Pindah Panas Lecture Note Principles of Food Engineering (IP 330) Dept of Food Science & echnology Faculty of Agricultural echnology Bogor Agricultural University BOGOR Pindah Panas ujuan Pembelajaran Mengerti prinsip dasar pindah panas untuk mengetahui bagaimana bahan pangan dipanaskan dan/atau didinginkan mengerti bagaimana pindah panas diukur menentukan laju pemanasan dan pendinginan bahan pangan Mengerti faktor-faktor apa saja yang mempengaruhi (dan bagaimana pengaruhnya) aplikasi pindah panas dalam proses penanganan, pengolahan, distribusi dan pemanfaatan pangan Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 1
Pindah Panas Heat transfer - movement of energy due to a temperature difference Can only occur if a temperature difference exists Occurs through: 1. conduction, 2. convection, and 3. radiation, or 4. combination of above Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB Heat ransfer (1) May be indicated as total transfer Identified by total heat flow (Q) with units of Btu Identified by rate of heat flow (q) or ΔQ/ Q/Δt with units of watts ot Btu/hr Also, may be expressed as heat transfer per unit area heat flux or q/a Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 2
Heat ransfer (2) Heat transfer can be classified as: 1. Steady-state: o all factors are stabilized with respect to time o temperatures are constant at all locations o steady-state state is sometimes assumed if little error results 2. Unsteady-state state (transient) heat transfer occurs when: o temperature changes with time o thermal processing of foods is an important example o must know time required for the coldest spot in can to reach set temperature Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB CONDUCION HEA RANSFER Occurs when heat moves through a material (usually solid or viscous liquid) due to molecular action only HEA Heat/energy is transferred at molecular level No physical movement of material Heating/cooling of solid Heat flux is directly proportional to the temperature gradient, and inversely proportional to distance (thickness of material). Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 3
CONDUCION HEA RANSFER May occur simultaneously in one, or two, or three directions Many practical problems involve heat flow in only one or two directions Conduction along a rod heated at one end is an example of two dimensional conduction Heat flows along the length of the rod to the cooler end (one direction) If rod is not insulated, heat is also lost to surroundings Center warmer than outer surface Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB CONDUCION HEA RANSFER - one dimensional (unidirectional) One dimensional conduction heat transfer is a function of: 1. temperature difference 2. material thickness 3. area through which heat flows 4. resistance of the material to heat flow Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 4
CONDUCION HEA RANSFER - one dimensional Fourier s Law Of Heat Conduction: ΔQ Δt q x - ka d dx q x X 1 X ΔQ Q otal heat flow 2 q x rate of heat flow in x direction by conduction, W k thermal conductivity, W/mC A area (normal to x-direction) through which heat flows, m 2 temperature, C x distance increment, variable, m Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB SIGN CONVENION EMPERAURE direction of heat flow Δ Δ x d slope - dx emperature profile DISANCE Purwiyatno Hariyadi/IP/ /IP/Fateta Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 5
