Module 8: Solved Problems

Module 8: Solved Problems . 1. Electric current passes through a long wire of diameter 1-mm, and dissipates 3150W/m. The wire reaches surface temperat...

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Module 8: Solved Problems

1. Electric current passes through a long wire of diameter 1-mm, and dissipates 3150W/m. The wire reaches surface temperature of 126C when submerged in water at 1 bar. Calculate the boiling heat transfer coefficient and find the value of the correlation coefficient Cs,f. Known: long wire, 1–mm-diameter, reaches a surface temperature of 126C in water at 1atm while dissipating 3150W/m. Find: (1) Boiling heat transfer coefficient and (2) correlation coefficient Cs,f, if nucleate boiling occurs Schematic:

Assumptions: (1) Steady-state conditions, (2) Nucleate boiling. From property table: Water (saturated, 1atm) Ts=100C, f=1/vg=0.5955kg/m3, cp,l=4217 J/kg.K, l=1/vf=957.9kg/m3, -6 2 -3 l=279*10 N.s/m , prl=1.76, hfg=2257 KJ/kg, =58.9*10 N/m. Analysis: (a) For the boiling process, the rate equation can be rewritten as _

h

_

h

q s" q'  s /(Ts  Tsat ) (Ts  Tsat ) D

3150W / m W /(126  100)C  1.00  10 6 2 / 26  C  38,600W / m 2 .K   0.001m m

Note that heat flux is very close to q”max, and nucleate boiling exists. (b) For nucleate boiling, the Rohsenow correlation may be solved for Cs,f, to give

C s,f

1 3

  h   g (     )     'f ,g      qs  

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 c p , Te   n   h f , g Pr 

Assuming the liquid surface combination is such that n=1 and substituting numerical values with ΔTe=Ts-Tsat, find 1/ 3

C s, f

 279  10 6 N .s / m  22257  10 3 J / kg    1.00  10 6 W / m 2  

 9.8m / s 2 (957.9  0.5955kg / m 3    58.9  10 3 N / m  

 4217 J / kg.K  26 K     3 2257  10 J / kg  1 . 76   C s , f  0.017

Comments: By comparison with the values Cs , f for other watersurface combinations (given in standard tables), the Cs , f value for the wire is quite large suggesting that its surface must be highly polished. Note that the value of the boiling heat transfer coefficient is much larger that for other convection processes previously encountered.

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2. A copper pan has a diameter of 150 mm. The bottom of the pan is maintained at 115C when placed on an electric cooking range. Estimate the heat required to boil the water in this pan. Determine the evaporation rate. What is the ratio of the surface heat flux to the critical heat flux? What pan temperature is required to achieve the critical heat flux? Known: copper pan 150 mm in diameter and filled with water at 1atm, maintained at 115C. Find: the power required to boil the water and the evaporation rate; ratio of the heat flux to the critical heat flux; pan temperature is required to achieve the critical heat flux. Schematic: .

m

Water, 1 atm

Pan, D=150mm Electric range element

qboil

Assumptions: (1) Nucleate pool boiling, (2) Copper pan is polished surface. Properties: Table: Water (1atm) Tsat=100C, l=957.9kg/m3, v=0.5955kg/m3, cp,l=4217 J/kg.K, l=279*10-6N.s/m2, prl=1.76, hfg=2257 KJ/kg, =58.9*10-3N/m. Analysis: the power requirement for boiling and the evaporation rate can be expressed as follows, qboil  q s" . As

.

m  qboil / h f , g

The heat flux for nucleate pool boiling can be estimated using the Ronsenow Correlation.

 g (     )  q s   h f , g     

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"

 c p ,Te   n   C s , f h f , g Pr 

3

Selecting Cs , f =0.013 and n=1 from standard table for the polished copper finish, find  279  10 6 N .s / m  2257  10 3 J / kg  9.8m / s 2 (957.9  0.5955kg / m 3  C s, f     58.9  10 3 N / m     4217 J / kg.K  26 K     3  2257  10 J / kg  1.76 

