유체역학 및 열전달
Chapter 12.
Heat transfer to fluids without Phase Change
-대류열전달
부산대학교 화공생명공학부
현 규 (Kyu Hyun)
Convection
• Nu (Nusselt Number) !
- =
=
Nu number at sphere
q
rr
r r + D
R
0
= ) (r2qr dr
d
3 / 1 2
/
1
Pr Re
6 . 0 + 2
= Nu
-With convection
(4)
−(4)
∆ = 0
2 0
= ) ( dr
r dT dr
d
T0
T r
T T R
r R
=
¥
®
=
=
에서 에서
0
0 T
T r T
T = R R - +
\ ( )
) ( 0
2 T T
r R dr
dT
R - -
=
) (
)
( 0 0
2 T T h T T
R
q kR R R
R
r = - = -
=
2
=
=
=
\
= hR
hD k
Nu hD R h
k ,
Boundary Layer Flow on a Flat Plate
• Velocity at interface between fluids is not zero.
• Velocity at solid is zero.
• Consider the flow of a fluid (B) over a thin, flat parallel with the direction of flow of the fluid upstream of the plate.
• A number of possibilities for mass transfer of another species (A) into B exist:
• (1) The plate might consist of material A, which is slightly soluble in B.
• (2) Component A might be held in the pores of an inert solid plate, from which it evaporates or dissolves into B.
• (3) The plate might be an inert, dense polymeric membrane, through which species A can pass into fluid B.
- fluid stream as two parts, the boundary layer and the
remaining fluid.
Heat transfer to fluids without Phase change
Internal Energy balance equation (1)
g τ
V VV
r r r
+ Ñ -
× Ñ -
× -Ñ
¶ =
¶ p
t ) ( ) [ ]
(
l Momentum balance equation l Internal energy balance equation
V V
q
V - Ñ× - Ñ× - Ñ
× -Ñ
¶ =
¶( ˆ) ( ˆ ) [ ] ( ) :
t r r
p t U
U
이 부분은 우리가 배우는 범위를 넘는 내용이지만 알아두면 유용하기 때문에 첨부하였습니다
4학년 과목인 “이동현상”내용을 발췌했습니다.
두 가지 식을 비교해보면 공통점이 있습니다.
1. Momentum (ρV)와 internal energy (ρÛ)를 이용한 balance 식이라는 점이다
2. 유동이 존재하는 상황에서 momentum과 내부에너지가 전달된다.
3. 특히 3가지 (빨, 초, 파) 부분은 특히나 유사한 형식의
방정식입니다.
Heat transfer to fluids without Phase change
Prandtl Number
It can be related to the thickness of the thermal and velocity boundary
layers. It is actually the ratio of velocity boundary layer to thermal boundary
layer. When Pr=1, the boundary layers coincide. When Pr is small, it means
that heat diffuses very quickly compared to the velocity (momentum). This
means the thickness of the thermal boundary layer is much bigger than the
velocity boundary layer for liquid metals.
In laminar boundary layer flow (1)
• Analogy
g V V
r m
r = -Ñ p + Ñ
2+ Dt
D
Dt Dp T T
Dt k
C
pDT ÷
ø ç ö
è æ
¶ - ¶ Ñ
- Ñ
= ln
: ln
ˆ r
t
r
2V
÷÷ ø ö çç è
æ
¶ + ¶
¶
= ¶
÷÷ ø ö çç è
æ
¶ + ¶
¶
= ¶
¶ + ¶
¶
¶
2 2 2
2
2 2 2
2
y u x
u
y u x
u y
u u x
u u
x x
x x
x y
x x
n r m
÷÷ ø ö çç è
æ
¶ + ¶
¶
= ¶
÷÷ ø ö çç è
æ
¶ + ¶
¶
= ¶
¶ + ¶
¶
¶
2 2 2
2
2 2 2
2
y T x
T
y T x
T C
k y
u T x
u T
p y
x
a r ˆ
-Two-dimensional laminar boundary layer -Momentum Balance equation
-Momentum Balance equation
-Energy balance equation -Energy balance equation
In laminar boundary layer flow (2)
• Two-dimensional laminar boundary layer
÷÷ ø ö çç è
æ
¶ + ¶
¶
= ¶
¶ + ¶
¶
¶
2 2
2 2
y u x
u y
u u x
u
xu
x y xn
x x÷÷ ø ö çç è
æ
¶ + ¶
¶
= ¶
¶ + ¶
¶
¶
2 2
2 2
y T x
T y
u T x
u
xT
ya
Momentum Balance equation Momentum Balance equation
Energy balance equation Energy balance equation
-Velocity
-Temperature
¥
=
=
=
=
¥
¥
y u at
and u y
u at
u
x x1 0
0
¥
= - =
= - - =
-
¥
¥
y T at
T T and T
y T at
T T T
s s s
s
0 0 1
y
x
¥
= - =
= - - =
-
¥
¥
y u at
u
u and u
y u at
u
u u
s s x s
s
x
0 0 1
¥
¥ T u ,
s
s T
u , B.C
In laminar boundary layer flow (3)
• Two-dimensional laminar boundary layer
2 2
y V y
u V x
u
xV
y¶
= ¶
¶ + ¶
¶
¶ n
2 2
y u y
u
xx
y¶
= ¶
¶ + ¶
¶
¶ q
q a q
Momentum Balance equation Momentum Balance equation
Energy balance equation Energy balance equation
¥
=
=
=
= 0 at y 0 and q 1 at y q
¥
=
=
=
= at y and V at y
V 0 0 1
s s x
u u
u V u
-
= -
¥
s s
T T
T T
-
= -
¥
q
In laminar boundary layer flow –Blasius’s solution
• Two-dimensional laminar boundary layer
2 2
y V y
u V x
u
xV
y¶
= ¶
¶ + ¶
¶
¶ n
Momentum Balance equation Momentum Balance equation
¥
=
=
=
= at y and V at y
V 0 0 1
s s x
u u
u V u
-
= -
¥
-Velocity y
x u¥
) ( noslip us = 0 Q
0
¶ = + ¶
¶
¶
y u x
u
x yContinuity equation
2 1
0
332 0
/
. ÷
ø ç ö è
= æ
¶
¶
¥¥
=
x
u u y
u
y x
n
x
xu x
u x
u x
u u . . Re
.
