1.96 x 10-3) = 0.85 m s-1,
• Now (Re) = (Dvρ/µ)
• = [(0.05 x 0.85 x 910)/(84 x 10-3)]
= 460
• so that the flow is streamline, and from Fig. moody, for (Re) = 460
• f = 0.03.
• Alternatively for streamline flow from f = 16/(Re) = 16/460 = 0.03 as before.
• And so the pressure drop in 170 m,
• DPf = (4frv2/2) x (L/D)
• = [4 x 0.03 x 910 x (0.85)2 x 1/2] x [170 x 1/0.05]
= 1.34 x 105 Pa
= 134 kPa.

68. Thermal
conductivity
Specific heat Density Viscosity Temperature
(J m-1 s-1 °C-1) (kJ kg-1 °C-1) (kg m-3) (N s m-2) (°C)
Water 0.57 4.21 1000 1.87 x 10-3 0
4.21 987 0.56 x 10-3 50
0.68 4.18 958 0.28 x 10-3 100
Sucrose 20% soln. 0.54 3.8 1070 1.92 x 10-3 20
0.59 x 10-3 80
60% soln. 6.2 x 10-3 20
5.4 x 10-3 80
Sodium chloride 22%
soln.
0.54 3.4 1240 2.7 x 10-3 2
Olive oil 0.17 2.0 910 84 x 10-3 20
Rape-seed oil 900 118 x 10-3 20
Soya-bean oil 910 40 x 10-3 30
Tallow 900 18 x 10-3 65
Milk (whole) 0.56 3.9 1030 2.12 x 10-3 20
Milk (skim) 1040 1.4 x 10-3 25
Cream 20% fat 1010 6.2 x 10-3 3
30% fat 1000 13,8 x 10-3 3

69. Energy Losses in Bends and Fittings
• energy losses due to altering the direction of flow,
fittings of varying cross-section
• This energy is dissipated in eddies and additional
turbulence and finally lost in the form of heat.
• Eƒ = kv2/2 Losses in fittings
• Ef = (v1 - v2)2/2 Losses in sudden enlargements
• Ef = kv2
2/2 Losses in sudden contraction

70. FRICTION LOSS
FACTORS IN FITTINGS
k
Valves, fully open:
gate 0.13
globe 6.0
angle 3.0
Elbows:
90° standard 0.74
medium sweep 0.5
long radius 0.25
square 1.5
Tee, used as elbow 1.5
Tee, straight through 0.5
Entrance, large tank to pipe:
sharp 0.5
rounded 0.05
LOSS FACTORS IN
CONTRACTIONS
D2/D1 0.1 0.3 0.5 0.7 0.9
k 0.36 0.31 0.22 0.11 0.02

71. FLUID-FLOW APPLICATIONS
• Two practical aspects of fluid flow in food technology :
• measurement in fluids: pressures and flow rates, and
• production of fluid flow by means of pumps and fans.
• Pumps and fans are very similar in principle and usually
have a centrifugal or rotating action
• a gas : moved by a fan,
• a liquid: moved by a pump.

72. MEASUREMENT OF PRESSURE IN A
FLUID
• Method :
– Piezometer ("pressure measuring") tube
– U-tube
– Pitot tube
– Pitot-static tube
– Bourdon-tube
• P = Z1r1g

73. EXAMPLE. Pressure in a vacuum evaporator
The pressure in a vacuum evaporator was measured by using a U-tube containing
mercury. It was found to be less than atmospheric pressure by 25 cm of mercury.
Calculate the extent by which the pressure in the evaporator is below atmospheric
pressure (i.e. the vacuum in the evaporator) in kPa, and also the absolute pressure in
the evaporator. The atmospheric pressure is 75.4 cm of mercury and the specific
gravity of mercury is 13.6, and the density of water is 1000 kg m-3.
• We have P = Zrg
= 25 x 10-2 x 13.6 x 1000 x 9.81
= 33.4 kPa
• Therefore the pressure in the evaporator is 33.4 kPa below atmospheric pressure
and this is the vacuum in the evaporator.
• For atmospheric pressure:
• P = Zrg
• P = 75.4 x 10-2 x 13.6 x 1000 x 9.81
= 100.6 kPa
Therefore the absolute pressure in the evaporator
= 100.6 - 33.4
= 67.2 kPa

74. MEASUREMENT OF VELOCITY IN A FLUID
• Pitot tube and manometer :
• Z1g + v1
2/2 + P1/r1 = Z2g + v2
2/2 + P2/r1
• Z2 = Z + Z'
• Z' be the height of the upper liquid surface in the pipe above the datum,
• Z be the additional height of the fluid level in the tube above the upper liquid
surface in the pipe;
• Z' may be neglected if P1 is measured at the upper surface of the liquid in the pipe,
or if Z' is small compared with Z
• v2 = 0 as there is no flow in the tube
• P2 = 0 if atmospheric pressure is taken as datum and if the top of the tube is open
to the atmosphere
• Z1 = 0 because the datum level is at the mouth of the tube.
• v1
2/2g + P1/r1 = (Z + Z')g  Z.
• Pitot-static tube
• Z = v2/2g

