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1. 1. Unsteady State Conduction Evelyn R. Laurito Lani Pestano Ch.E. 206
2. 2. Lecture Objectives <ul><li>To understand the concept of unsteady state conduction </li></ul><ul><li>To study the case of unidirectional unsteady state conduction </li></ul><ul><li>To understand how to use Geankoplis Charts in solving unidirectional unsteady state conduction problems </li></ul><ul><ul><li>Gurney and Lurie Charts </li></ul></ul><ul><ul><li>Heisler Chart </li></ul></ul><ul><ul><li>Chart for Average Temperature </li></ul></ul><ul><ul><li>Chart for Semiinfinite solid </li></ul></ul><ul><li>To understand how to use Numerical Methods in solving unidirectional unsteady state conduction problems </li></ul>
3. 3. Unsteady State Conduction <ul><li>This happens when the temperature gradient across the solid changes with time. </li></ul><ul><li>This may be due to unstable boundary temperatures at startup, sudden temperature fluctuations during steady state conditions, or internal generation of heat. </li></ul><ul><li>Sample Cases: </li></ul><ul><ul><li>Startup of a Furnace </li></ul></ul><ul><ul><li>Heat Treatment of Solids </li></ul></ul><ul><ul><li>Deep Oil Frying </li></ul></ul><ul><ul><li>Change of Weather </li></ul></ul>
4. 4. Unidirectional Unsteady State Case  x  y  z x x+  x q in q out A = V = Solid properties:  , c P Mass = Heat Balance Across  x: q in - q out = Rate of heat accumulation Rate of heat accumulation = Using Fourier’s Law: q in = q out = c P  y  z  x  y  z  x  y  z   x  y  z   T  t - k  y  z  T  x x - k  y  z  T  x x +  x
5. 5. Unidirectional Unsteady State Case The Heat Balance becomes: - = Simplifying: -  x From Calculus: Final Equation:  but - k  y  z  T  x x - k  y  z  T  x x +  x  x  y  z  c P  T  t  T  t = k  c P  T  x x +  x  T  x x  T  x x +  x -  T  x x  x =  2 T  x 2  T  t = k  c P =  2 T  x 2 
6. 6. Unidirectional Unsteady State Case <ul><li>Depends on solid geometry </li></ul><ul><li>Requires PDE solution methods that results into Fourier series solutions that are tedious to evaluate </li></ul><ul><li>May be simplified by the use of charts or numerical methods </li></ul>Use of Charts: <ul><li>Gurnie-Lurie Charts – to determine point temperatures </li></ul><ul><li>Heisler Charts – to determine central temperatures </li></ul><ul><li>Average Temperature Chart </li></ul><ul><li>Chart for Semiinfinite Solids </li></ul> T  t =  2 T  x 2  Solution of:
7. 7. Geankoplis Charts <ul><li>Gurney-Lurie Charts </li></ul><ul><ul><li>Fig. 5.3-5/340 for large flat plate </li></ul></ul><ul><ul><li>Fig. 5.3-7/343 for long cylinder </li></ul></ul><ul><ul><li>Fig. 5.3-9/345 for sphere </li></ul></ul><ul><li>Heisler Chart </li></ul><ul><ul><li>Fig. 5.3-6/341 for large flat plate </li></ul></ul><ul><ul><li>Fig. 5.3-8/344 for long cylinder </li></ul></ul><ul><ul><li>Fig. 5.3-10/346 for sphere </li></ul></ul><ul><li>Fig. 5.3-13/349 for Ave. Solid Temperature </li></ul><ul><li>Fig. 5.3-3/337 for Semi-infinite solid </li></ul>
8. 8. Nomenclature <ul><li>Gurney-Lurie and Heisler Charts: </li></ul><ul><ul><li>T o = temperature at t(time)= 0 (uniform) </li></ul></ul><ul><ul><li>T 1 = new and constant surface temperature </li></ul></ul><ul><ul><li>x 1 = ½ plate thickness, outer radius of cylinder or sphere </li></ul></ul><ul><ul><li> = constant thermal diffusivity </li></ul></ul><ul><ul><li>X =  t/ x 1 2 : relative time </li></ul></ul><ul><ul><li>x = distance from plate center or any radius of a cylinder or a sphere </li></ul></ul><ul><ul><li>n = x/x 1 : relative position </li></ul></ul><ul><ul><li>T = point temperature at position x and time t </li></ul></ul><ul><ul><li>Y = (T 1 -T)/(T 1 -T o ) :unaccomplished temp. change </li></ul></ul><ul><ul><li>h = convective heat transfer coefficient </li></ul></ul><ul><ul><li>m = k/(hx 1 ) : relative resistance </li></ul></ul>
9. 9. Nomenclature <ul><li>Average Temperature Chart </li></ul><ul><ul><li>T o = temperature at t(time)= 0 (uniform) </li></ul></ul><ul><ul><li>T 1 = new and constant surface temperature </li></ul></ul><ul><ul><li>T av = average solid temperature at time t </li></ul></ul><ul><ul><li>E = (T 1 -T av )/(T 1 -T o ) </li></ul></ul><ul><ul><li>a = ½ plate thickness, outer radius of cylinder or sphere </li></ul></ul><ul><ul><li>b = ½ plate width </li></ul></ul><ul><ul><li>c = ½ plate length, ½ cylinder length </li></ul></ul>
