Pindah Panas Pindah Panas - Prof. Purwiyatno Hariyadi

Pindah Panas
8/24/2011
Pindah
Panas
Lecture Note
Principles of Food Engineering (ITP 330)
Dept of Food Science & Technology
Faculty of Agricultural Technology
Bogor Agricultural University
BOGOR
Pindah Panas
Tujuan 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
g
faktor
faktor--faktor apa
p saja
j yang
y g
mempengaruhi (dan bagaimana pengaruhnya)
aplikasi pindah panas dalam proses penanganan,
pengolahan, distribusi dan pemanfaatan pangan
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
1
Pindah Panas
8/24/2011
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
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Heat Transfer (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/Δ
Δt with units
of watts ot Btu/hr
y be expressed
p
as heat transfer per
p unit
• Also,, may
area = heat flux or q/A
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
2
Pindah Panas
8/24/2011
Heat Transfer (2)
• Heat transfer can be classified as:
1. SteadySteady-state:
o all factors are stabilized with respect to time
o temperatures are constant at all locations
o steadysteady-state is sometimes assumed if little error
results
2. UnsteadyUnsteady-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
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
CONDUCTION HEAT TRANSFER
• Occurs when heat moves through a material
(usually solid or viscous liquid) due to molecular
action only
HEAT
• 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
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
3
Pindah Panas
8/24/2011
CONDUCTION HEAT TRANSFER
•
•
•
•
•
•
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/ITP/
Hariyadi/ITP/Fateta
Fateta/IPB
/IPB
CONDUCTION HEAT TRANSFER
- 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/ITP/
Hariyadi/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
4
Pindah Panas
8/24/2011
CONDUCTION HEAT TRANSFER - one dimensional
Fourier’s Law Of Heat Conduction:
ΔQ = qx = - kA dT
dx
Δt
qx
X1
X2
ΔQ = Total heat flow
qx = rate of heat flow in x direction by conduction
conduction, W
k = thermal conductivity, W/mC
A= area (normal to xx-direction) through which heat flows, m2
T = temperature, C
x = distance increment, variable, m
Purwiyatno Hariyadi/ITP/
Hariyadi/ITP/Fateta
Fateta/IPB
/IPB
SIGN CONVENTION
TEMPERATURE
direction of heat flow
slope = -
ΔT
Δx
dT
dx
Temperature profile
DISTANCE
Purwiyatno Hariyadi/ITP/
Hariyadi/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
5
Pindah Panas
8/24/2011
PH/TPG/Fateta/IPB
USING FOURIER’S LAW
Integrating :
T1
T2
X1
q
X2
=- k A
x
qx
A
∫
X2
dT
dx
X = X1 ...........> T = T1
X = X2 ...........> T = T2
qx
X1
dx = -kdT
qx
A
d =dx
T
∫ kdT
T1
qx
A ( x1 - x 2 ) = − k(T1 - T 2 )
T1 = T2 −
q1
kA
q x = − kA
(x1 - x2 )
(T1 - T2 )
(X1 - X 2 )
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
6
Pindah Panas
8/24/2011
HEAT CONDUCTION IN MULTILAYERED SYSTEMS
Composite Rectangular Wall (In Series)
kA
kB
kC
q
Tempeerature
q
T1
xA
xB
xC
Temperature profile in
a multilayered system
T2
X
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
kA
kB
kC
q
q
xA
xB
USING FOURIER’S LAW :
q = -kA
Δ T = -q
dT
dX
Δx
kA
xC
Δ T A = -q
Δ T B = -q
Δ T C = -q
Δx A
kAA
Δx B
k BA
Δx C
k CA
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
7
Pindah Panas
8/24/2011
kA
q
kB
kC
T1
q
T2
xA
xB
xC
Δ T A = -q
Δ T = T1 − T 2
Δ T B = -q
Δ T = Δ TA + Δ TB + Δ TC
T1 − T 2 = -
q ⎛ ΔX A
ΔX B
ΔX C
⎜
+
+
A ⎜⎝ k A
kB
kC
⎞
⎟
⎟
⎠
Δ T C = -q
Δx A
kAA
Δx B
k BA
Δx C
k CA
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
ro
CONDUCTION IN CYLINDRICAL OBJECTS
dr
Fourier’s law in cylindrical coordinates
qr
= - kA
dT
Integrating
g
g:
q
r dr
dr
qr = -k 2 π r L
dT
dr
Boundary Conditions :
T = Ti
at
r = ri
T = To
at
r = ro
2πL
∫
ri
o
r
ri
= −k
∫
To
dT
Ti
ro
To
Ln r = − kT
T
2πL
ri
i
2π Lk(T i − T o )
q =
⎛r ⎞
ln ⎜ o ⎟
⎜ r ⎟
⎝ i⎠
q
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
8
Pindah Panas
8/24/2011
COMPOSITE CYLINDRICAL TUBE
r3
r1
Ti
r2
To
FROM FOURIER’S LAW:
qr =
2πLk(Ti − To )
⎛r ⎞
ln⎜ o ⎟
⎜ ri ⎟
⎝ ⎠
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
A =?
