Operational Principle of Thermal Bubble Jet

injected droplet

heaterbubble

Chamber neck

liquid

• Current pulse heats up liquid• Bubble as a pump pushes out droplet• Refill by capillary force

Boiling heat transfer

Satellite Droplets

Satellite Droplets

Inkjet Droplet injection sequence

LiquidEntrance

ManifoldNarrowheater

Wideheater

Nozzle

Chamber

Commonline Electrode

Liquid

Micro-Machined Thermal Bubble Jet

Heater

Silicon Wafer

1-D Heat Conduction with GenerationThin heater deposited on siliconcan be modeled as heat generation

Silicon Waferthermal conductivity k

T,1, h1

L

Ts

T,1

T,2

1/(h1A)

T

T,2, h2

L/(kA) 1/(h2A)

q

q

q1q2

/ ( ) / ( ) / ( )

, ,q i R q qT T

h A

T T

L kA h A

21 2

1

1

2

21 1

T

1-D, steady Heat Transfer in Cylindrical Coordinate

Radial direction

qr=-kAr(dT/dr)

qr+dr=qr+(dqr/dr)dr

Net heat transfer: since no generation, steady state

-d

dr assume k = constant

integrate twice and apply boundary conditions

rq q

kAdT

dr

d

drk rL

dT

drd

drrdT

dr

r dr

0

2 0

0

( ) [ ( ) ] ,

( ) ,

Cylindrical Heat Conduction

d

drrdT

dr

dT

drC

dT

dr

C

rT C r C

T T

CT T

r rC T

T T

r rr

T rT T

r rr r T

q kAdT

dr

kL T T

r r

T T

Rth

( ) , ,

ln

) )

ln( / ),

ln( / )ln( )

( )ln( / )

ln( / )

( )

ln( / )

( )

0

2

11

1 2

1 2

11 2

1 22 2

1 2

1 22

1 2

1 22 2

1 2

2 1

1 2

integrate once: r

Integrate again:

Apply boundary conditions, T(r and T(r

,

temperature distribution along the radial direction

Thermal resistance in cylindrical coordinate R

1 2

th

ln( / )r r

kL2 1

2

Heat Loss from a Cylindrical Pipe (no insulation)

Steam at 300°C flow in a cast iron circular pipe (k=80W/m.K). Determine the heat loss per unit length of the pipe to the surroundings at T=20°C, with a combined (radiation & convection) heat transfer coefficient of h2=20 W/m2.K). The convection coefficient inside the pipe is h1=50 W/m2.K)r1=1 cm

r2=1.25 cm Rconv,1 RcondRconv,2

qT T

R

T T

R R R

T T

h A

r r

kL h Ath conv cond conv

, ,ln( / )

1 2

1

2 1

2

1

2

1

TT

Heat Loss from a Cylindrical Pipe (no insulation) (cont.)

Rh A r

K W

Rr r

kLK W

Rh A r

K W

qT T

RW

conv

cond

conv

th

,

,

( )( )( ). ( / )

ln( / ) ln( . / )

(80)( ). ( / )

( )( )( ). ( / )

. . .. ( )

11 1 1

2 1

22 2 2

1 1

50 2 10 318

2

1 25 1

2 10 000807

1 1

20 2 10 637

300 20

0 318 0 000807 0 637292 9

Note: the cast iron pipe is a good conductor, therefore, it has a small thermal resistance. To prevent heat loss, it is recommended that insulation be used outside the pipe.

Insulation

It is suggested that a thick layer of insulation (k=0.5 W/m.K) be wrapped outside the pipe. Determine the heat loss as a function of the thickness of the fiber glass wrap.

Rconv,1 RcondRconv,2T

TRinsulation

r1=1 cmr2=1.25 cm

Assume the fiber thickness is tr3=r2+tRconv,1 and Rcond stay the same, unchanged

Rr r

k L

t rt

Rh A h r L t t

insulinsul

conv

ln( / ) ln( / )

( . )( ). * ln( )

( ) ( )( )( . )( ) . ( . ),

3 2 2

22 3 2 3

2

1

2 0 5 10 318 1 80

1 1

2

1

20 2 0 0125 1

1

125 6 0 0125

r3

r2

r1

Insulation-2

qT T

R

T T

R R R R

T T

h A

r r

kL

r r

k L h A

tt

tt

th conv cond insul conv

insul

, ,

ln( / ) ln( / )

. . . ln( ). ( . )

. . ln( ). ( . )

1 2

1

2 1 3 2

2 3

1

2 2

1

300 20

0 318 0 000807 0 318 1 801

125 6 0 0125

280

0 319 0 318 1 801

125 6 0 0125

• Plot this function in the following slide

Critical Radius of Insulation

0 0.02 0.04 0.06 0.08 0.1250

270

290

310

330

350

t (insulation thickness)

q (h

eat

loss

)

q( )t

t

• Heat loss increases initially when one adds insulation (why?)• Maximum heat loss at critical radius of insulation that rcritical =kinsul/houtside

• In the present case, rcritical=0.5/20=0.025• Its critical thickness is t=0.025-0.0125=0.0125(m)

Critical thickness

Note: similar derivation can be made for a spherical container. The critical radius for a spherical shell is rcritical=2k/h