Module18: Internal Flow Flow Flow and and Heat Heat Tra Trans nsfe ferr thro throug ugh h Pipes and Ducts
Introduction to Internal Flow • Characteristics follow I&D, Chapter 8 – primary difference from external flow is the presence of an opposing, confining surface that constrains the boundary layer growth – entry (entrance) length exists (B.L. is developing) developing) – fully developed region eventually forms
r 0 ≡ R ref. Incr Incropera opera & DeWit DeWitt, t, Chap. 8
Hydrodynamic Boundary Layer Development
Introduction to Internal Flow • Characteristics follow I&D, Chapter 8 – primary difference from external flow is the presence of an opposing, confining surface that constrains the boundary layer growth – entry (entrance) length exists (B.L. is developing) developing) – fully developed region eventually forms
r 0 ≡ R ref. Incr Incropera opera & DeWit DeWitt, t, Chap. 8
Hydrodynamic Boundary Layer Development
Ter ms and Nota ot ation ti on • Criti Critica call Reyn Reynol olds ds num numbe ber r Re D ,c
=
umD
ν
≈
2300
onset of turbulence
• Hydr Hydrod odyn ynam amic ic ent entry ry len lengt gth h Laminar: x fd ≈ 0.05 Re D
D
Turbulent:
10
≤
x fd D
≤
60
• Mean velocity, um such that
& m
= ρu m A c
constant for steady incompressible flow
• Mean Mean temp temper erat atur ure e Tm (to be defined later)
General Assumptions We will assume: – steady, incompressible, Newtonian,constant properties – Kn (= λ/L) << 1 (continuum) for air @ STP λ ≈ 0.1 µm continuum breaks down for – very low pressure – very small dimensions Careful using for microchannels with gases !
Hydrodynamics - Laminar Flow ∂u 1 ∂ (rv) + = 0 continuity ∂x r ∂r ∂u ∂u ⎞ ∂p µ ∂ ⎛ ∂u ⎞ ⎛ ρ⎜ u + v ⎟ = − + r x − momentum ⎜ ⎟ ∂r ⎠ ∂x r ∂r ⎝ ∂r ⎠ ⎝ ∂x ∂v ∂v ⎞ ∂p µ ∂ ⎛ ∂ v ⎞ ⎛ ρ⎜ u + v ⎟ = − + r r-momentum ⎜ ⎟ ∂r ⎠ ∂r r ∂ r ⎝ ∂ r ⎠ ⎝ ∂x
Entrance Region:
B.C.s:
u(x, R) = 0
∂u = ∂r r =0
v(x, R) = 0
0
u(x = 0, r) = uo(r)
Needs numerical solution …
Assume axisymmetry
Hydrodynamics - Laminar Flow •Fully Developed Region – Poiseuille Flow (parallel flow): ∂u = 0 ⇒ u(r) v=0 ⇒ ∂x • v-momentum equation yields ∂ p = 0 ⇒ p = p ( x) only
∂r
•x-momentum reduces to balance between pressure & shear forces:
µ
du ⎞ ⎛ r ⎟ = ⎜ dx r dr ⎝ dr ⎠ dp
•B.C.s
u(R) = 0 ∂u =0 ∂r r =0
d
•
Integrate twice and apply B.C.s to get
∴ •
•
Mean velocity
µ r
⎛ r du ⎞ = dp ⎜ ⎟ dr ⎝ dr ⎠ dx d
2 ⎡ r ⎤ ⎛ ⎞ 2 u(r) = − R ⎢1 − ⎜ ⎟ ⎥ ⎜ ⎟ 4µ ⎝ dx ⎠ ⎣ ⎝ R ⎠ ⎦
1
Velocity distribution um
•
Velocity profile
can integrate easily since dp/dx is independent of r
u(r) um
⎛ dp ⎞
=
& m
ρπR 2
=∫
R
0
ρu(2πr)dr R 2 =− 2 8µ ρπR
⎡ ⎛ r ⎞ ⎤ = 2 ⎢1 − ⎜ ⎟ ⎥ ⎣ ⎝ R ⎠ ⎦ 2
dp dx
Notice dimensionless velocity distribution not a functions of Re – why?