Let us define logarithmic mean area Am
such that
q r = kA m
where A m
( Ti − To )
(ro − ri )
(ro − ri )
= 2π L
⎛r ⎞
ln ⎜⎜ o ⎟⎟
⎝ ri ⎠
q (r − r )
Ti − To = r o i
kA m
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
r1
Ti
r3
r2
To
T1 − T 2 =
T 2 − T3 =
q r (r 2 − r1 )
(kA
m
) 12
q r (r3 − r2 )
(kA
m
) 23
adding above two equations
( T1 − T 3 )
qr =
⎛ Δr ⎞
⎛Δ r ⎞
⎜
⎟ +⎜
⎟
⎜ kA ⎟
⎜ kA ⎟
m ⎠ 12
m ⎠
⎝
⎝
23
9
Pindah Panas
8/24/2011
Convection Heat Transfer
• Transfer 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/ITP/Fateta/IPB
CONVECTIVE HEAT TRANSFER : heat transfer to fluid
q
Ta < Ts
Surface area = A
Ts
q = h A(Ts - Ta)
q = rate of heat transfer
h = convective heat transfer coefficient, W/m2.oC
Ts= surface temperature
Ta= surrounding fluid temperature
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
10
Pindah Panas
8/24/2011
Natural
Convection
Colder fluid
(higher
(hi
h d
density)
it )
Fluid absorbs heat
(temperature increase:
density decrease)
PH/TPG/Fateta/IPB
HEAT TRANSFER TO 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
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
11
Pindah Panas
8/24/2011
HEAT TRANSFER TO FLUID……………> h?
q = h A (Ts - Ta)
y, velocity,
y, diameter,, viscosity,
y, specific
p
h = f ((density,
heat, thermal conductivity, viscosity of fluid at
wall temperature
The convective heat transfer coefficient is
determined by dimensional analysis.
A series of experiment are conducted to determine
relationships between following dimensionless
numbers.
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
HEAT TRANSFER TO FLUID……………> h?
