Unsteady State Basics

Unsteady State Conduction

Evelyn R. LauritoLani Pestano

Ch.E. 206

Lecture Objectives

To understand the concept of unsteady state conduction

To study the case of unidirectional unsteady state conduction

To understand how to use Geankoplis Charts in solving unidirectional unsteady state conduction problems

Gurney and Lurie Charts Heisler Chart Chart for Average Temperature Chart for Semiinfinite solid

To understand how to use Numerical Methods in solving unidirectional unsteady state conduction problems

Unsteady State Conduction

This happens when the temperature gradient across the solid changes with time.

This may be due to unstable boundary temperatures at startup, sudden temperature fluctuations during steady state conditions, or internal generation of heat.

Sample Cases: Startup of a Furnace Heat Treatment of Solids Deep Oil Frying Change of Weather

Unidirectional Unsteady State Case

x

y

z

x x+x

qin qout

A = yz

V = xyz

Solid properties: , cP

Mass = xyz

Heat Balance Across x:

qin - qout = Rate of heat accumulation

Rate of heat accumulation = xyz Tt

Using Fourier’s Law:

qin = - k yzTx x

qout = - k yzTx x +x

cP


The Heat Balance becomes:

- k yzTx x

- - k yzTx x +x

= xyz cP

Tt

Simplifying:

Tt =

k cP

Tx x +x

- Tx x

x

From Calculus: Tx x +x

- Tx x

x=

2Tx2

Final Equation: Tt =

k cP

= but

2Tx2


Tt =

2Tx2Solution of:

a) Depends on solid geometryb) Requires PDE solution methods that results into

Fourier series solutions that are tedious to evaluatec) May be simplified by the use of charts or numerical methods

Use of Charts:

a) Gurnie-Lurie Charts – to determine point temperaturesb) Heisler Charts – to determine central temperaturesc) Average Temperature Chartd) Chart for Semiinfinite Solids

Geankoplis Charts Gurney-Lurie Charts

Fig. 5.3-5/340 for large flat plate Fig. 5.3-7/343 for long cylinder Fig. 5.3-9/345 for sphere

Heisler Chart Fig. 5.3-6/341 for large flat plate Fig. 5.3-8/344 for long cylinder Fig. 5.3-10/346 for sphere

Fig. 5.3-13/349 for Ave. Solid Temperature Fig. 5.3-3/337 for Semi-infinite solid

Nomenclature Gurney-Lurie and Heisler Charts:

To = temperature at t(time)= 0 (uniform) T1 = new and constant surface temperature x1 = ½ plate thickness, outer radius of cylinder or sphere = constant thermal diffusivity X = t/ x1

2 : relative time x = distance from plate center or any radius of a

cylinder or a sphere n = x/x1: relative position T = point temperature at position x and time t Y = (T1-T)/(T1-To) :unaccomplished temp. change h = convective heat transfer coefficient m = k/(hx1) : relative resistance

Nomenclature Average Temperature Chart

To = temperature at t(time)= 0 (uniform) T1 = new and constant surface temperature Tav = average solid temperature at time t E = (T1-Tav)/(T1-To) a = ½ plate thickness, outer radius of

cylinder or sphere b = ½ plate width c = ½ plate length, ½ cylinder length

Nomenclature Chart for Semi-infinite solid

Semi-infinite solid – solid where the unidirectional conductive heat transfer is infinite (Ex. Ground)

To = initial uniform solid temperature T1 = constant ambient temperature to which solid

surface is exposed T = temperature of solid at position x 1- Y = (T-To)/(T1-To): Ordinate h(t)0.5/k : convective parameter x/[2 (t)0.5]: Abscissa

T T 0

T 1 T 01 Y erfc

x

2 t

exph tk

x

t

h tk

erfc

x

2 t

h tk

Problems from Geankoplis Exercises:

5.3-5 (Plate) 5.3-7 (Long Cylinder) 5.3-9 (Sphere) Find the average solid temperature for all of

the above cases 5.3-3

Homework: Geankoplis: 5.3-2;5.3-4;5.3-6;5.3-8;5.3-10

(pages 375-377) Foust: 11.18;11.20;11.21(page 231)

5.3-5 Cooling a Slab of Meat

A slab of meat 25.4 mm thick originally at a uniform temperature of 10oC is to be cooked from both sides until the center reaches 121oC in an oven at 177oC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h.

Solution for 5.3-5Given:

t= 25.4 mmTo=10oC

T=121oC (Center T) at x1

T1=177oC

h=25.6 W/m2-Kk=0.69 W/m-K=5.85x10-4m2/h

Required: t= ?

Solution for 5.3-5x1=12.7mm x=25.4m

m

x

tX

21

TT

TTY

o1

1

10177

121177

3350Y .

hx

km

1

01270625

690

..

.

1222m .

1000712

t10x85523

2

4

.

..

h8820t .

5.3-7 Cooling of a Steel Rod

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/m2-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 =0.0381m2/h


D= 0.305 mTo=588 K

T1=311 K

h=125 W/m2-Kt=1 hk=38 W/m-K=0.0381 m2/h

Required: T at the center

Solution for 5.3-7

x

tX

21

TT

TTY

o1

1

K391T

hx

km

1

01270625

690

..

. 1222.

2m3050

h1h

m03810

2

2

.

.

x

xn

1

203810

0

. 0

63831.

588311

T311290

.

5.3-9 Temp. of Oranges on Trees During Freezing Weather

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 21.1oC, calculate the center temperature of the orange if exposed to air at –3.9oC for 6 h. The oranges are 102 mm in diameter and the convective coefficient is estimated as 11.4W/m2-K. The thermal conductivity k is 0.431 W/m-K and =4.65x10-

4m2/h. Neglect any latent heat effects.


D= 102 m x=102/2=51mmTo=21.1oC=294.1K

T1=-3.9oC=269.1K

h=11.4 W/m2-Kt=6 hk=0.431 W/m-K=4.65x10-4 m2/h

Required: T at the center

Solution for 5.3-9

x

tX

21

TT

TTY

o1

1

K35270T .

hx

km

1

m

100051

Km

W411

mKW

4310

2

.

. 740.

100051

h6h

m10x654

2

24

.

071.

12941269

T1269050

..

..

From Fig. 5.3-10:Y = 0.05

C652T o.

5.3-3 Cooling a Slab of Aluminum

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/m2-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 =0.340m2/h and k=208W/m-K.


D= 0.305 mTo=505.4 K

T1=338.8 K

T=388.8K when x=25.4mmh=455 W/m2-Kk=208 W/m-K=0.304 m2/h

Required: time in hours for the temperature to reach 388.8K at a depth of 25.4 mm

Solution for 5.3-9

t2

xX

TT

TT

o1

o

K35270T .

k

th

mKW

208

ts

m3400

Km

W455

2

2

.

740.

70.

12941269

T1269050

..

..

From Fig. 5.3-10:Y = 0.05

C652T o.

45058338

45058388

..

..

t34002

m1000

425

X.

.

Unsteady State Basics

Education

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