PH/PG/Fateta/IPB USING FOURIER S LAW 1 X 1 X 2 q -k A x 2 X X... 1 > 1 X X... 2 > 2 q x A dx d dx -kd q x Integrating : q X 1 x - kd A dx 1 X2 q x A ( x 1 - x ) k( 1-2 2 q1 (x - x 1 1 ) 2 ka 2 ( 1 - ) 2 q ka x (X 1 - X ) 2 ) Purwiyatno Hariyadi/IP/Fateta/IPB 6
HEA CONDUCION IN MULILAYERED SYSEMS Composite Rectangular Wall (In Series) k A k B k C q q empe erature 1 x A x B x C emperature profile in a multilayered system 2 X k A k B k C q q q -ka x A x B x C USING FOURIER S LAW : Δ -q d dx Δ x ka Δ A Δ Δ B C -q -q -q Δ x k A A Δ x k k B A A Δ x C A B C Purwiyatno Hariyadi/IP/Fateta/IPB 7
k A k B k C q 1 2 q 1 x A x B x C Δ 1 2 Δ 2 Δ - A + Δ q Δ X A k A B A + Δ C Δ X + k B B + Δ X k C C Δ A Δ Δ B C -q -q -q Δ x k A A Δ x k k B A A Δ x C A B C CONDUCION IN CYLINDRICAL OBJECS Fourier s law in cylindrical coordinates q r - ka q r -k d dr 2 πr r L d dr Boundary Conditions : i at o at r r i r r o dr r o Integrating g : q r dr o k o d 2πL r r i i q r o o Ln r k 2πL r i i 2π Lk( i o ) q r ln o ri r i Purwiyatno Hariyadi/IP/Fateta/IPB 8
COMPOSIE CYLINDRICAL UBE r 2 r 3 i r 1 o FROM FOURIER S LAW: q r 2πLk( i o ) ro ln ri A? Let us define logarithmic mean area A m such that (i o ) q r ka m (ro ri ) (ro ri ) where A m 2πL ro ln r i i o q r (ro ri ) ka m q r 1 2 i 2 3 q o r q r r 1 (r (ka 2 (r (ka 3 m r 2 m ) 12 ) 1 r 2 23 r 3 r ) adding above two equations ( ) 1 3 Δ r Δ r + ka m ka m 12 ) 23 Purwiyatno Hariyadi/IP/Fateta/IPB 9
Convection Heat ransfer ransfer of energy due to the movement of a heated fluid Movement of the fluid (liquid or gas) causes transfer of heat from regions of warm fluid to cooler regions in the fluid Natural Convection occurs when a fluid is heated and moves due to the change in density of the heated fluid Forced Convection occurs when the fluid is moved by other methods (pumps, fans, etc.) Purwiyatno Hariyadi/IP/Fateta/IPB CONVECIVE HEA RANSFER : heat transfer to fluid q a < s Surface area A s q h A( s - a ) q rate of heat transfer h convective heat transfer coefficient, W/m 2. o C s surface temperature a surrounding fluid temperature Purwiyatno Hariyadi/IP/Fateta/IPB 10
Natural Convection Colder fluid (higher h density) Fluid absorbs heat (temperature increase: density decrease) PH/PG/Fateta/IPB HEA RANSFER O FLUID (Forced Convection) FLUID FLOW IN A PIPE Fluid flow can occur as - laminar flow - turbulent flow - transition between laminar and turbulent flow - direction of flow.. > parallel or perpendicular to the solid object Purwiyatno Hariyadi/IP/Fateta/IPB 11
HEA RANSFER O FLUID > h? q h A ( s - a ) h f (density, velocity, diameter, viscosity, specific heat, thermal conductivity, viscosity of fluid at wall temperature he convective heat transfer coefficient is determined by dimensional analysis. A A series of experiment are conducted to determine relationships between following dimensionless numbers. HEA RANSFER O FLUID > h? Dimensionless Numbers In Convective Heat ransfer Nusselt Number N nu (hd)/k Prandtl Number N Pr μc p /k Reynolds Number Re (ρvd)/ vd)/μ Where D characteristic dimension k thermal conductivity it of fluid v velocity of fluid C p specific heat of fluid ρ density of fluid μ viscosity of fluid Purwiyatno Hariyadi/IP/Fateta/IPB 12