Dimensionless Numbers In Convective Heat
Transfer
Nusselt Number = Nnu = (hD)/k
Prandtl Number = NPr = μCp/k
Reynolds Number = Re = (ρ
(ρvD)/
vD)/μ
μ
Where
D = characteristic dimension
k = th
thermall conductivity
d ti it off fluid
fl id
v = velocity of fluid
Cp= specific heat of fluid
ρ= density of fluid
μ= viscosity of fluid
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
12
Pindah Panas
8/24/2011
HEAT TRANSFER TO FLUID
….>
FORCED CONVECTION
Nnu = f (NRe, NPr)
Laminar flow in pipes: If NRe<2100
For
(NRe
R x NPr
P x D/L) < 100
N Nu
D
0.14
0.085⎛⎜ NRe xNPr x ⎞⎟
⎛
⎞
L
μ
⎠ ⎜ b⎟
⎝
= 3.66 +
0.66
D ⎞ ⎜⎝ μ w ⎟⎠
⎛
1 + 0.045⎜ NRe xNPr x ⎟
L⎠
⎝
For (NRe x NPR x D/L) > 100
N Nu
D⎞
⎛
= 1.86⎜ N RE xN PR x ⎟
L⎠
⎝
0.33
⎛μ ⎞
⎜⎜
⎟⎟
w
μ
⎝
⎠
0.14
All physical properties are evaluated at bulk fluid
temperature, except μw (viscosity at the wall)
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
HEAT TRANSFER TO FLUID
….>
FORCED CONVECTION
Transition Flow in Pipes Æ NRE between 2100 and 10,000:
Æ use chart to determine h :
Æ diagram J Colburn factor (J) vs Re.
2
Æ
⎛ h ⎞ ⎛ Cp.
Cp μ ⎞ 3 ⎛ μ w
⎟⎟ ⎜
⎟ ⎜⎜
J = ⎜⎜
⎝ ρCpV ⎠ ⎝ k ⎠ ⎝ μ
⎞
⎟⎟
⎠
0.14
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
13
Pindah Panas
8/24/2011
HEAT TRANSFER TO FLUID
….>
FORCED CONVECTION
Turbulent Flow in Pipes: ………….> NRE > 10,000:
N NU = 0.023 N
0.33
Pr
⎛μ
x ⎜ μw
⎜
⎝
⎞
⎟⎟
⎠
0.14
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
HEAT TRANSFER TO FLUID
….>
FREE CONVECTION
Free convection involves the dimensionless
number called Grashof Number, NGr
N
Gr
=
(d
3
2
ρ gβ Δ T
µ
2
hD
N
Nu
= a(N
=
k
)
N
Gr
m
)
Pr
d= Dimension of the system; ρ= density; β = koeff ekspansi volumetrik
(koef muai volumetrik; 1/K); µ=viscosity;
µ=viscosity; g= gravity
a and m = constant
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
14
Pindah Panas
8/24/2011
HEAT TRANSFER TO FLUID
N
hD
Nu
=
k
= a(N G r N Pr )
….>
FREE CONVECTION
m
V l off a and
Value
d m =f(physical
f( h i l configuration)
fi
ti )
Vertical surface
D=vertical dim. < 1
Horizontal cylinder
D = dia < 20 cm
Horizontal flat surface
Facing Upward
Facing downward
mNGrNPr<104
a=1.36
m=1/5
NGrNPr<10-5
10-5<NGrNPr<1
1<NGrNPr<104
a=0.49
a=0.71
a=1,09
m=0
m=1/25
m=1/10
105< NGrNPr<2x107 a=0.54
2x107< NGrNPr<3x1010 a=0.14
3x105< NGrNPr<3x1010 a=0.27
m=1/4
m=1/3
m=1/4
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Equations for calculating heat transfer coefficient (h) in free convection
from water to air (Toledo, p. 271, Table 7.3)
Kondisi p
permukaan
Silinder horisontal
yang
dipanaskan/didinginkan
Fluida di atas plate
horizontal yang
dipanaskan
Persamaan
Nilai untuk C
Udara/Uap
Air
h = C(ΔT/D)0.25
1.3196
291.1
h = C(ΔT)0.25
2.4492
…. dst
dt
Purwiyatno Hariyadi/ITP/Fateta/IPB
15
Pindah Panas
8/24/2011
HEAT TRANSFER TO FLUID……………> U?
Temperature profile : conductive and convective heat transfer
through a slab
T1
Ta
hi
ho
T2
Tb
Q = UA(Ta-Tb)
where
U = Overall heat transfer coefficient [=] W/m2C
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
HEAT TRANSFER TO FLUID……………> U?
