高等传热学课程作业1

Homework 5.2

1. Slug flow in a tube(u(r) = V) with a fully developed temperature profile:

- Constant heat flux - What is T(r)?

·Remember dT m /dx = constant

·Use dT/dr = 0 at r = 0 and T w at r = R to get ingration constants - What is Nu? A: 根据能量方程，

22

1p p T T T T c u c v k r k r x r r r x

ρρ???????+=+ ????????

(1)

对于充分发展的管内弹状流，u=0, 且由于

m

dT T x dx

?=?= constant ，因此(1)式可化简为： 1m p dT T

c V

k r dx r r r

ρ????

= ?????

(2) 解之可得，

2

12(r)ln 4p m V c dT T r C r C k dx

ρ=

??++ (3) 根据边界条件，

0,

, T(r)=T w T

r r

r R ?==?= 代入(3)中，得到C 1=0, 2

24p m w V c dT C T R k dx

ρ=-

? 平均温度201

()2R

m T T r rdr R ππ=? 28()m w m p dT k

T T dx V c R ρ=- 因此，得到

222

22(r)(1)22()4m w m w w m dT VR r r T T T T T T dx R R

α=+??-=-+-

22

88q''h()()()222p p m w m w m w m p mc c R V dT k k

T T T T T T R dx R V c R R ρπππρ=-=?=?-=- 28h d h R Nu k k

??===

2. Fully developed Poiseuille flow between parallel plates

- Constant heat flux

- Top and bottom at the same temperature - Neglect viscous dissipation - What is T(y)?

·Remember dT m /dx = constant

·Use T w at y = 0 and y = h to get integration constants - What is Nu? A: 根据能量方程，

2222()p p T T T T c u c v k x y x y ρρ????+=+???? (1)

对于x 方向充分发展的Poiseuille 流，v=w=0 由于

m

dT T x dx

?=?= constant ，(1)式可化为： 22p T T c u k x y

ρ??=?? (2)

根据01()h

m u u y dy h =?，且2

2()6()m u y y y u h h

=-，代入得到

22226()m m p dT y y T

u c k h h dx y ρ?-=?

解得温度分布为

3

4

122

()()2p m m c u dT y y T y C y C k dx h h ρ=

?-++ (3)

根据Poiseuille 流边界条件：

0,w y T T == ,w y h T T ==

带入(3)式中解得，12p m m

c u dT C h k

dx

ρ=-, 2w C T =

因此，得到温度分布为

3

4

2()()22p m p m m m w c u c u dT dT y y T y y T k dx h h k dx

ρρ=?--?+ 根据

342220011()()6()22h h

p m p m m m m w m m m c u c u dT dT y y y y T T y udy y T u dy hu hu k dx h h k dx h h ρρ????

==?--+-?????

?????∴

2140()17m w m m p dT k

T T dx u c h

ρ=- ∴140q''h()()17p m w m w m mc

dT k T T T T L dx L =-=

?=- Nu 140

17

=

Homework 5.3

1. Find the equation for the boundary layer thickness δ and for C f for:

23()

u y a b c d U

ηηη=+++ Compare to the exact values and the fourth order equation solutions. A: 对于

23()

u y a b c d U

ηηη=+++，根据边界条件， 0, u=0y =，∴ a = 0

,y u U δ==，∴ 1 = b + c + d 对于平板，有22

0y u

y =?=?，∴c = 0

根据

0y u y

δ

=?=?，∴b + 3d = 0

由此可得，a = 0, b = 1.5, c = 0, d = -0.5.

3()31

22

u y U ηη=- 220

0333()222w y y u y U

U y

μτμ

μδδδ

==?=-=-?-=-?

1039

(1)280

i u u y d U U δδδ=-=? 对于U = constant ，2

3-2i w U

U x δδμρτδδ

?==?

积分后得到，

x

δ

=

与精确解相差7.2%，与4级近似解相差20.5%

02

223=1122

y f u

y U C U U U μ

τ

μδρρρ=??==

，且x δ=

f C =

=

与精确解相差2.6%，与4级近似解相差5.5%

2. Derive the ordinary differential equation and the boundary conditions for the Blasius solution energy equation for flow over a flat plate. (See pages 29-30) A:根据能量方程，

22()k p T T T c u v x y y

ρ???+=???

w w T T T T θ∞-=-，22()k p c u v x y y θθθρ???+=???

常壁温边界条件：

0,0

,1

y y θθ===∞=

根据

Blasius solution δη≈

===

3/22222221122x x x x y y U y y y x θθηθθηηηηθθηθηηθθηθθ

ηηνη????-?????===?-? ? ??????????