HEA RANSFER O FLUID. > FORCED CONVECION N nu f (N Re, N Pr ) Laminar flow in pipes: If N Re <2100 For (N Re xn Pr x D/L) < 100 0.085085 D N Re xn Prx 3. 66 L μ N Nu + 0.66 1 0. 045 D + μ N Re xn Prx L For (N Re x N PR x D/L) > 100 0. 14 N Nu 1. 86 N RE xn PR D x L 0. 33 μ μw All physical properties are evaluated at bulk fluid temperature, except μ w (viscosity at the wall) b w 0.14 HEA RANSFER O FLUID. > FORCED CONVECION ransition Flow in Pipes N RE between 2100 and 10,000: use chart to determine h : diagram J Colburn factor (J) vs Re. 2 0.14 h Cp. μ 3 μ w J CpV k ρ μ Purwiyatno Hariyadi/IP/Fateta/IPB 13
HEA RANSFER O FLUID. > FORCED CONVECION urbulent Flow in Pipes:. > N RE > 10,000: N NU 0. 023 N 0.33 Pr x μ μ w 0.14 HEA RANSFER O FLUID. > FREE CONVECION Free convection involves the dimensionless number called Grashof Number, N Gr N G r N Nu Gr 3 ( d ρ 2 gβ Δ ) h D k µ a 2 ( N N ) m G r Pr d Dimension of the system; ρ density; β koeff ekspansi volumetrik (koef muai volumetrik; 1/K); µviscosity; g gravity a and m constant Purwiyatno Hariyadi/IP/Fateta/IPB 14
HEA RANSFER O FLUID. > FREE CONVECION N Nu h D k a ( N N ) m G r Pr Vl Value of a and m f(physical f(h configuration) Vertical surface Dvertical dim. < 1 mn Gr Gr N Pr Pr <10 4 a1.36 m1/5 Horizontal cylinder D dia < 20 cm N Gr N Pr <10-5 a0.49 m0 10-5< N Gr N Pr <1 a0.71 m1/25 1<N Gr N Pr <10 4 a1,09 m1/10 Horizontal flat surface Facing Upward 10 5 < N Gr N Pr <2x10 7 a0.54 m1/4 2x10 7 < N Gr N Pr <3x10 a0.14 m1/3 Facing downward 3x10 5 < N Gr N Pr <3x10 a0.27 m1/4 Equations for calculating heat transfer coefficient (h) in free convection from water to air (oledo, p. 271, able 7.3) Kondisi permukaan Silinder horisontal yang dipanaskan/didinginkan Fluida di atas plate horizontal yang dipanaskan. dt dst Nilai untuk C Persamaan Udara/Uap Air h C(Δ/D) 0.25 1.3196 291.1 h C(Δ) 0.25 2.4492 Purwiyatno Hariyadi/IP/Fateta/IPB 15
HEA RANSFER O FLUID > U? emperature profile : conductive and convective heat transfer through a slab a 1 h i h o Q UA( a - b ) where U Overall heat transfer coefficient [] W/m 2 C 2 b HEA RANSFER O FLUID > U? Steady State : q i q x q o q q UA( a - b ) a 1 q i qh i A( a - 1 ) q x qka(1-2)/ 2)/Δx q o qh o A( 2 - b ) a - b ( a - 1 )+( 1-2 )+( 2 - b ) h i h o 2 b q U A 1 U Δ q + q x + h A ka 1 h i Δ + x + k 1 i h O q h A O Atau, umum : 1 U A i i A i A 1 h A A lm A i i + A o A Δ x ka lm + 1 h A O O Purwiyatno Hariyadi/IP/Fateta/IPB 16
HEA RANSFER O FLUID > U? a r 2 1 a r 1 2 b h i h o Surrounding fluid temp; b < a 1 U 1 h Δ + r k + 1 i h O Atau, umum : 1 1 Δ + r U A h A ka i A lm i i A o ln i A i - A o A i lm + 1 h A O O HEA RANSFER O FLUID > U? r i r 1 inside radius of cylinder r 3 r o r 3 outside radius of cylinder i r 1 r 2 h i inside heat transfer coefficient i h o outside heat transfer coefficient o Calculating U based on outside radius of cylinder: 1 U 1 h + r o ln(r 2 /r 1 ) o k 1 + r o ln(r 3 /r 2 ) r o ln(r n /r n-1 ) r + + + o r i h i k 2 K n-1 r o Purwiyatno Hariyadi/IP/Fateta/IPB 17