T1
Steady State :
Ta
qi = qx =qo=q
q = UA(Ta-Tb)
qi=q=hiA(Ta-T1)
=q=kA(T1--T2)/Δ
T2)/Δx
qx=q=kA(T1
hi
qo=q=hoA(T2-Tb)
ho
T2
Tb
Ta-Tb = (Ta-T1)+(T1-T2)+(T2-Tb)
q = q + qΔ x + q
U A
hiA
kA
h OA
1 = 1 + Δx + 1
U
hi
k
hO
Atau,
umum :
1
U iA
=
i
1 + Δx +
1
hiA i
kA lm h O A
O
Ai=Alm=Ao=A
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
16
Pindah Panas
8/24/2011
HEAT TRANSFER TO FLUID……………> U?
T1
Ta
r2
Ta
r1
Surrounding fluid temp; Tb < Ta
1 = 1 + Δr + 1
U
hi
k
hO
1
U iA
Atau,
umum :
hi
ho
=
1 + Δr +
1
hiA i
kA lm h O A
i
Alm =
Tb
T2
O
Ao - Ai
Ao
ln
Ai
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
HEAT TRANSFER TO FLUID……………> U?
ri = r1= inside radius of cylinder
r1
Ti
r2
r3
ro= r3= outside radius of cylinder
hi = inside heat transfer coefficient
ho = outside heat transfer coefficient
To
Calculating U based on outside radius of cylinder:
1
U
=
1 + roln(r2/r1) + roln(r3/r2)
r
r ln(r /r )
+ …+ o n n-1 + o
ho
k1
r
Kn-1
k2
i hi
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
17
Pindah Panas
8/24/2011
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 21oC adalah 2.84 W/m2K dan pada sisi
yangg lain bersuhu -15oC adalah 11.4 W/m2K.
y
Konduktivitas panas gelas adalah 0.52 W/mK dan
pada lapisan udara adalah 0.031 W/mK.
Jawab:
1.6mm 0.8mm 1.6mm
h1 = 2.84 W/m2K
T1 = 21oC
h2 = 11.4 W/m2.K
h1
h2
T2= -15oC
k1
k2
k1 = 0.52 W/mK
k2 = 0.031 W/mK
k1
Heat flow
Terdapat 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/m2K
q/A = U ΔT = 2.12 (21-(-15) = 76.32 W/m2
Purwiyatno Hariyadi/ITP/Fateta/IPB
18
Pindah Panas
8/24/2011
Soal 2
• Hitung overall heat transfer coefficient (U) untuk
heat exchanger dengan koefisien pindah panas
568/m2K di bagian dalam dan 5678 W/m2K di
bagian luar. Dinding tube mempunyai
konduktivitas panas (k) 55.6 W/mK. Tube
mempunyai
p y diameter dalam 2.21 cm dan ketebalan
1.65 mm. Jika suhu fluida di dalam tube 80oC dan
di luar 120oC, hitung juga suhu pada dinding tube
sebelah dalam.
ro = 1.2701 cm
(a) Menghitung U:
T1 =
80oC
ri = 1.105 cm
hi = 568 W/m2K
ho = 5678 W/m2.K
k = 55.6 W/mK
T2 = 120oC
ri = 2.21/2 = 1.105 cm
ro = 1.105 + 0.1651 = 1.2701 cm
Terdapat 3 hambatan yang dialami selama proses pindah panas melalui heat
exchanger: 2 konveksi, 1 konduksi
1/U = ro/rihi + + ro ln(ro/r1)/k + 1/ho
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/m2K
Purwiyatno Hariyadi/ITP/Fateta/IPB
19
Pindah Panas
8/24/2011
(b) Suhu pada dinding tube sebelah dalam:
q(overall) = q (yang melewati konveksi) = q (melewati konduksi)
L t Tf = suhu
Let:
h fluida
fl id di dalam
d l tube;
t b Tw = suhu
h pada
d dinding
di di tube)
t b )
UAoΔT = hiAi(Tw – Tf)
448(2πroL)(120-80) = 568(2πr1L)(Tw-80)
(Tw – 80) = 448(ro)(40)/568(r1)
= 80 + 448(1.2701)(40)/568(1.105)
448(1 2701)(40)/568(1 105)
= 80 + 36.3 = 116.3oC
Uap panas
125oC, hi = 11400 W/m2K
PR….