75. EXAMPLE . Velocity of air in a duct
Air at 0°C is flowing through a duct in a chilling system. A Pitot-static
tube is inserted into the flow line and the differential pressure head,
measured in a micromanometer, is 0.8 mm of water. Calculate the
velocity of the air in the duct. The density of air at 0°C is 1.3 kg m-3.
• Z = v1
2/2g
 r1Z1 = r2Z2.
• Now 0.8 mm water = 0.8 x 10-3 x 1000
1.3
= 0.62 m of air
• v1
2 = 2Zg
= 2 x 0.62 x 9.81
= 12.16 m2s-2
• Therefore v1 = 3.5 m s-1

76. Venturi and orifice meters
• v1
2/2 + P1/r1 = v2
2/2+ P2/r2 (Bernouilli's equation)
• A1v1 = A2v2 (mass balance, eqn)
 r1 = r2 = r
 v1
2/2 + P1/r = (v1A1/A2)2/2 + P2/r
v1
2 = [2(P2 -P1)/r] x A2
2/(A2
2 -A1
2)
 (P2 -P1)/r = gZrm /r
 Z = (P2 -P1)/rm g
 v1 = C √[2(P2 -P1 )/r]x A2
2/(A2
2 -A1
2 )
 In a properly designed Venturi meter,
C lies between 0.95 and 1.0.

77. Pompa dan Fan
• mechanical energy from some other source is converted
into pressure or velocity energy in a fluid.
• The food technologist is not generally much concerned
with design details of pumps, but should know what
classes of pump are used and something about their
characteristics.
• The efficiency of a pump is the ratio of the energy supplied
by the motor to the increase in velocity and pressure
energy given to the fluid.

78. Jenis Pompa
• Positive Displacement Pumps
• the fluid is drawn into the pump and is then forced
through the outlet
• Positive displacement pumps can develop high-
pressure heads but they cannot tolerate throttling or
blockages in the discharge.

79. Jet Pumps
• a high-velocity jet is produced in a Venturi nozzle, converting the
energy of the fluid into velocity energy.
• This produces a low-pressure area causing the surrounding fluid to
be drawn into the throat
• Jet pumps are used for difficult materials that cannot be satisfactorily
handled in a mechanical pump.
• They are also used as vacuum pumps.
• Jet pumps have relatively low efficiencies but they have no moving
parts and therefore have a low initial cost.

80. Air-lift Pumps
• air or gas can be used to impart energy to the liquid
• The air or gas can be either provided from external sources or produced by boiling within the
liquid. Examples of the air-lift principle are:
• Air introduced into the fluid as shown in Fig. 4.3(e) to pump water from an artesian well.
Air introduced above a liquid in a pressure vessel and the pressure used to discharge the
liquid.
Vapours produced in the column of a climbing film evaporator.
In the case of powdered solids, air blown up through a bed of powder to convey it in a
"fluidized" form.
• A special case of this is in the evaporator, where boiling of the liquid generates the gas
(usually steam) and it is used to promote circulation. Air or gas can be used directly to
provide pressure to blow a liquid from a container out to a region of lower pressure.
• Air-lift pumps and air blowing are inefficient, but they are convenient for materials which will
not pass easily through the ports, valves and passages of other types of pumps.

81. Propeller Pumps and Fan
• Propellers can be used to impart energy to fluids
• They are used extensively to mix the contents of tanks and
in pipelines to mix and convey the fluid.
• Propeller fans are common and have high efficiencies.
• They can only be used for low heads, in the case of fans
only a few centimetres or so of water

82. Centrifugal Pumps and Fans
• The centrifugal pump converts rotational energy into velocity and
pressure energy
• The fluid to be pumped is taken in at the centre of a bladed rotor and
it then passes out along the spinning rotor, acquiring energy of
rotation. This rotational energy is then converted into velocity and
pressure energy at the periphery of the rotor.
• Centrifugal fans work on the same principles. These machines are
very extensively used and centrifugal pumps can develop moderate
heads of up to 20 m of water. They can deliver very large quantities
of fluids with high efficiency.

83. Gambar jenis-jenis pompa

84. • EXAMPLE Centrifugal pump for raising water
Water for a processing plant is required to be stored in a reservoir to
supply sufficient working head for washers. It is believed that a
constant supply of 1.2 m3 min-1 pumped to the reservoir, which is 22
m above the water intake, would be sufficient. The length of the pipe
is about 120 m and there is available galvanized iron piping 15 cm
diameter. The line would need to include eight right-angle bends.
There is available a range of centrifugal pumps whose characteristics
are shown in Fig. 4.4. Would one of these pumps be sufficient for the
duty and what size of electric drive motor would be required?