Contoh soal Hitung heat flux (q/a) yang melewati glass pane yang terbuat dari lapisan gelas dengan ketebalan 1.6 mm yang dipisahkan oleh 0.8 mm lapisan insulator. Koefisien pindah panas pada sisi yang satu pada 21 o C adalah 2.84 W/m 2 K dan pada sisi yang lain bersuhu -15 o C adalah 11.4 W/m 2 K. Konduktivitas panas gelas adalah 0.52 W/mK dan pada lapisan udara adalah 0.031 W/mK. Jawab: 1 21 o C h1 1.6mm 0.8mm 1.6mm h2 h1 2.84 W/m 2 K h2 11.4 W/m 2.K k1 0.52 W/mK 2-15 o C k2 0.031 W/mK k1 k2 k1 Heat flow erdapat 5 hambatan yang dialami selama proses pindah panas: 2 konveksi, 3 konduksi 1/U 1/h1 + x1/k1 + x2/k2 + x3/k3 + 1/h2 1/U 1/2.84 + 1.6 x 10-3 /0.52 + 0.8x10-3 /0.031 + 1.6x10-3 /0.52 + 1/11.4 0.352 + 0.0031 + 0.0258 + 0.0031 + 0.0877 0.4718 U 2.12 W/m 2 K q/a U Δ 2.12 (21-(-15) 76.32 W/m 2 Purwiyatno Hariyadi/IP/Fateta/IPB 18
Soal 2 Hitung overall heat transfer coefficient (U) untuk heat exchanger dengan koefisien pindah panas 568/m 2 K di bagian dalam dan 5678 W/m 2 K di bagian luar. Dinding tube mempunyai konduktivitas panas (k) 55.6 W/mK. ube mempunyai diameter dalam 2.21 cm dan ketebalan 1.65 mm. Jika suhu fluida di dalam tube 80 o C dan di luar 120 o C, hitung juga suhu pada dinding tube sebelah dalam. (a) Menghitung U: 1 80 o C r i 1.105 cm r o 1.2701 cm h i 568 W/m 2 K h o 5678 W/m 2.K k 55.6 W/mK 2 120 o C ri 2.21/2 1.105 cm r o 1.105 + 0.1651 1.2701 cm erdapat 3 hambatan yang dialami selama proses pindah panas melalui heat exchanger: 2 konveksi, 1 konduksi 1/U r o /r i h i ++r + o ln(r o /r 1 )/k + 1/h o 1/U 1.2701x10-2 /[(1.105x10-2 )(568)] + 1.2701x10-2 ln(1.2701x10-2 /1.105x10-2 )/55.6 + 1/5678 20.236x10-4 + 0.318x10-4 + 1.76x10-4 22.315x10-4 U 448 W/m 2 K Purwiyatno Hariyadi/IP/Fateta/IPB 19
(b) Suhu pada dinding tube sebelah dalam: q(overall) q (yang melewati konveksi) q (melewati konduksi) Lt Let: f suhu fluida di dalam tube; w suhu pada dinding tube) UA o Δ h i A i ( w f ) 448(2πr o L)(120-80) 568(2πr 1 L)(w-80) (w 80) 448(r o )(40)/568(r 1 ) 80 + 448(1.2701)(40)/568(1.105) 105) 80 + 36.3 116.3 o C PR. Ke dalam sebuah pipa baja berinsulasi (ID0.04089 m; OD0.04826 m) dialirkan uap air r (125 o C, hi11400 W/m 2 1 K). i k1 Ketebalan insulator adalah 5 cm k2 (k20.1 W/mK). Diketahui baja memiliki k145 W/mK, koefisien pindah panas di lingkungan/luar pipa (ho) adalah 6 W/m 2 K, dan suhu lingkungan g 20 o C. Hitunglah nilai U dan panas yang berpindah ke lingkungan (QUAΔ)). Uap panas 125 o C, hi 11400 W/m2K r 2 r 3 Lingkungan ( o 20C ho6 W/m 2 K) Baja Insulator Purwiyatno Hariyadi/IP/Fateta/IPB 20
Soal 3 Air mengalir dengan laju 0.02 kg/s di dalam pipa penukar panas horizontal (ID 2.5 cm, k0.633 W/m o C), dan dipanaskan dari 20 o C menjadi 60 o C (µ air 658.026x10-6 Pas dan µ w 308.909x10-6 Pas). Suhu permukaan dalam pipa adalah 90 o C. Perkirakan koefisien pindah panas (h) pada permukaan dalam pipa yang panjangnya 1 m. C p air 4175 J/kg o C Jawab Soal 3 Air mengalir pada pipa horizontal, berarti ada proses pemompaan (forced convection). Maka h akan dipengaruhi nilai N Re, N Pr, D/L. N Re untuk Newtonian fluida: Re ρdv/µ Diketahui : µ air 658.026x10-6 Pas; µ w 308.909x10-6 