Ke dalam sebuah pipa baja
berinsulasi (ID=0.04089 m;
OD=0.04826 m)) dialirkan uapp air
(125oC, hi=11400 W/m2K).
Ketebalan insulator adalah 5 cm
(k2=0.1 W/mK). Diketahui baja
memiliki k1=45 W/mK, koefisien
pindah panas di lingkungan/luar pipa
(ho) adalah 6 W/m2K, dan suhu
lingkungan
g
g 20oC. Hitunglah
g nilai U
dan panas yang berpindah ke
lingkungan (Q=UAΔT)).
Purwiyatno Hariyadi/ITP/Fateta/IPB
r3
Ti
k1
k2
r1
r2
Lingkungan (To=20C
ho=6 W/m2K)
Baja
Insulator
20
Pindah Panas
8/24/2011
Soal 3
• Air mengalir dengan laju 0.02 kg/s di dalam
pipa penukar panas horizontal (ID= 2.5
2 5 cm,
cm
k=0.633 W/moC), dan dipanaskan dari 20oC
menjadi 60oC (µair 658.026x10-6 Pas dan µw
= 308.909x10-6 Pas). Suhu permukaan
dalam pipa adalah 90oC. Perkirakan
koefisien pindah panas (h) pada permukaan
dalam pipa yang panjangnya 1 m. Cp air =
4175 J/kgoC
Jawab Soal 3
•
•
•
Air mengalir pada pipa horizontal, berarti ada proses pemompaan (forced convection). Maka h akan
dipengaruhi nilai NRe, NPr, D/L.
NRe untuk
t k Newtonian
N t i fluida:
fl id Re
R = ρDV/µ
DV/
Diketahui :
µair = 658.026x10-6 Pas; µw = 308.909x10-6 Pas; Cp air = 4175 J/kgoC
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/m3)(π(D/2)2 (m2),
maka = NRe = 4ρD(ṁ/ρπD2 µ) = 4ṁ/πµD = 4(0.02)/π (658.026x10-6)(2.5x10-2)
= 1457.9 (Laminar)
NPr = µCp/k = (658.026x10-6)(4175)/(0.633) = 4.34
D/L = 2.5x10-2/1 = 2.5x10-2
•
Maka NRe x NPr x D/L = 1457.9 x 4.34 x 2.5x10-2 = 168 (>100)
•
Maka gunakan persamaan empiris: Nu = 1.86(NRexNPrxD/L)0.33 (µ/µw)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 Æ h = 2355 w/m2C
•
•
Purwiyatno Hariyadi/ITP/Fateta/IPB
21
Pindah Panas
8/24/2011
PR…
• Hitung berapa nilai h,
h bila kecepatan aliran
dinaikkan menjadi 1.5 kg/s
Soal 3
• Hitunglah
tu g a kecepatan
ecepata kehilangan
e a ga panas
pa as kee
lingkungan dari pipa baja (ID=0.04089 m) berisi
uap pada suhu 130oC. Koefisien konveksi (h) pada
sisi uap adalah 11400 W/m2K dan di luar pipa ke
udara adalah 5.7 W/m2K. Suhu lingkungan ratarata 15oC, konduktivitas panas (k) dinding pipa
baja adalah 45 W/mK.
• Bila
Bil pipa
i diberi
dib i hambatan
h b t (insulator)
(i l t ) setebal
t b l 5 cm
yang memiliki k=0.07 W/mK (h uap dan udara
seperti di atas), berapa energi yang dapat
diselamatkan?