Fully Developed Laminar Flow Pressure Drop : Expressed in terms of the Moody (or Darcy) friction factor D = 2R
⎛ dp ⎞
f
−⎜ ⎟D dx ⎠ ⎝ ≡ ρu 2m / 2
∆ pfd = f
ρu 2m 2D
f =
L
64 Re D
• fully developed • laminar
ref. Incropera & DeWitt, Chap. 8
Note: dP/dx is constant, but f is not, due to funny nondimensionaliza tion
For turbulent flow the analysis is not as simple as above, and the pressure drop is very sensitive to roughness (unlike in laminar flow). For smooth surfaces, f turb = 0.316 ReD-0.25 (ReD < 20000) = 0.184 ReD-0.20 (ReD > 20000)
Thermal Considerations - Laminar Flow Characteristics:
ref. Incropera & DeWitt, Chap. 8
r 0 ≡ R
Terms and Notation •Thermal entrance length
x fd,t D
≈
0.05 ReD Pr
(unlike in laminar flow, the entrance length is nearly independent of Pr in turbulent flow, with L t / D ~ 10) • The shape of the fully developed profile T(r, x) is different depending on whether T s or q″ is a constant • Example: for engine oil (Pr
≈ 6000), say D = 1 cm, u m = 1 m/s,
ν = 550 x 10-6 m2/s: ReD = (1) (0.01)/(550 x 10 -6) = 18 (laminar) xfd,t = 0.05 Re Pr = 5455 tube diameters! i.e., a tube length of 54 m!! (δt never reaches the centerline in pipes of reasonable length)
Bulk Mean Temperature • Bulk mean temperature: rate of thermal energy transport
& E t
& v Tm = = mC
∫
ρuCv TdAc
Ac
Tm
∫ =
ρuCvTdA c
Ac
& v mC
Weighted w.r.t. mass flow rate • For a circular cross section, with constant-property flow Tm
=
2 um
T(r) u(r) r dr ∫ R 2
Thermally Fully Developed Flow − T(x, r) Ts − Tm (x) • The relative shape of the temperature profile no longer changes if field is “fully developed”
• Define a dimensionless temperature
∂ ⎡ Ts − T ⎤ =0 ⎢ ⎥ ∂x ⎣ Ts − Tm ⎦
Ts
−∂T/∂r r=R ∂ ⎡ Ts − T ⎤ = = constant ≠ f (x) ⎢ ⎥ ∂r ⎣ Ts − Tm ⎦ r =R Ts − Tm
∂T ⇒ ′′ q s = h(Ts − Tm ) = k ∂r r = R
h k
= constant
In thermally f.d. flow with const. props., local h is independent of x ! That is, Nux is independent of x
Can this happen without hydrodynamically fullydeveloped flow?
Energy Balance Temperature Distribution - Energy Balance
Energy Balance:
& E
in
=
& E
out
specific volume
v =
d(C v Tm + pv) ⎤ ⎡ & & & dq conv + m(C dx ⎥ = 0 v Tm + pv) − ⎢ m(C v Tm + pv) + m dx ⎣ ⎦ thermal energy
flow work
1
ρ
Energy Balance (con’td) d(Cv Tm + pv) ⎤ ⎡ & & dq conv + dx ⎥ = 0 + pv) − ⎢ m(Cv Tm + pv) + m dx ⎣ ⎦ Perfect gas ⇒ pv= RTm ; also dqconv = q′′s (x)Pdx d(C v + R)Tm ′′ & q s (x)Pdx − m dx = 0 Perimeter & m(C v Tm
dx
But C v
+ R = Cp
and d ( Cp Tm ) = di
For constant properties: & p q′′s (x)P = mC
dTm
dx For a circular pipe P dTm dx
=
πDq′′s (x) & p mC
=
πD
What if it’s not a perfect gas ? Neglect pressure work, set Cp =Cv to get the same result.
Constant Heat Flux Boundary Conditions dTm dx
=
πDq ′′s & p mC
= constant
Furthermore, q′′s = h ( Ts − Tm ) ⇒ ( Ts − Tm ) = constant
⇒
dTs dx
=
dTm dx
= constant
Also, recall θ = Since Tm Thus,
Ts -T(x,r) Tm
− Ts
=constant with x
− Ts =constant, Ts -T(x,r) is also constant
∂T dTs dTm = = ∂x dx dx
All temperatures rise at the same rate axially!
Constant Temperature Boundary Conditions dTm
=
dx
πDq′′s (x) & p mC
= not constant
However, q′′s (x) = h ( Ts d ( Ts
− Tm )
dx
=
πD & p mC
− Tm ) .Thus :
h ( Ts − Tm )
( Ts − Tm ) = ( Ts − Tm )inlet exp(−
πD & p mC
Careful! h is constant only in fully-developed region!
hx)
Bulk temperature varies exponentially! Furthermore,since θ=
∂T Ts − T = ∂x Ts − Tm
dTm dx
Ts -T(x,r)
=constant with x
− Ts 1 ∂T 1 ⇒ = Ts − T ∂x Ts − Tm Tm
dTm dx
= constant
All temperatures tend towards Ts exponentially with x!
Axial Temperature Variation
Is bulk temperature variation linear (or exponential) through the FD region? What about the temperature at a point (x,r)?
Hydrodynamically and Thermally Fully Developed Flow Solution Energy Equation: (f.d. vel. Profile, v=0)
∂T α ∂ ⎛ ∂T ⎞ = u r ⎟ ⎜ ∂x r ∂r ⎝ ∂r ⎠
• Constant Surface Flux
Constant Surface
Temperature
→ q′′s =
constant = h(Ts − Tm )
∂T dTm → = ∂x dx See Sec. 8.4.1 I&D for solution
→ →
dTs dx ∂T
∂x
=0 =
Ts − T
dTm
Ts − Tm
dx
Solution for Constant Heat Flux BC ∂T α ∂ ⎛ ∂T ⎞ = u r ⎟ ⎜ ∂x r ∂r ⎝ ∂r ⎠
• LHS known •Integrate twice to obtain T(r) = Ts
−
2u m R 2 dTm
α
dx
⎡ 3 1 ⎛ r ⎞4 1 ⎛ r ⎞ 2 ⎤ ⎢ + ⎜ ⎟ − ⎜ ⎟ ⎥ ⎢⎣16 16 ⎝ R ⎠ 4 ⎝ R ⎠ ⎥⎦
Is this known?