Pas; C p air 4175 J/kg o C D 2.5 cm 2.5x10-2 m laju masa (ṁ) 0.2 kg/s, L pipa 1 m v (m/s) ṁ (kg/s) /ρ(kg/m 3 )(π(d/2) 2 (m 2 ), maka N Re 4ρD(ṁ/ρπD 2 µ) 4ṁ/πµD 4(0.02)/π (658.026x10-6 )(2.5x10-2 ) 1457.9 (Laminar) N Pr µcp/k (658.026x10-6 )(4175)/(0.633) 4.34 D/L 2.5x10-2 /1 2.5x10-2 Maka N Re x N Pr x D/L 1457.9 x 4.34 x 2.5x10-2 168 (>100) Maka gunakan persamaan empiris: Nu 1.86(N Re xn Pr xd/l) 0.33 (µ/µ w ) 0.14 Nu 1.86(1547.9x4.34x2.5x10-2)0.33(658.026x10-6 /308.909x10-6 ) 93 Nu hd/k 93 2.5x10-2/0.633 h 2355 w/m 2 C Purwiyatno Hariyadi/IP/Fateta/IPB 21
PR Hitung berapa nilai h, bila kecepatan aliran dinaikkan menjadi 1.5 kg/s Soal 3 Hitunglah kecepatan kehilangan panas as ke lingkungan dari pipa baja (ID0.04089 m) berisi uap pada suhu 130 o C. Koefisien konveksi (h) pada sisi uap adalah 11400 W/m 2 K dan di luar pipa ke udara adalah 5.7 W/m 2 K. Suhu lingkungan ratarata 15 o C, konduktivitas panas (k) dinding pipa baja adalah 45 W/mK. Bila pipa diberi ihambatan bt (insulator) lt setebal tbl5 cm yang memiliki k0.07 W/mK (h uap dan udara seperti di atas), berapa energi yang dapat diselamatkan? Purwiyatno Hariyadi/IP/Fateta/IPB 22
RANSIEN (UNSEADY-SAE) HEA RANSFER Purwiyatno Hariyadi/IP/ /IP/Fateta/IPB PH/PG/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 23
PH/PG/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Boiling water 100 o C Solid food material s, initial 35 o C r Change in temperature?? s f(t,r) Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 24
RANSIEN (UNSEADY-SAE) HEA RANSFER Importance of internal and external resistance to heat transfer relative importance of conductive and convective heat transfer Biot number, N Bi hd/k D / k N Bi 1/ h or NBi Internal External resistance resistant Boiling water 100 o C r to heat transfer to heat transfer Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Negligible internal resistance. >N Bi < 0.1 q ρ V Cp d/dt h A ( a -) d h A d t h A t ln( ) a - a ρ CpV ρ C V a a - - o i e p t 0 - (h A/ ρ C V) t p Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 25
RANSIEN (UNSEADY-SAE) HEA RANSFER Finite Surface and Internal Resistance o Heat ransfer. > 0.1<N Bi < 40.. > m1/n Bi Negligible Surface Resistance o Heat ransfer. > N Bi > 40.. > m1/n Bi 0 Infinite Slab, infinite cylinder and sphere Use Gurnie-Lurie Chart and/or Heisler Chart > temperature-time time (-t) chart Dimensionless number : Fourier number (N Fo) N Fo kt ρ C D α t 2 D p 2 D characteristic dimension D sphere radius D inf cylinder radius D inf slab half thickness RANSIEN (UNSEADY-SAE) HEA RANSFER he physical meaning of Fourier Number : 1 k 2 D α t D N Fo 2 3 D ρ CpD t N Fo Rate of heat conduction Rate of heat storage across D in volume in volume D 3 D 3 (W/C) (W/C) Large value of N Fo indicates deeper penetration of heat into solid in a given period of time Purwiyatno Hariyadi/IP/Fateta/IPB 26
RANSIEN (UNSEADY-SAE) HEA RANSFER Prosedur pengunaan diagram -t 1. Untuk silinder tak berbatas R - Suhu pusat (sumbu) silinder setelah pemanasan selama t? a. hitung N Fo, gunakan R sebagai D b. hitung N Bi, gunakan R sebagai D > hitung 1/N Bi mk/hd c. gunakan diagram untuk silinder tak berbatas, dari N Fo dan N Bi cari ratio Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER 1/N bi m N Fo Diagram -t : hubungan antara suhu di sumbu silinder dan N Fo Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 27