85. Reynold number
• Assume properties of water at 20°C are density 998 kg m-3, and viscosity 0.001 N
s m-2
• Cross-sectional area of pipe A = (π/4)D2
= π /4 x (0.15)2
= 0.0177 m-2
Volume of flow V = 1.2 m3 min-1
= 1.2/60 m3 s-1
= 0.02 m3 s-1.
• Velocity in the pipe = V/A
= (0.02)/(0.0177)
= 1.13 ms-1
• Now (Re) = Dvρ/µ
• = (0.15 x 1.13 x 998)/0.001
= 1.7 x 105
so the flow is clearly turbulent.

86. friction loss of energy
From Table 3.1, the roughness factor ε is 0.0002 for galvanized
iron
and so
roughness ratio ε /D = 0.0002/0.15 = 0.001
So from Fig. 3.8,
ƒ = 0.0053
Therefore the friction loss of energy
= (4ƒv2/2) x (L/D)
= [4ƒv2L/2D]
= [4 x 0.0053 x (1.13)2 x 120]/(2 x 0.15)
= 10.8 J.

87. TABLE 3.1
RELATIVE ROUGHNESS FACTORS FOR PIPES
Material Roughness factor (e) Material Roughness factor (e)
Riveted steel 0.001- 0.01 Galvanized iron 0.0002
Concrete 0.0003 - 0.003
Asphalted cast
iron
0.001
Wood staves 0.0002 - 0.003
Commercial
steel
0.00005
Cast iron 0.0003 Drawn tubing Smooth

88. Friction factors in pipe

89. TABLE 3.2
FRICTION LOSS FACTORS IN FITTINGS
k
Valves, fully open:
gate 0.13
globe 6.0
angle 3.0
Elbows:
90° standard 0.74
medium sweep 0.5
long radius 0.25
square 1.5
Tee, used as elbow 1.5
Tee, straight through 0.5
Entrance, large tank to pipe:
sharp 0.5
rounded 0.05

90. • For the eight right-angled bends, from Table 3.2 we would
expect a loss of 0.74 velocity energies at each, making (8 x
0.74) = 6 in all.
velocity energy = v2/2
= (1.13)2/2
= 0.64 J
• So total loss from bends and discharge energy
= (6 + 1) x 0.64
= 4.5 J
There would be one additional velocity energy loss
because of the unrecovered flow energy discharged into
the reservoir.
Energy loss from bends and discharge

91. Energy to move 1 kg water
• Energy to move 1 kg water against a head of 22 m
of water is
E = Zg
= 22 x 9.81
= 215.8 J.
• Total energy requirement per kg:
Etot = 10.8 + 4.5 + 215.8
= 231.1 J

92. energy requirement of pump
• and theoretical power requirement
= Energy x volume flow x density
= (Energy/kg) x kgs-1
= 231.1 x 0.02 x 998
= 4613 J s-1.
• Now the head equivalent to the energy requirement
= Etot/g
= 231.1/9.81
= 23.5 m of water,
• and from Fig. 4.4 this would require the 150 mm impeller pump to be
safe, and the pump would probably be fitted with a 5.5 kW motor.

94. Energy Balance

95. Gas and Vapour
• naturally associated with foods and food-processing systems:
– Equilibrium between food and water vapor determines temperatures achieved during
processing.
– Dissolved gases in foods such as oxygen affect shelf life.
– Gases are used to flush packages to eliminate oxygen and prolong shelf life.
– Modified atmospheres in packages have been used to prolong shelf life of packaged
foods.
– Air is used for dehydration.
– Gases are used as propellants in aerosol cans and as refrigerants.
• The distinction between gases and vapors is very loose because
theoretically all vapors are gases.
• The term “vapor” is generally used for the gaseous phase of a
substance that exists as a liquid or a solid at ambient conditions.

96. Kinetika Gas
• The postulates of the kinetic theory
– Gases are composed of discreet particles called molecules, which are in
constant random motion,colliding with each other and with the walls of the
surrounding vessel.
– The force resulting from the collision between the molecules and the walls of
the surrounding vessel is responsible for the pressure of the gas.
– The lower the pressure, the farther apart the molecules, thus, attractive forces
between moleculeshave reduced influence on the overall properties
of the gas.
– The average kinetic energy of the molecules is directly proportional to the
absolute temperature

97. Absolute Temperature and Pressure
• pressure : force of collisions of gas molecules against a surface in contact with the
gas.
• pressure is proportional to the number of gas molecules and their velocity
(absolute pressure).
• Pressure is often expressed as gauge pressure when the measured quantity is
greater than atmospheric pressure, and as vacuum when below atmospheric.
• Unit: psig, psia, kPa absolute, kPa above atmospheric, atmospheres (atm)
• standard atmosphere, the mean atmospheric pressure at sea level, equivalent to
760 mm Hg, 29.921 in. Hg, 101.325 kPa, or 14.696 lbf/in.2
• Temperature (T) is a thermodynamic quantity related to the velocity of motion of
molecules