Purwiyatno Hariyadi/ITP/Fateta/IPB
22
Pindah Panas
8/24/2011
TRANSIENT
(UNSTEADY
UNSTEADY--STATE)
STATE)
HEAT TRANSFER
Purwiyatno Hariyadi/ITP/
Hariyadi/ITP/Fateta
Fateta/IPB
/IPB
PH/TPG/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
23
Pindah Panas
8/24/2011
PH/TPG/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Boiling water
100oC
Solid
food
material
Ts,initial=35oC
r
Change in temperature??
Ts = f(t,r)
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
24
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
z Importance of internal and
Boiling water
100oC
external resistance to heat
transfer
relative importance of
conductive and convective
heat transfer
r
Biot number, NBi = hD/k
NBi =
or NBi =
D/k
1/ h
Internal resistance to heat transfer
External resistant to heat transfer
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
z Negligible internal resistance
………….>N
Bi
< 0.1
q = ρ V Cp dT/dt = h A (Ta - T)
dT
∫ Ta - T =
hAdt
∫ ρ CpV
t
T
ln(T
a
− T) =
Ti
Ta - T
=e
h At
ρ C pV
0
- (h A/ ρ Cp V) t
Ta - To
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
25
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
z Finite Surface and Internal Resistance To Heat Transfer
………….>
0.1<NBi < 40
………..>
m=1/NBi
z Negligible Surface Resistance To Heat Transfer ………….>
NBi > 40 ………..> m=1/NBi = 0
Infinite Slab, infinite cylinder and sphere
Use GurnieGurnie-Lurie Chart and/or Heisler Chart
…………> temperaturetemperature-time (T(T-t) chart
Dimensionless number : Fourier number (NFo)
N Fo =
kt
2
ρ C pD
=
α t
D2
D = characteristic dimension
Dsphere = radius
Dinf cylinder = radius
Dinf slab = half thickness
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
The physical meaning of Fourier Number :
1
k ⎛⎜ ⎞⎟D 2
D
αt
N Fo = 2 = ⎝ ⎠ 3
D
⎛ ρ Cp D ⎞
⎜
⎟
⎜
⎟
t
⎝
⎠
NFo =
Rate of heat conduction
across D in volume D 3 (W/C)
Rate of heat storage in volume D 3
(W/C)
Large value of NFo indicates deeper penetration of heat
into solid in a given period of time
Purwiyatno Hariyadi
Hariyadi/ITP/
/ITP/Fateta
Fateta/IPB
/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
26
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Prosedur pengunaan diagram TT-t
1. Untuk silinder tak berbatas
R
∞Suhu pusat (sumbu) silinder setelah pemanasan selama t?
a. hitung NFo, gunakan R sebagai D
b. hitung NBi, gunakan R sebagai D ………> hitung 1/NBi=m=k/hD
c. gunakan diagram untuk silinder tak berbatas,
dari NFo dan NBi cari ratio T
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
1/Nbi = m
NFo
Diagram T-t : hubungan antara suhu di sumbu silinder dan NFo
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
27
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
2. Untuk lempeng tak berbatas
ketebalan, X = 2D
lebar = ∞; panjang =∞
=∞
Tebal=X
Suhu di tengah (midplane) lempeng tak berbatas setelah pemanasan
selama t ??