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C s , f  4.619  10 5 W / m 2

The power and evaporation rate are qboil  4.619  10 5 W / m 2 

 4

(0.150m) 2  8.16kW

.

m boil  8.16kW / 2257  10 3 J / kg  3.62  10 3 kg / s  13kg / h

The maximum or critical heat flux was found as Q”max=1.26MW/m2. Hence, the ratio of the operating to maximum heat flux is q s"  4.619  10 5 W / m 2 / 1.26MW / m 2  0.367 " q max

From the boiling curve, ΔTe 30C will provide the maximum heat flux

3. A silicon chip has a thickness L=25 mm and has thermal conductivity ks=135W/m.K. The chip is cooled by boiling a saturated fluorocarbon liquid (Tsat=57C) on its surface. The electronic circuits on the bottom of the chip are perfectly insulated. Properties of the saturated fluorocarbons are cp,l=110J?kg.K, hfg=84,400J/kg, l=1619.2kg/m3, =13.4kg/m3,=8.1*103 kg/s2,l=440*10-6kg/m.s and prl=9.01. The nucleate boiling constants are Cs,f=0.005 and n=1.7. What is the steady-state temperature at the bottom of the chip? If the chip bottom heat flux is raised to 90% of the critical heat flux, what is the new steady-state value of the chip bottom temperature? Known: Thickness and thermal conductivity of a silicon chip. Properties of saturated fluorocarbon liquid on top side. Find: (a) Temperature at bottom surface of chip for a prescribed heat flux, (b) Temperature of bottom surface at 90% of CHF. Schematic:

Assumptions: (1) steady-state conditions, (2) uniform heat flux and adiabatic sides, hence one-dimensional conduction in chip, (3) Constant properties, (4) Nucleate boiling in liquid.

Properties:

Saturated

fluorocarbon

(given):

c p ,  1100 J / kg.K ,

hf,g=84,400 J/kg,   =1619.2kg/m3, =13.4kg/m3,=8.1*10-3kg/s2,   =440*10-6kg/m-s, Pr =9.01. "  k s (To  Ts ) / L  qb" . Analysis: (a) Energy balances yield qo"  qcond

Obtain Ts from the Rohsenow correlation. Ts  Tsat

C s , f h f , g Prn  q s'    h c p ,   f ,g

     g (   )     

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1/ 3 Ts  Tsat 

0.005(84,400 J / kg )9.011.7 1100 J / kg.K

  8.1  10 3 kg / s 2  2 3  9.807 m / s (1619.2  13.4)kg / m 

  5  10 4 W / m 2    6  440  10 kg / m.s  84,400 J / kg 

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 15.9C

Ts  (15.9  57)  C  72.9C

From the rate equation, q o" 5  10 4 W / m 2  0.0025m To  Ts   72.9  C   73.8C ks 135W / m.K

(b) With the heat rate 90% of the critical heat flux (CHF)

q

" max

g (  l   v )   0.149hfgv   v 2  

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 0.149  84,400 J / kg  13.4kg / m 3

 8.1  10 3  9.807 m / s 2 (1619.2  13.4)kg / m 3    13.4kg / m 3   " q max  15.5  10 4 W / m 2

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"  13.9  10 4 W/m 2 q "o  0.9q max

Te  Te ) a (q "o /q "o,a )1/3  15.9C  1.41  22.4  C

To

 79.4C 

Ts  22.4C  57C  79.4  C

13.9  10 4 W/m 2  .0025  82.0  C 135W/m.K

Comments: Pool boiling is not adequate for many VLSI chip design

4. Strip steel is made by rolling the strip on a set of rollers in a hot rolling mill. As strip steel leaves the last set of rollers, it is quenched by water jets before being coiled. Because the plate temperatures are high, film boiling is achieved as a result. Consider conditions for which the strip steel beneath the vapor blanket is at a temperature of 907K and has an emissivity of 0.35. Neglecting the effects of the strip and jet motions and assuming convection within the film to be approximated by that associated with a large horizontal cylinder of 1-m-diameter, estimate the rate of heat transfer per unit surface area from the strip to the wall. Known: Surface temperature and emissivity of strip steel. Find: heat flux across vapor blanket.