98. Absolute and gage pressure

99. Conversion factor

102. Calculate the absolute pressure inside an evaporator operating under
20 in. Hg vacuum. Atmospheric pressure is 30 in. Hg. Express this
pressure in SI and in the American Engineering System of units.
• Pabsolute =Patmospheric − Pvacuum =(30 − 20) in. Hg=10 in. Hg
• From the table of conversion factors, the following conversion factors
are obtained:

103. The Ideal Gas Equation
• Pressure, the force of collision between gas molecules and a surface, is directly
proportional to temperature and the number of molecules per unit volume.
• PV = nRT the ideal gas equation.
• R is the gas constant and has values of 0.08206 L(atm)/(gmole.K); or 8315
N(m)/(kgmole.K) or 1545 ft(lbf)/(lbmole.◦R).
• a fixed quantity of a gas that follows the ideal gas equation undergoes a process
where the volume, temperature, or pressure is allowed to change, the product of
the number of moles n and the gas constant R is a constant

104. Calculate the quantity of oxygen entering a package in 24 hours if the packaging
material has a surface area of 3000 cm2 and an oxygen permeability 100 cm3/(m2)(24
h) STP (standard temperature and pressure = 0oC and 1 standard atmosphere of
101.325 kPa).
• Jawaban:

105. Calculate the volume of CO2 in ft3 at 70oF and 1 atm, which
would be produced by vaporization of 1 lb of dry ice.

106. Calculate the density of air (M = 29) at 70◦F and 1 atm in (a)
American Engineering
and (b) SI units.

107. Suatu proses memerlukan debit udara bertekanan 2 atm
sebesar 10 m3/s pada suhu 20◦C. Hitung debit kompresor
pada STP yang harus diberikan.
• Kondisi STP adalah suhu (T) = 0oC (273 K),
tekanan (P) = 1 atm atau 101,325 kPa
• Debit 1 (V1) = 10 m3/s; T1 = 293 K; P1 = 2 atm atau
202,65 kPa;
• V = (P1V1T)/(T1P) = (2 x 293 x 273)/(293 x 1) =
18,64 m3/s

108. An empty can was sealed in a room at 80oC and 1 atm pressure. Assuming that only
air is inside the sealed can, what will be the vacuum after the can and contents cool to
20oC?
• Solution:

109. Gas Mixtures
• If components of a gas mixture at constant volume
are removed one after the other, the drop in
pressure accompanying complete removal of one
component is the partial pressure of that
component
• Pt = Pa + Pb + Pc + . . . Pn (Dalton’s law of partial
pressures)
• PaV = naRT

110. Hitung kuantitas udara pada headsapce kaleng yang bersuhu 20oC jika tekanan pada
headspace sebesar 10 in Hg. Tekanan atmosfer sebesar 30 in Hg. Volume
headspace sebesar 15 ml berisi uap jenuh dan udara.
• vapor pressure of water at 20oC = 2336.6 Pa.
• Pt (tekanan absolut dalam kaleng) = 30-10 in Hg = 20 in Hg
• = 20 x 3386,38 = 67727,6 Pa
• Pudara = Pt - Puap = 67727,6 – 2336,6 = 65391 Pa
• V = 15 x 10-6 m3
• T = 20 + 273 = 293 K
• Nudara = (Pudara V)/(RT)
• = (65391 x 1,5 x 10-5)/(8315 x 293)
• =4,03 x 10-7 kgmol

112. • Assume there are no dissolved gases in the product at the
time of sealing, therefore the only gases in the headspace
are air and water vapor. The vapor pressure of water at
20◦C and 80◦C are 2.3366 and 47.3601 kPa, respectively.
In the gas mixture in the headspace, air is assumed to
remain at the same quantity in the gaseous phase, while
water condenses on cooling

113. soal
• Proses penutupan kaleng dilakukan pada suhu
80oC dan tekanan 1 atm. Di bagian headspace
hanya ada udara dan uap. Setelah dilakukan
sterilisasi, kaleng lalu didinginkan hingga suhu
20oC. Hitung berapa tekanan di headspace?
Diasumsikan jumlah udara di headspace tetap dan
uap air mengkondensasi pada saat pendinginan.