????==???????????==?=? ???????

代入能量方程中得到，

)1'''''2p p U c f U c f f k x vx ηρθρηθθ??

-

+-= ?

??

即2''Pr '0f θθ+= 边界条件为

00;

1

ηθηθ===∞=，，

3. Start from the local friction coefficient for flow over a flat plate, C f , on slide 31

and derive the average friction coefficient over the entire plate, C L. Show your work.

A:

2

(x)

1

2

f

C

U

τ

ρ

===

1L

L

C

L

==

?

4. A projectile in the form of a bluff-ended cylinder 20 cm in diameter and 60 cm

long, moves through the air in the direction of its long axis at a velocity U of 100 m/s.

2

1

2

D

frontal

F

C

A U

ρ

=

The drag coefficient, C, is equal to 1.0 for this object.

Frontal area=πD2/4, F D= total drag

Assuming that the boundary layer thickness over the cylindrical surface of the

projectile at a distance x from the leading edge is given by:

x

δ

=

and that the momentum thickness of the boundary layer is given by:

7

72

i

δ

δ

=

Find what proportion of the total drag on the projectile is attributable to skin friction over the curved surface, assuming no pressure gradient in the boundary layer in the streamline direction.

Data. Air kinematic viscosity is 0.15 cm2/s

Air density is 1.15 Kg/m3

Assumptions: Assume this is like a flat plate, then use the equation on slide 15. A:

2

2*

()

i

w

U dU

U

x dx

δ

ρρδτ

?

+=

?

若将圆柱表面看成平板，U = constant,

0dU

dx

=，则

22()772i w U U x x

δδτρρ??==??

因为

x

δ

=

∴4155x x x δ-???????=== ? ?????

?

12

5

772w U τρ-=

11

25

500

070.131372L

L

L w f dx U dx x dx τρ--====??

?12.16N

221

42

D F D U π

ρ=

?=180.64N D

f

F =6.7%

Homework 5.4

1. Derive the equation relating q” to the temperature difference for natural convection driven flow in a round tube (like on slide 19 for parallel plates). - water properties at 25 C - tube length = 100 cm - tube diameter = 1 mm - constant heat flux

- fully developed flow in a tube (from the first homework)

a) Plot ?T and Q (m 3/s) versus heat flux for fluxes of 500 to 5000 W/m 2 b) Is the assumption of fully developed flow valid? - hint: is L/(D Re) > 0.05

c) Is the Boussinesq approximation valid here? A:圆管内充分发展的流体，其速度分布为 22

214R p r u z R μ???=- ????

平均速度

2222

20

1(1)2r 48R

R p r R p u dr R z R z

ππμμ??=

??-=-????

体积流量为

442

88R p R p

Q R u z L

πππμμ??===

? ()()()0

022

L

L

L L L

p gdy g T T dy g T T g T ρρρβρβρβ∞∞∞∞∞∞?=-=-=-=??

? 根据能量平衡，

''2p Qc T q RL ρπ?=

即4''28p R p

c T q RL L πρπμ??=

将p ?代入，得到

()232

''32p c gR q T L

ρβμ∞=?

Pr ''

''32R cond Gr q q =，''cond T q R λ?=

a) 根据以上推导过程可以得到，

T ?=

Q =

分别作出当2''500~5000/q W m =时T ?与Q 随q’’变化的曲线，如下图所示。

b)

max max

Re L L L RL Ru D D Ru D Q D μπμ

ρρρμ

=≥==?125.4>0.05 ∴满足充分发展的假设 c) 热膨胀系数

1T T ρρβρ∞

∞∞

-=-

-

max T T ρρ

ββρ∞∞

-=?≤?=0.085 < 1

满足Boussinesq 假设

2. a) Derive the similarity solution equation for axisymmetric stagnation flow slide 40

b) Derive the similarity solution boundary equations for this problem A:根据连续性方程， 10ru v

r r y

??+=?? R 方向动量方程

22u u p u

u v r y r y ρρμ????+=-+???? (1) 边界层顶端，

2dp dU U C r dr dr

ρρ=-=-

又η=

，()f η=

211''df d u r Crf r y r d dy ψηη?=

=??==?

2112df d v r rf f r r r d dr ψη

η??=-

=-+=-???

∴

'u

Cf r

?=?

'u df d Cr Crf y d dy ηη?==?

2222'''u rC Crf f y y v ???== ??? 代入动量方程(1)中得到，

2

2

2'''''''C Crf Cf f C r rf ρρρμν

??-?=+?

即()

21

102

f ff f ''''''+--=

边界条件为

0 '0 ,=0

'1

f f f δηηη====