RANSIEN (UNSEADY-SAE) HEA RANSFER 2. Untuk lempeng tak berbatas ketebalan, X 2D lebar ; panjang ebalx Suhu di tengah (midplane) lempeng tak berbatas setelah pemanasan selama t?? a. hitung N Fo, gunakan (1/2)X sebagai D b. hitung N Bi, gunakan (1/2)X sebagai D > hitung 1/N Bi c. gunakan diagram untuk lempengtak berbatas, dari N Fo dan N Bi cari ratio Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Diagram -t : hubungan suhu di midplane lempeng tak berbatas dan N Fo Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 28
RANSIEN (UNSEADY-SAE) HEA RANSFER Diagram -t : hubungan antara suhu di pusat bola dan N Fo Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Diagram Gurnie-Lurie untuk LEMPENG : 1. Menentukan suhu setelah pemanasan/pendinginan cari nilai N Fo αt/δ 2 cari nilai N bi dan m1/n bi tentukan posisi dimana suhu ingin diketahui, n x/δ cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt cari rasio suhu, pada posisi ttt yang diketahui, n r/r cari nilai N Bi dan m1/n bi cari N Fo αt/δ 2 ; dan hitung t Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 29
RANSIEN (UNSEADY-SAE) HEA RANSFER Diagram Gurnie-Lurie untuk SILINDER : 1. Menentukan suhu setelah pemanasan/pendinginan cari nilai N Foαt/R 2 cari nilai N bi dan m1/n bi tentukan posisi dimana suhu ingin diketahui, n r/r cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt cari rasio suhu, pada posisi ttt yang diketahui, n r/r cari nilai N bi dan m1/n bi cari N fo αt/r 2 ; dan hitung t Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Diagram Gurnie-Lurie untuk BOLA : 1. Menentukan suhu setelah pemanasan/pendinginan cari nilai N Fo αt/r 2 cari nilai N bi dan m1/n bi tentukan posisi dimana suhu ingin diketahui, n r/r cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt cari rasio suhu, pada posisi ttt yang diketahui, n r/r cari nilai N bi dan m1/n bi cari N fo αt/r 2 ; dan hitung t Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 30
RANSIEN (UNSEADY-SAE) HEA RANSFER Diagram Gurnie-Lurie :(oledo) Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 31
RANSIEN (UNSEADY-SAE) HEA RANSFER Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Finite object.. > finite slab (bentuk bata, panjangl, lebarw, tinggih) a a a a x x a i a i Inf. Slab a i Inf slab, a i Inf slab, Finite slab, l,w,h l w h length depth width Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 32
RANSIEN (UNSEADY-SAE) HEA RANSFER Finite object. > finite slab (bentuk kaleng, jari-jarir, jarir, tinggih) Infinite cylinder, radius R Infinite slab, thicknessh a a i Finite cylinder R, h a a i x Infinite cylinder R a a i Infinite slab (h) Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER Penentuan posisi pada benda berbatas R? Lokasi : tengah tutup kaleng - ditengah silinder : n0 - dipermukaan lempeng: n1 δ X1/2δ r 1/2R X? Lokasi x - n silinder r/r1/2 - n lempeng x/δ 1/2 Purwiyatno Hariyadi/IP/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 33