115. A gas mixture used for controlled atmosphere storage of vegetables contains 5%
CO2, 5% O2, and 90% N2. The mixture is generated by mixing appropriate quantities
of air and N2 and CO2 gases. 100 m3 of this mixture at 20oC and 1 atm is needed per
hour. Air contains 21% O2 and 79% N2. Calculate the volume at which the component
gases must be metered into the system in m3/h at 20oC and 1 atm.
• All percentages are by volume. No volume changes occur on mixing of ideal
gases. Because volume percent in gases is the same as mole percent, material
balance equations may be made on the basis of volume and volume percentages.
Let X = volume O2, Y = volume CO2, and Z = volume N2, fed into the system per
hour.
• Oxygen balance: 0.21(X) = 100(0.05); X = 23.8 m3
• CO2 balance: Y = 0.05(100); Y = 5 m3
• Total volumetric balance: X + Y + Z = 100
• Z = 100 − 23.8 − 5 = 71.2 m3

116. Ruangan penyimpanan buah segar diatur sehingga komposisi gas yang masuk ke ruang
penyimpanan menjadi 88% N2, 6% O2 dan 6% CO2. Udara yang bersuhu 25oC tekanan 1 atm
mengalir ke dalam ruang penyimpanan dengan debit 80 m3/jam. Hitung kebutuhan udaranya
jika komposisi udara adalah 79% N2 dan 21% O2.

117. Campuran gas terdiri dari 5% CO2, 5% O2, and 90% N2 digunakan pada CAS untuk buah. Gas
campuran dibuat dengan mencampur udara, N2 dan CO2. Komposisi udara adalah 21% O2
and 79% N2. Gas campuran dibutuhkan sebanyak 100m3/jam. Hitung volume komponen gas
yang harus diatur kedalam CAS pada suhu 20oC dan tekanan 1 atm.
• Semua persentase dalam bentuk volume, tidak ada perubahan volume dalam
pencampuran gas.
• Persen volume sama dengan persen mol
• Neraca massa total U +

118. Udara 1 m3 bertekanan 5 atm dijenuhkan dengan uap air pada suhu 50◦C. Jika udara
tersebut diturunkan tekanannya menjadi 1 atm dan suhu 20◦C, hitung jumlah uap air
yang mengkondensasi.
• The vapor pressure of water at 50◦C and 20◦C are 12.3354 and 2.3366 kPa,
respectively.
• Basis:1 m3 air at 5 atm pressure and 50◦C. The number of moles of air will remain
the same on cooling
Moles water condensed =
0.004593 - 0.004344 = 0.000249
kg moles.

119. Tekanan parsial uap air di udara pada 25◦C dan 1 atm adalah 2,520 kPa. Jika udara
ditekan hingga 5 atm pada suhu 35◦C, hitung tekanan parsial uap air di udara.
• Increasing the total pressure of a gas mixture will proportionately increase the
partial pressure of each component
• for the mixture and for the water vapor, let V1 = the volume of the gas mixture at
25◦C and 1 atm; Pt = total pressure; Pw = partial pressure of water vapor.
• The total number of moles of air and water vapor is
Assuming no condensation, the ratio, nt/nw will be
the same in the low-pressure and high-pressure air,
therefore:

120. Temperature Pressure
Enthalpy
(sat. vap.)
Latent heat
Specific
volume
(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)
20 2.34 2538 2454 57.8
22 2.65 2542 2449 51.4
24 2.99 2545 2445 45.9
26 3.36 2549 2440 40.0
28 3.78 2553 2435 36.6
30 4.25 2556 2431 32.9
40 7.38 2574 2407 19.5
50 12.3 2592 2383 12.0
60 19.9 2610 2359 7.67
70 31.2 2627 2334 5.04
80 47.4 2644 2309 3.41
90 70.1 2660 2283 2.36
100 101.35 2676 2257 1.673
105 120.8 2684 2244 1.42
110 143.3 2692 2230 1.21
115 169.1 2699 2217 1.04
120 198.5 2706 2203 0.892
125 232.1 2714 2189 0.771
130 270.1 2721 2174 0.669
135 313.0 2727 2160 0.582
140 361.3 2734 2145 0.509
150 475.8 2747 2114 0.393
160 617.8 2758 2083 0.307

123. PROPERTIES OF SATURATED AND
SUPERHEATED STEAM
• Steam and water are the two most used heat transfer mediums in food processing.
• Saturated Liquid:. Liquid water in equilibrium with its vapor. If the total pressure above a liquid equals
the vapor pressure, the liquid is at the boiling point.
• Saturated Vapor: saturated steam and is vapor at the boiling temperature of the liquid. Lowering the
temperature of saturated steam at constant pressure by a small increment will cause vapor to condense
to liquid. The phase change is accompanied by a release of heat. If heat is removed from the system,
temperature and pressure will remain constant until all vapor is converted to liquid. Adding heat to the
system will change either temperature or pressure or both.
• Vapor-Liquid Mixtures: Steam with less than 100% quality. Temperature and pressure correspond to the
boiling point; therefore, water could exist either as saturated liquid or saturated vapor. Addition of heat
will not change temperature and pressure until all saturated liquid is converted to vapor. Removing heat
from the system will also not change temperature and pressure until all vapor is converted to liquid.
• Steam Quality: The percentage of a vapor-liquid mixture that is in the form of saturated vapor.
• Interpolation: data