Since both bc are Neumann-type, temperature can only be determined up to an additive constant. What is the physical meaning of this?
Nusselt Number Steps:
∂T = h ( Ts − Tm ) ∂r r = R
1. Recall
k
2. Find
∂T from temperature solution ∂r r = R
3. Find bulk temperature: Tm
=
2 um
T(r) u(r) r dr ∫ R 2
4. Hence find h, and thus NuD Nu D
=
hD k
=
4.36 for q′′s
Known
= constant
Notice Nusselt number not a function of Re or Pr!
Solution for Constant Temperature BC • Solution is a bit more complicated because LHS is not constant. • Solution obtained numerically • Can show that
Nu D
=
3.66
Ts
= constant
Other Useful Relationships • For the entire tube (i - inlet, o - outlet), overall energy & p (Tm,o − Tm,i ) q conv = mC balance: • Also
dTm dx
=
q′′s P
=
& p mC
P & p mC
h(Ts
− Tm ) q′′s = constant
Ts=constant
⎡ Px ⎤ − Tm (x) = exp ⎢ − h⎥ & p ⎥ Ts − Tm,i ⎢⎣ mC ⎦
Ts
q = h A s ∆Tlm
As
= P⋅L
∆To − ∆Ti ∆Tlm ≡ ln( ∆To / ∆Ti )
Tm (x) = Tm,i + q = q′′s P ⋅ L
q′′s P & p mC
x
Developing Flow Terminology • “Thermal entry length problem” – Flow is fully developed, temperature is not
• “Combined entry length problem” – Both flow and temperature are developing
• “Unheated starting length” – There is an insulated length of duct at the entrance so that the flow has a chance to develop while the temperature does not – Synonymous with “Thermal entry length” T=Ts
T=Ts
Convection Correlations • Refer to the course text to find correlations for NuD for: •
– entry region (section 8.4.2) – Hansen formula (Eq. 8.56 I&D) - assumes only thermal entry length; for constant surface temperature – Seider-Tate formula (Eq. 8.57 I&D) - for combined entry length; less accurate; evaluate properties at mean temperature defined as average between inlet and outlet
•
– turbulent flow (section 8.5) – Colburn relation for friction factor for smooth circular tubes (Eq. 8.58 I&D); Dittus Boelter (Eq. 8.59) and Seider-Tate (Eq. 8.61) correlations for Nusselt number
•
– non-circular tubes (section 8.6) – (Laminar flow - use Table 8.1) – Turbulent flow, use correlations for circular tubes with hydraulic diameter: Dh = 4Ac/P, where P is the wetted perimeter
•
concentric tubes (section 8 7)
Entry Length in Circular Pipes
• • • •
NuD at x = 0 = ? Why the difference between thermal and combined entry lengths? Notice that curves are independent of Re, Pr if x axis is scaled as shown Graetz number = Re Pr/(x/D) (some texts use inverse)
Example: •Water at 280 K enters a 1-inch diameter tube kept at a constant surface temperature of 360 K. The tube is 2 m long and water velocity = 1 m/s. •Find the heat transfer coefficient & exit temperature. •Solution: First, estimate the exit temperature to evaluate properties. Try 350 K T = T + T = 350 + 280 = 315 K mi
m
µ 631 × 10− ν= = = 6.36 × 10− m ρ 991 Pr = 4.16 ; k = 0.634 W/mK
2
6
7
Re D
=
1* 0.0254 6 36 10
7
= 3.99 × 10
4
mo
2
⎛Q ρ = 1 = 10 = 991 kg / m ⎞ ⎜ ⎟ v 1.009 ⎝ ⎠ 3
2
/s
turbulent
3
L D
=
2 0.0254
= 78.7 > 10
⇒ f .d.
• Dittus-Boelter correlation (8.60) Nu D
= 0.023(Re
D
)
0.8
( Pr )
0.4
(n = 0.4 for heating,Q Ts
>T
m
)
= 195 ∴ h = 4867 W/m K 2
• Use Eq. (8.43) to get exit temperature (Eq. 8.41 would be for q s” = constant) •
⎛ Px h ⎞ − T (x) = exp ⎜ − ⎟ & C T −T m ⎝ ⎠
Ts
m
s
m ,i
p
for Ts
= constant
@
x = L,
Tm (x) = Tm ,L
P = πD; PL h & C p m
=
4L D
St = PL h & C p m
∴
=T
& = ρu m m
h
ρu
m
Cp
Nu
=
4L D
=
m ,o
πD
2
4 where St = Stanton #
St
195
Re Pr 3.99×10
4
×4.16
= 0.0017
= − 0.3685 360 −Tm,o 360− 280
= 0.692
not 350K as assumed!
⇒
now recalculate with new T
m
Tm,o
=
= 304.6 K
304.6 + 280 exercise, not done here 2