a. hitung NFo, gunakan (1/2)X sebagai D
b. hitung NBi, gunakan (1/2)X sebagai D ………> hitung 1/NBi
c. gunakan diagram untuk lempengtak berbatas,
dari NFo dan NBi cari ratio T
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Diagram T-t : hubungan suhu di “midplane” lempeng tak berbatas dan NFo
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
28
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Diagram T-t : hubungan antara suhu di pusat bola dan NFo
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Diagram Gurnie-Lurie untuk LEMPENG :
1. Menentukan suhu setelah
pemanasan/pendinginan
• cari nilai NFo=αt/δ2
• cari nilai Nbi dan m=1/Nbi
• tentukan posisi dimana suhu ingin
diketahui, n = x/δ
• cari ratio suhu
2. Menentukan waktu
pemanasan/pendinginan untuk
mencapai suhu ttt
• cari rasio suhu
suhu, pada posisi ttt
yang diketahui, n = r/R
• cari nilai NBi dan m=1/Nbi
• cari NFo= αt/δ2; dan hitung t
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
29
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Diagram Gurnie-Lurie untuk SILINDER :
1. Menentukan suhu setelah
pemanasan/pendinginan
• cari nilai NFo=αt/R2
• cari nilai Nbi dan m=1/Nbi
• 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 Nbi dan m=1/Nbi
• cari Nfo =αt/R2; dan hitung t
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Diagram Gurnie-Lurie untuk BOLA :
1. Menentukan suhu setelah
pemanasan/pendinginan
• cari nilai NFo=αt/R2
• cari nilai Nbi dan m=1/Nbi
• 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 Nbi dan m=1/Nbi
• cari Nfo =αt/R2; dan hitung t
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
30
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Diagram Gurnie-Lurie :(Toledo)
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
31
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Finite object ….> finite slab (bentuk bata, panjang=l, lebar=w, tinggi=h)
⎛T T⎞
⎛T T⎞
⎛T T⎞ ⎛T T⎞
⎜⎜ a − ⎟⎟ = ⎜⎜ a − ⎟⎟ x ⎜⎜ a − ⎟⎟ x ⎜⎜ a − ⎟⎟
⎝ Ta − Ti ⎠Finite⎝ Ta − Ti ⎠Inf. Slab ⎝ Ta − Ti ⎠ Inf slab, ⎝ Ta − Ti ⎠ Inf slab,
l
slab,,
l,w,h
w
h
length
depth
width
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
32
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Finite object …….> finite slab (bentuk
(bentuk kaleng, jarijari-jari=R, tinggi=h)
Infinite cylinder,
radius R
Infinite slab,
thickness=h
⎛ Ta − T ⎞
⎜⎜
⎟⎟
T
T
−
⎝ a i⎠
=
⎛ Ta − T ⎞
⎜⎜
⎟⎟
T
T
−
⎝ a i⎠
Finite cylinder
R, h
x
⎛ Ta − T ⎞
⎜⎜
⎟⎟
T
T
−
⎝ a i ⎠Infinite
slab
(h)
Infinite cylinder
R
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Penentuan posisi pada benda berbatas
? Lokasi : tengah tutup kaleng
- ditengah silinder : n=0
- dipermukaan lempeng: n=1
R
δ
r=
1/2R
X=1/2δ
X? Lokasi x
- n silinder = r/R=1/2
- n lempeng = x/δ
x/δ = 1/2
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
33
Pindah Panas
8/24/2011
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
CONTOH SOAL
Apel didinginkan dari suhu 20oC menjadi 8oC, dengan menggunakan air
dingin mengalir (5oC). Aliran air dingin ini memberikan koef. Pindah panas
konvensi sebesar 10 M/m2.K. Asumsikan apel sebagai bola dengan
diamater 8 cm. Nilai k apel = 0.4 W/m/K, Cp apel
apel= 3.8 kJ/kg.K dan
densitasnya=960 kg/m3. Untuk pusat geometri apel mencapai suhu 8oC,
berapa lama harus dilakukan pendinginan?
Jawab :
1. Cek NBi ;
apakah nilainya <0.1?
0,1<NBi<40?
atau NBi >40??