/ /
¥
¥
¥
¥
¥
÷÷ =
ø ö çç è
= æ
÷÷ ø ö çç è
= æ 0 332 0 332
332 0
2 1 2
1
m r m
r
Heat transfer in boundary Laminar Flow (1)
x
Ts
T T
. Re )
( 0 332
-
¶ =
¥
-Velocity -Temperature y u ,¥ T¥
s
s T
u ,
• Laminar flow heat transfer to flat plate with Pr=1
¥
= - =
= - -
-
= - =
= - -
-
¥
¥
¥
¥
y T at
T T T u
u u u
y T at
T T T u
u u u
s s x
s s x
1
0 0
0 0 0 0 2
2
y V y
u V x
u
xV
y¶
= ¶
¶ + ¶
¶
¶ n
2 2
y u y
u
xx
y¶
= ¶
¶ + ¶
¶
¶ q
q a q
x s
x y
x
u x x u
y u
u . Re
) (
.332 Re 0 332
0
0
-
=
¶ =
¶
¥
¥
=
332 3
0. Re Pr
)
(T Ts x
T = -
¶
¥ 3
Pr1/
TBL = HBL
q Laminar flow heat transfer to flat plate
Heat transfer in boundary Laminar Flow (2)
0
=
¥ ¶
- ¶
= -
=
y s
x y
y k T T
T A h
q ( )
-Local heat transfer coefficient at any distance x :
ΔT차이때문에 열전달이 이루어진다고 고려하고 계산되는 열전달계수
0
=
x x =0.5L x = L
0
=
hx hx=0.5L hx=L
-
¥=
D T T
sT
-
¥=
D T T
sT D T = T
s- T
¥• Laminar flow heat transfer to flat plate
3
0
332
0. Re Pr
)
( s x
y T T x
y
T = -
¶
¶
¥
=
3
0
332
0. Re Pr
x s y
x x
k y
T T T
h k =
¶
¶ - -
=
\
¥ =
332
30 . Re
xPr
x
x
Nu
k x
h = =
Heat transfer by forced Convection in Laminar Flow
.) (
Pr .
Pr . Re
const x C
h C
x k u
h
x h k
x x x
=
=
\
=
=
¥ 3
3
332 0
332 0
m r
-The average value of Nu over the entire length of the plate x
Lk Nu = hx
LL L
o L
o x
av
h
L L C
L dx C
L x dx C
L h
h 1 1 2 2 2
=
=
=
=
= ò ò
L x
av
Nu
Nu 0 664
1 3 1 22
1
=
=
\ . (Pr)
/(Re )
/Mass transfer by in Turbulent Flow on a Flat Plate
Transition point
10
52 ´
x
>
5
Re 10 2 ´
x
<
Re
L L
av
Nu
Nu = 0 . 0365 (Pr)
1/3(Re )
4/5= 1 . 25
\
3 1 5
0292
40 . Re
x/Pr
/x
x
Nu
k x
h = =
3 1 2
332
10 . Re
x/Pr
/x
x
Nu
k x
h = =
L x
av
Nu
Nu 0 664
1 3 1 22
1
=
=
\ . (Pr)
/(Re )
/Laminar flow heat transfer in Tubes (1)
T
aT
bLaminar flow heat transfer in Tubes (2)
-Heating case
-Cooling case 1
v >
f
1
f
<Heat transfer by forced Convection in Turbulent flow
Geometry Laminar flow Turbulent flow Flat plate
Pipe or tube
Summaries for Nu Number
3 1 2
332
10 . Re
x/Pr
/Nu =
xNu =
x0 . 0292 Re
4x/5Pr
1/33 1 8
023
00 . Re
.Pr
/=
= k Nu hD
10
52 ´
x
<
Re Re
x> 2 ´ 10
5100 2, Re >
3
2 Gz
1/k Nu = hD =
100 2
10 < Re < ,
k
vNu = hD = 0 . 023 Re
0.8Pr
1/3f Gz
vk
Nu = hD = 2
1/3f
3 1 3
1 3
85 1
1
/ /
/ Pr Re
. ÷
ø ç ö è æ
L D