124. If 1 lb of water at 100 psig and 252◦F is allowed to expand to 14.7 psia, calculate
(a) the resulting temperature after expansion and (b) the quantity of vapor produced.
• The absolute pressure= 100 + 14.7 = 114.7 psia. At 252◦F, water will not boil until the pressure is
reduced to 30.9 psia. The water therefore is at a temperature much below the boiling point at 114.7 psia
and it would have the properties of liquid water at 252◦F.
• (a) After expansion to 14.7 psia, the boiling point at 14.7 psia is 212◦F. Part of the water will flash
• to water vapor at 212◦F and the remaining liquid will also be at 212◦F.
• (b) The enthalpy of water at 252◦F is (hf at 252◦F) 220.62 BTU/lb.
• Basis: 1 lb H2O. Heat content = 220.62 BTU. When pressure is reduced to 14.7 psia, some vapor will
be formed, but the total heat content of both vapor and liquid at 212◦F and 14.7 psia will still be 220.62
BTU.

125. How much heat would be given off by cooling steam at 252◦F and
30.883 psia to
248◦F, at the same pressure?
• First, check the state of water at 30.883 psia and 252◦F and 248◦F. From steam tables, the
boiling point of water at 30.883 psia is 252◦F. Therefore, steam at 252◦F and 30.883 psia is
saturated vapor.
• At 30.883 psia and 248◦F, water will be in the liquid state, because 248◦F is below the
boiling temperature at 30.883 psia.
• Heat given off = q = hg at 252◦F − hf at 248◦F
• From steam tables,
• hg at 252◦F = 1164.78 BTU/lb
• hf at 248◦F = 216.56 BTU/lb
• q = 1164.78 − 216.56 = 948.22 BTU/lb
• Saturated steam is a very efficient heat transfer medium. Note that for only a 4◦F change in
temperature, 948 BTU/lb of steam is given off. The heat content of saturated vapors come
primarily from the latent heat of vaporization, and it is possible to extract this heat simply by
causing a phase change at constant temperature and pressure.

126. Superheated Steam Tables
• Superheated Steam: Water vapor at a temperature higher than the boiling point.
The number of degrees the temperature exceeds the boiling temperature is the
degrees superheat. Addition of heat to superheated steam could increase the
superheat at constant pressure or change both the pressure and temperature at
constant volume. Removing heat will allow the temperature to drop to the boiling
temperature where the temperature will remain constant until all the vapor has
condensed.
• A superheated steam table: Both temperature and absolute pressure must be
specified to accurately define the degree of superheat.
• From the temperature and absolute pressure, the specific volume v in ft3/lb and the
enthalpy h in BTU/lb can be read from the table

128. How much heat is required to convert 1 lb of water at 70◦F to
steam at 14.696 psia
and 250◦F?
• First determine the state of steam at 14.696 psia
and 250◦F. At 14.696 psia, the boiling point is
212◦F. Steam at 250◦F and 14.696 psia is
superheated steam. From the superheated steam
table, h at 250◦F is 1168.8 BTU/lb.
• Heat required = hg at 250◦F and 14.696 psia − hf at
70◦F
• = 1168.8 BTU/lb − 38.05 BTU/lb
• = 1130.75 BTU/lb

129. How much heat would be given off by cooling superheated steam at
14.696 psia
and 500◦F to 250◦F at the same pressure?
• Basis: 1 lb of steam.
• Heat given off = q = h at 14.696 psia and 500◦F −
hg at 14.696 psia and 250◦F
• = 1287.4 − 1168.8 = 118.6 BTU/lb
• Superheated steam is not a very efficient heating
medium. Note that a 250◦F change in temperature
is accompanied by the extraction of only 118.6
BTUs of heat.

130. Soal
• Campuran gas terdiri dari 5% CO2, 5% O2, and 90% N2 digunakan
pada CAS untuk buah. Gas tersebut dibuat dengan mencampur
udara, N2 dan CO2. Komposisi udara adalah 21% O2 and 79% N2.
CAS membutuhkan gas campuran sebanyak 100m3/jam. Hitung
volume komponen gas yang harus diatur kedalam CAS pada suhu
20oC dan tekanan 1 atm.
• Campuran Udara dan uap air 1 m3 bertekanan 5 atm absolut
bersuhu 50oC. Jika udara tersebut diturunkan tekanannya menjadi 1
atm dan suhu 20oC, hitung jumlah uap air yang mengkondensasi