NBi= (hR/k)=1 …………> 0.1<NBi<40 :
Æ gunakan diagram TT-t (m=1/NBi=1)
Purwiyatno Hariyadi/ITP/Fateta/IPB
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
2. Hitung rasio suhu yang
dikehendaki :
(Ta-T)/(Ta-Ti) =
(58)/(5-20) = 0.2
θ=0.2
3. Posisi?
Di pusat geometri
Æ n=0
4. Cari nilai NFo, dan
tentukan t
n=0
m=1
NFo=αt/R2=0.78
NFo=αt/R2=0.78
t = 0.78R
0 78R2/α
/
2
t = 0.78R /[k/(ρ.Cp)]
t = 0.78(0.04)2/[0.4/(960)(3800)]
t = 11,381 s
t = 3.16 h
Purwiyatno Hariyadi/ITP/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
34
Pindah Panas
8/24/2011
PH/TPG/Fateta/IPB
PH/TPG/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
35
Pindah Panas
8/24/2011
PH/TPG/Fateta/IPB
PH/TPG/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
36
Pindah Panas
8/24/2011
PH/TPG/Fateta/IPB
PH/TPG/Fateta/IPB
Purwiyatno Hariyadi/ITP/Fateta/IPB
37
Pindah Panas
8/24/2011
PH/TPG/Fateta/IPB
Selesai ….……….
NEXT
Heat Exchangers
Purwiyatno Hariyadi/ITP/Fateta/IPB
38
Pindah Panas
8/24/2011
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 121oC. Udara luar
bersuhu 25oC. Retort dibuat dari baja (k =
42 W/mK) dan mempunyai ketebalan 0.635
m.
Jawaban Soal 3
• Diketahui
Diketahui, udara melewati silinder
horisontal, maka h = 1.3196(ΔT/Do)0.25
Purwiyatno Hariyadi/ITP/Fateta/IPB
39
Pindah Panas
8/24/2011
Soal 3
• Hitung overall heat transfer coefficient (U) untuk
saus tomat (densitas 995 kg/m3, viskositas 0.676
Pas) yang dipanaskan dari suhu 20oC ke 80oC
dalam stainless steel tube dengan panjang 5 m
dengan inside diameter 1.034 cm dan ketebalan
2.77 mm. Uap
p panas
p
di luar tube bersuhu 120oC.
Koefisien pindah panas steam di dalam tube 6000
W/m2K. Laju aliran (v) adalah 0.1 m/s.
Soal 3
• Hitung
tu g nilai
a koefisien
oe s e ppindah
da panas
pa as (h)
( ) dan
da overall
ove a heat
eat
transfer coefficient (U) untuk saus tomat (densitas 995
kg/m3, n=0.34 dan K= 10.42 Pas; equivalent Newtonian
viscosity (µ)= 0.5 Pa.s, µw= 0.45 Pa.s, panas jenis=3817
J/kg.K) yang dipompa dan dipanaskan dalam stainless steel
tube. Saus masuk pada suhu 20oC dan keluar pada suhu
80oC. 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 120oC. Koefisien pindah panas steam di dalam
tube 6000 W/m2K. Laju aliran rata-rata (v) adalah 0.1 m/s.
Purwiyatno Hariyadi/ITP/Fateta/IPB
40
Pindah Panas
8/24/2011
Jawaban Soal 3
n
8 ( v ) 2 -n ( R ) ρ
•
•
Dipompa, berarti forced convection
Reynolds Number:
•
•
•
NRe = 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 = NPr = μCp/k = 0.5*3817/17.3 = 110.3
D/L = 1.034E-2/5 = 0.02
•
Maka: (NRe x NPr x D/L) = 11
11.02
02 * 110.3
110 3 * 0.02
0 02 = 24.31
24 31 < 100
•
0 . 085
Maka,
N = 3. 66 +
Nu
⎛⎜
⎝ N
NRe =
D ⎞
xN x
⎠
Re Pr L
D
⎛
1 + 0 . 045 ⎜N xN x
⎝ Re Pr
L
0.66
⎞
⎟
⎠
⎡ 3 n + 1⎤
K⎢
⎣ n ⎥⎦
⎛ μ
⎞
b ⎟
⎜
⎜
⎟
⎝ μ w ⎠
n
0 .14
= (hD)/k
h = ….
Overall heat transfer coeffiecient:
1/U = ro/rihi + + ro ln(ro/r1)/k + 1/ho
Maka : U???
PR…..
Purwiyatno Hariyadi/ITP/Fateta/IPB
41