RANSIEN (UNSEADY-SAE) HEA RANSFER CONOH SOAL Apel didinginkan dari suhu 20 o C menjadi 8 o C, dengan menggunakan air dingin mengalir (5 o C). Aliran air dingin ini memberikan koef. Pindah panas konvensi sebesar 10 M/m 2.K. Asumsikan apel sebagai bola dengan diamater 8 cm. Nilai k apel 0.4 W/m/K, Cp apel 3.8 kj/kg.k dan densitasnya960 kg/m 3. Untuk pusat geometri apel mencapai suhu 8 o C, berapa lama harus dilakukan pendinginan? Jawab : 1. Cek N Bi ; apakah nilainya <0.1? 0,1<N Bi <40? atau N Bi >40?? N Bi (hr/k)1 > 0.1<N Bi <40 : gunakan diagram -t (m1/n Bi 1) Purwiyatno Hariyadi/IP/Fateta/IPB RANSIEN (UNSEADY-SAE) HEA RANSFER θ0.2 2. Hitung rasio suhu yang dikehendaki : (a-)/(a-i) (5-8)/(5-20) 0.2 3. Posisi? Di pusat geometri n0 4. Cari nilai N Fo, dan tentukan t n0 m1 N Fo αt/r 2 0.78 Purwiyatno Hariyadi/IP/Fateta/IPB N Fo αt/r 2 0.78 t 0.78R 2 /α t 0.78R 2 /[k/(ρ.cp)] t 0.78(0.04) 2 /[0.4/(960)(3800)] t 11,381 s t 3.16 h Purwiyatno Hariyadi/IP/Fateta/IPB 34
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PH/PG/Fateta/IPB PH/PG/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 36
PH/PG/Fateta/IPB PH/PG/Fateta/IPB Purwiyatno Hariyadi/IP/Fateta/IPB 37
PH/PG/Fateta/IPB Selesai.. NEX Heat Exchangers Purwiyatno Hariyadi/IP/Fateta/IPB 38
Soal 3 Hitung laju kehilangan panas (q) dari sebuah retort horizontal dengan diameter dalam 1.524 m dan panjang 9.144 m. Uap di dalam retort bersuhu 121 o C. Udara luar bersuhu 25 o C. Retort dibuat dari baja (k 42 W/mK) dan mempunyai ketebalan 0.635 m. Jawaban Soal 3 Diketahui udara melewati silinder Diketahui, udara melewati silinder horisontal, maka h 1.3196(Δ/Do) 0.25 Purwiyatno Hariyadi/IP/Fateta/IPB 39
Soal 3 Hitung overall heat transfer coefficient (U) untuk saus tomat (densitas 995 kg/m 3, viskositas 0.676 Pas) yang dipanaskan dari suhu 20 o C ke 80 o C dalam stainless steel tube dengan panjang 5 m dengan inside diameter 1.034 cm dan ketebalan 2.77 mm. Uap ppanas di luar tube bersuhu 120 o C. Koefisien pindah panas steam di dalam tube 6000 W/m 2 K. Laju aliran (v) adalah 0.1 m/s. Soal 3 Hitung nilai koefisien e pindah panas as (h) dan overall heat transfer coefficient (U) untuk saus tomat (densitas 995 kg/m 3, n0.34 dan K 10.42 Pas; equivalent Newtonian viscosity (µ) 0.5 Pa.s, µ w 0.45 Pa.s, panas jenis3817 J/kg.K) yang dipompa dan dipanaskan dalam stainless steel tube. Saus masuk pada suhu 20 o C dan keluar pada suhu 80 o C. Stainless steel tube berdimensi panjang 5 m, inside diameter 1.034 cm, dan ketebalan 2.77 mm, konduktivitas panas (k) tube wall 17.3 W/mK. Uap panas di luar tube bersuhu 120 o C. Koefisien pindah panas steam di dalam tube 6000 W/m 2 K. Laju aliran rata-rata (v) adalah 0.1 m/s. Purwiyatno Hariyadi/IP/Fateta/IPB 40
Dipompa, berarti forced convection Reynolds Number: Jawaban Soal 3 N Re n ρ 8( v ) 2-n ( R ) 3n + 1 K n N Re 8(0.1) 2-0.34 +(1.034x10-2 ) 0.34 *995 11.02 (Laminar) 10.42 [(3*0.34+1)/0.34] 0.34 Prandtl Number N Pr μcp/k 0.5*3817/17.3 110.3 D/L 1.034E-2/5 0.02 Maka: (NRexNPrxD/L)1102* NPr x 11.02 110.3 * 0.02 02 24.31 < 100 Maka, 0. 085 N 3. 66 + Nu h. 1 + 0. 045 N xn x Re Pr N xn x Re Pr D L D 0.66 L μ b μ w n 0. 14 (hd)/k Overall heat transfer coeffiecient: 1/U r o /r i h i + + r o ln(r o /r 1 )/k + 1/h o Maka : U??? PR.. Purwiyatno Hariyadi/IP/Fateta/IPB 41