131. Harap dikerjakan
• Proses produksi Sodium Sitrat (Na2C6H6O7) dilakukan dengan mereaksikan larutan asam
sitrat (C6H8O7) 10%(berat) dengan NaOH. Larutan Sodium sitrat yang terbentuk dipekatkan
sehingga diperoleh larutan dengan konsentrasi 30% berat. Larutan lalu didinginkan pada
suhu 15oC untuk mengkristalkan sodium sitrat. Jika kelarutan sodium sitrat pada suhu 15oC
sebesar 20% berat, hitung kristal sodium sitrat yang diperoleh, untuk setiap 100 kg larutan
asam sitrat yang digunakan.
– Asumsi : - Sodium sitrat dalam bentuk anhydrous
– Berat atom O : 16; C : 12; H : 1; Na : 23
– Reaksi berlangsung secara sempurna
• Ekstraksi menggunakan supercritical CO2 beroperasi pada tekanan 30 MPa dan suhu 60oC
di wadah ekstraksi (extraction chamber). Debit gas CO2 meninggalkan ekstraktor pada
tekanan 101,3 kPa dan suhu 20oC sebesar 10 L/menit. Hitung waktu tinggal (residence
time) dari CO2 dalam extraction chamber jika diketahui chamber berbentuk tabung dengan
diameter 5 cm dan tinggi 45 cm. Waktu tinggal adalah volume chamber/debit gas dalam
chamber.

132. Temperature Pressure
Enthalpy
(sat. vap.)
Latent heat
Specific
volume
(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)
20 2.34 2538 2454 57.8
22 2.65 2542 2449 51.4
24 2.99 2545 2445 45.9
26 3.36 2549 2440 40.0
28 3.78 2553 2435 36.6
30 4.25 2556 2431 32.9
40 7.38 2574 2407 19.5
50 12.3 2592 2383 12.0
60 19.9 2610 2359 7.67
70 31.2 2627 2334 5.04
80 47.4 2644 2309 3.41
90 70.1 2660 2283 2.36
100 101.35 2676 2257 1.673
105 120.8 2684 2244 1.42
110 143.3 2692 2230 1.21
115 169.1 2699 2217 1.04
120 198.5 2706 2203 0.892
125 232.1 2714 2189 0.771
130 270.1 2721 2174 0.669
135 313.0 2727 2160 0.582
140 361.3 2734 2145 0.509
150 475.8 2747 2114 0.393
160 617.8 2758 2083 0.307

133. Heat
• Sensible heat is defined as the energy transferred between two bodies at different
temperatures, or the energy present in a body by virtue of its temperature.
• Latent heat is the energy associated with phase transitions, heat of fusion, from
solid to liquid, and heat of vaporization, from liquid to vapor.
• Enthalpy, is an intrinsic property, the absolute value of which cannot be measured
directly.
• However, if a reference state is chosen for all components that enter and leave a
system such that at this state the enthalpy is considered to be zero, then the
change in enthalpy from the reference state to the current state of a component
can be considered as the value of the absolute enthalpy for the system under
consideration.
• The reference temperature (Tref) for determining the enthalpy of water in the steam
tables is 32.018◦F or 0.01◦C.

134. Specific Heat
• The specific heat (Cp) is the amount of heat that
accompanies a unit change in temperature for a
unit mass.
• The specific heat, which varies with temperature, is
more variable for gases compared with liquids or
solids.
• Most solids and liquids have a constant specific
heat over a fairly wide temperature range.

135. specific heat J/(kg K)

136. Estimation of Cp
• Cavg = 3349M+ 837.36 in J/(kg K) for fat free plant material
• Cavg = 1674.72 F + 837.36 SNF + 4l86.8M in J/(kg K)
• the mass fraction fat (F), mass fraction solids non-fat
(SNF), and mass fraction moisture (M)
• Example: Calculate the heat required to raise the
temperature of a 4.535 kg roast beef containing 15%
protein, 20% fat, and 65% water from 4.44◦C to 65.55◦C
• Solution:
• Cavg = 0.15(837.36) + 0.2(1674.72) + 0.65(4186.8) = 3182
J/(kg K)
• q = 4.535 kg[3182 J/(kg K)] (65.55 − 4.44)K = 0.882 MJ

137. Specific heat of gas and vapor
• whereCpm is mean specific heat from the reference temperature To
to T1. Tabulated values for the mean specific heat of gases are
based on ambient temperature of 77◦F or 25◦C, as the reference
temperature.