Schematic: Wall jet q”s

Water vapor

Steel strip, Ts=907K,

Assumptions: (1) Steady-state conditions, (2) Vapor/jet interface is at Tsat for P=1atm, (3) Negligible effect of jet and strip motion.

Properties: Table: saturated Water (100C 1atm) l=1/vf=957.9kg/m3, hfg=2257 KJ/kg: saturated water vapor (Tf=640K): v=175.4kg/m3, cp,v=42 J/kg.K, v=32*10-6N.s/m2,k=0.155W/m.K,v=0.182*10-6m2/s. Analysis: The heat flux is _

q "x  h ΔTe Where ΔTe  907K  373K  534K - 4/3

_

- 4/3

h  h

and

conv

1/3

 h rad h

with h 'fg  h fg  0.80c p, v (Ts  Tsat )  2.02  10 7 J/kg  9.8m / s 2 (957.9  175.4)kg / m 3 (2.02  10 7 J / kg )(1m) 3  N u D  0.62   0.182  10 6 m 2 / s (0.155W / m.K )(907  373) K   _

1/ 4

hence, _

_

h conv  N u D k v / D  6243W / m 2 .K (0.155W / m.K / 1m)  968W / m 2 .K _

h rad 

 (Ts4  Tsat4 ) Ts  Tsat



0.35  5.67  10 8 W / m 2 .k (907 4  373 4 ) K 4 (907  373) K

_

h rad  24W / m2.K

hence, and

_

h  968W / m2.K  (3 / 4)(24W / m2.K )  986W / m2.K q "s  986W / m 2 .k (907  373) K  5.265  10 5 W / m 2

 6243

5. Saturated steam at 0.1 atm condenses with a convection coefficient of 6800 W/m2.K on the outside of a brass tube having inner and outer diameters of 16.5 and 19mm, respectively. The heat transfer coefficient for water flowing inside the tube is 5200W/m2.K. Assuming that the mean water temperature is 30C, calculate the steam condensation rate per unit tube length. Known: saturated steam condensing on the outside of a brass tube and water flowing on the inside of the tube; convection coefficients are prescribed. Find: Steam condensation rate per unit length of the tube. Schematic:

Assumptions: (1) Steady-state conditions. Properties: Table: Water, vapor (0.1 bar): Tsat320K, hfg=2390*103J/kg; _

Table: Brass ( T  (Tm  Tsat ) / 2  300 K ); k=110W/m.K. Analysis: The condensation rate per unit length is written as . '

m  q ' / h fg

'

(1)

Where the heat rate follows from equation using overall heat transfer coefficient q '  U o . D o ( T sat  T m )

(2)

 1 D /2 D D 1 U o    o n o  o  k Di Di hi   ho

1

(3)

1 0.0095m 19 19 1   n Uo     2 2   6800W / m .K 110W / m.K 16.5 16.5 5200W / m .K 

1

U o  147.1  10 6  12.18  10 6  192.3  10 6 W / m 2 .K  2627W / m 2 .K

Combining equations, (1) and (2) and substituting numerical values, find . '

m  U oDo (Tsat  Tm ) / h '

fg

. '

m  2627W / m 2 .K (0.019m)(320  303) K / 2410  10 3 J / kg  1.11  10 3 kg / s

Comments: (1) Note from evaluation of equation. (3) That the thermal resistance of the brass tube is not negligible. (2) With Ja  c p , (Tsat  Ts ) / h f , g , h ' fg  h fg [1  0.68 Ja]. note from expression for Uo, that the internal resistance is the largest. Hence, estimate Ts ,o  To  ( Ro / R)(To  Tm )  313K . Hence h 'fg  2390  10 3 J / kg[1  0.68  4179 J / kg.K (320  313) K / 2390  10 3 J / kg ] h 'fg  2410kJ / kg Where c p, for water(liquid) is evaluated at Tf  (Ts ,o  To ) / 2  317 K