10. 10. Nomenclature <ul><li>Chart for Semi-infinite solid </li></ul><ul><ul><li>Semi-infinite solid – solid where the unidirectional conductive heat transfer is infinite (Ex. Ground) </li></ul></ul><ul><ul><li>T o = initial uniform solid temperature </li></ul></ul><ul><ul><li>T 1 = constant ambient temperature to which solid surface is exposed </li></ul></ul><ul><ul><li>T = temperature of solid at position x </li></ul></ul><ul><ul><li>1- Y = (T-T o )/(T 1 -T o ): Ordinate </li></ul></ul><ul><ul><li>h(  t) 0.5 /k : convective parameter </li></ul></ul><ul><ul><li>x/[2 (  t) 0.5 ]: Abscissa </li></ul></ul>
11. 11. Problems from Geankoplis <ul><li>Exercises: </li></ul><ul><ul><li>5.3-5 (Plate) </li></ul></ul><ul><ul><li>5.3-7 (Long Cylinder) </li></ul></ul><ul><ul><li>5.3-9 (Sphere) </li></ul></ul><ul><ul><li>Find the average solid temperature for all of the above cases </li></ul></ul><ul><ul><li>5.3-3 </li></ul></ul><ul><li>Homework: </li></ul><ul><ul><li>Geankoplis: 5.3-2;5.3-4;5.3-6;5.3-8;5.3-10 (pages 375-377) </li></ul></ul><ul><ul><li>Foust: 11.18;11.20;11.21(page 231) </li></ul></ul>
12. 12. 5.3-5 Cooling a Slab of Meat <ul><li>A slab of meat 25.4 mm thick originally at a uniform temperature of 10 o C is to be cooked from both sides until the center reaches 121 o C in an oven at 177 o C. The convection coefficient can be assumed constant at 25.6 W/m 2 -K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10 -4 m 2 /h. </li></ul>
13. 13. Solution for 5.3-5 <ul><li>Given: </li></ul><ul><ul><li>t= 25.4 mm </li></ul></ul><ul><ul><li>T o =10 o C </li></ul></ul><ul><ul><li>T=121 o C (Center T) at x 1 </li></ul></ul><ul><ul><li>T 1 =177 o C </li></ul></ul><ul><ul><li>h=25.6 W/m 2 -K </li></ul></ul><ul><ul><li>k=0.69 W/m-K </li></ul></ul><ul><ul><li> =5.85x10 -4 m 2 /h </li></ul></ul><ul><li>Required: t= ? </li></ul>
14. 14. Solution for 5.3-5 <ul><ul><li>x 1 =12.7mm </li></ul></ul><ul><ul><li>x=25.4mm </li></ul></ul>
15. 15. 5.3-7 Cooling of a Steel Rod <ul><li>A long steel rod 0.305 m in diameter is initially at a temperature of 588K. It is immersed in an oil bath maintained at 311K. The surface convective coefficient is 125 W/m 2 -K. Calculate the temperature at the center of the rod after 1 h. The average physical properties of the steel are k=38 W/m-K and  =0.0381m 2 /h </li></ul>
16. 16. Solution for 5.3-7 <ul><li>Given: </li></ul><ul><ul><li>D= 0.305 m </li></ul></ul><ul><ul><li>T o =588 K </li></ul></ul><ul><ul><li>T 1 =311 K </li></ul></ul><ul><ul><li>h=125 W/m 2 -K </li></ul></ul><ul><ul><li>t=1 h </li></ul></ul><ul><ul><li>k=38 W/m-K </li></ul></ul><ul><ul><li> =0.0381 m 2 /h </li></ul></ul><ul><li>Required: T at the center </li></ul>
17. 17. Solution for 5.3-7
18. 18. 5.3-9 Temp. of Oranges on Trees During Freezing Weather <ul><li>In orange-growing areas, the freezing of the oranges on the trees during cold nights is economically important. If the oranges are initially at a temperature of 21.1 o C, calculate the center temperature of the orange if exposed to air at –3.9 o C for 6 h. The oranges are 102 mm in diameter and the convective coefficient is estimated as 11.4W/m 2 -K. The thermal conductivity k is 0.431 W/m-K and  =4.65x10 -4 m 2 /h. Neglect any latent heat effects. </li></ul>
19. 19. Solution for 5.3-9 <ul><li>Given: </li></ul><ul><ul><li>D= 102 m  x=102/2=51mm </li></ul></ul><ul><ul><li>T o =21.1 o C=294.1K </li></ul></ul><ul><ul><li>T 1 =-3.9 o C=269.1K </li></ul></ul><ul><ul><li>h=11.4 W/m 2 -K </li></ul></ul><ul><ul><li>t=6 h </li></ul></ul><ul><ul><li>k=0.431 W/m-K </li></ul></ul><ul><ul><li> =4.65x10 -4 m 2 /h </li></ul></ul><ul><li>Required: T at the center </li></ul>
20. 20. Solution for 5.3-9 From Fig. 5.3-10: Y = 0.05
21. 21. 5.3-3 Cooling a Slab of Aluminum <ul><li>A large piece of aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4K. The surface is suddenly exposed to an environment at 338.8K with a surface convection coefficient of 455W/m 2 -K. Calculate the time in hours for the temperature to reach 388.8 K at a depth of 25.4 mm. The average physical properties are  =0.340m 2 /h and k=208W/m-K. </li></ul>
22. 22. Solution for 5.3-3 <ul><li>Given: </li></ul><ul><ul><li>D= 0.305 m </li></ul></ul><ul><ul><li>T o =505.4 K </li></ul></ul><ul><ul><li>T 1 =338.8 K </li></ul></ul><ul><ul><li>T=388.8K when x=25.4mm </li></ul></ul><ul><ul><li>h=455 W/m 2 -K </li></ul></ul><ul><ul><li>k=208 W/m-K </li></ul></ul><ul><ul><li> =0.304 m 2 /h </li></ul></ul><ul><li>Required: time in hours for the temperature to reach 388.8K at a depth of 25.4 mm </li></ul>
23. 23. Solution for 5.3-9 From Fig. 5.3-10: Y = 0.05