138. Contoh
• Hitung kebutuhan panas untuk menaikkan suhu
udara pengering pd tekanan 1 atm dari suhu ruang
25oC ke suhu pengeringan 50oC jika tiap menit
dialirkan udara sebanyak 100m3
• q= mCp (50-25)
• m=PVM/RT
• R = 0.08206 m3 atm/kg mole K

139. PROPERTIES OF SATURATED AND
SUPERHEATED STEAM
• Steam and water are the two most used heat transfer mediums in food processing. Water is also a major component of
food products. The steam tables that list the properties of steam are a very useful reference when determining heat
exchange involving a food product and steam or water. At temperatures above the freezing point, water can exist in
either of the following forms.
• Saturated Liquid:. Liquid water in equilibrium with its vapor. The total pressure above the liquid must be equal to or be
higher than the vapor pressure. If the total pressure above the liquid exceeds the vapor pressure, some other gas is
present in the atmosphere above the liquid. If the total pressure above a liquid equals the vapor pressure, the liquid is
at the boiling point.
• Saturated Vapor: This is also known as saturated steam and is vapor at the boiling temperature of the liquid. Lowering
the temperature of saturated steam at constant pressure by a small increment will cause vapor to condense to liquid.
The phase change is accompanied by a release of heat. If heat is removed from the system, temperature and pressure
will remain constant until all vapor is converted to liquid. Adding heat to the system will change either temperature or
pressure or both.
• Vapor-Liquid Mixtures: Steam with less than 100% quality. Temperature and pressure correspond to the boiling point;
therefore, water could exist either as saturated liquid or saturated vapor. Addition of heat will not change temperature
and pressure until all saturated liquid is converted to vapor. Removing heat from the system will also not change
temperature and pressure until all vapor is converted to liquid.
• Steam Quality: The percentage of a vapor-liquid mixture that is in the form of saturated vapor.
• Superheated Steam: Water vapor at a temperature higher than the boiling point. The number of degrees the
temperature exceeds the boiling temperature is the degrees superheat. Addition of heat to superheated steam could
increase the superheat at constant pressure or change both the pressure and temperature at constant volume.
Removing heat will allow the temperature to drop to the boiling temperature where the temperature will remain constant
until all the vapor has condensed.

140. Steam table
• The saturated steam table consists of entries under the headings of
temperature, absolute pressure, specific volume, and enthalpy.

142. Temperature Pressure
Enthalpy
(sat. vap.)
Latent heat
Specific
volume
(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)
20 2.34 2538 2454 57.8
22 2.65 2542 2449 51.4
24 2.99 2545 2445 45.9
26 3.36 2549 2440 40.0
28 3.78 2553 2435 36.6
30 4.25 2556 2431 32.9
40 7.38 2574 2407 19.5
50 12.3 2592 2383 12.0
60 19.9 2610 2359 7.67
70 31.2 2627 2334 5.04
80 47.4 2644 2309 3.41
90 70.1 2660 2283 2.36
100 101.35 2676 2257 1.673
105 120.8 2684 2244 1.42
110 143.3 2692 2230 1.21
115 169.1 2699 2217 1.04
120 198.5 2706 2203 0.892
125 232.1 2714 2189 0.771
130 270.1 2721 2174 0.669
135 313.0 2727 2160 0.582
140 361.3 2734 2145 0.509
150 475.8 2747 2114 0.393
160 617.8 2758 2083 0.307

143. contoh
• Pada tekanan vakum berapa sehingga air mendidih pada suhu 80oC,
nyatakan dalam kPa dan dalam cm Hg
– Lihat tabel uap= 47,4 kPa abs
– Tekanan vakum = 101 - 47,4 = 53,6 kPa
– Tekanan vakum = (53,6/101) x 76 = 40,3 cmHg
• Sterilisasi dilakukan pada suhu 120oC, berapa tekanan yang terbaca
pada manometer yang menggunakan satuan psi?
– Dari tabel pada suhu 120oC tekanan uap = 198,5 kPa, maka tekanan pada
manometer = 198,5 – 101 = 97,5 kPa
– 1 atm = 14,7 psi = 101 kPa
– Tekanan pada manometer = (97,5/101) x 14,7 = 14,2 psig

145. Freezing Points of Food Products Unmodified
from the Natural State
• the heat to be removed during freezing of a food
product : sensible heat and latent heat.
• determining the amount of heat by calculating the
enthalpy change.
• calculating enthalpy change below the freezing
point (good only for moisture contents between
73% and 94%) is the procedure of Chang and Tao
(1981). In this correlation, it is assumed that all
water is frozen at 227 K (−50◦ F).

147. Calculate the freezing point and the amount of heat that must
be removed in order to freeze 1 kg of grape juice containing
25% solids from the freezing point to −30◦C.
• Solution:
• Y = 0.75.
• for juices: Tf = 120.47 + 327.35(0.75) − 176.49(0.75)2 = 266.7 K
• Hf = 9792.46 + 405,096(0.75) = 313, 614 J
• a = 0.362 + 0.0498(0.02) − 3.465(0.02)2 = 0.3616
• b = 27.2 − 129.04(0.1316) − 481.46(0.1316)2 = 1.879
• Tr = (−30 + 273 − 227.6)/(266.7 − 227.6) = 0.394
• H = 313,614[(0.3616)0.394 + (1 − 0.3616)(0.394)1.879 ]= 79, 457
J/kg
• The enthalpy change from Tf to −30 ◦C is
• ΔH = 313,614 − 79, 457 = 234, 157 J/kg