Heat and Mass Transfer

이 윤 우

서울대학교 화학생물공학부

Heat and Mass Transfer

18

STEADY STATE

HEAT CONDUCTION

Supercritical Fluid Process Lab

Overview

Steady-State Heat Conduction in a (1) Flat Wall

(2) Multilayer Flat Wall (3) Hollow Cylinder

(4) Multilayer Cylinder

(5) Thermal Contact Resistance

Heat Conduction in a Flat Wall

Δx 0

Distance, x

t

₂

t

₁

t

₂

dx kA dt q = −

t

₁

Temperature

q q

Δx

fIf heat is flowing normal to the principal surfaces, the area term A is constant.

fIf k is assumed to be constant, q at any cross section is proportional to dt/dx.

fIf energy is neither generated nor accumulated in the wall, q is identical at all cross sections and so is dt/dx.

Fourier conduction equation

Supercritical Fluid Process Lab

Heat Conduction in a Flat Wall

t

₂

dx kA dt q = −

t

₁

q q

Δx

fIf k varies with temp, dt/dx ≠ constant.

fIf q and A = constant, and k increases with decreasing temp(Al₂O₃), the temperature gradient must diminish in the direction of decreasing temperature. Thus the curve representing temperature in steady-state flow for this system is concave upward.

Fourier conduction equation

Δx t

₁

t

₂

0

Distance, x

Temperature

t

₁

t

₂

k increasing with increasing T

ex: SUS304, Fused quartz, mercury (l), H₂(g), CO₂(g), air(g), steam

k decreasing with increasing T

ex: copper, tungsten, Al₂O3, NH₃(l)

k independent of T

x

₁

x

₂

Influence of the temperature dependence of k on dt/dx

in a flat wall during heat transport by conduction

Supercritical Fluid Process Lab

t

₁

t

₂

k increasing with increasing T

ex: SUS304, Fused quartz, mercury (l), H2(g), CO₂(g), air(g), steam

k decreasing with increasing T

ex: copper, tungsten, Al₂O3, NH₃(l),

k independent of T

ex: Engine oil

x

₁

x

₂

Influence of the temperature dependence of k on dt/dx

in a flat wall during heat transport by conduction

dx kA dt q = −

Fourier conduction equation

Temperature Profile in a flat wall

(18-1)

The integration of Eq (18-1) is readily performed when q, k, and A are constant, And this gives

x t t

q kA

Δ

= (

₁

−

₂

)

(18-2) The equation can also be integrated and solved for the temperature at any point x.

t

1

kA x

t = − q +

^(18-3)

Supercritical Fluid Process Lab

Conduction in a multi-layer flat wall

c c

b b

a

a x

t At

x k t At

x k t A t

k

q Δ

= − Δ

= − Δ

= ¹ − ^{2 2 3 3 4} (18-4)

Distance, x Δx_a

t

₁

Temperature

t

₂

(a) (b) (c)

k

_a

A k

x A

k x A

k x

t q t

c c b

b a

a + Δ + Δ

Δ

= ¹ − ⁴

(18-5)

A k q x t

t

A k q x t

t

A k q x t

t

c c b

b a

a

= Δ

−

= Δ

−

= Δ

−

4 3

3 2

2 1

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Δ + Δ + Δ

=

− k A

x A

k x A

k q x t

t

c c b

b a

a 4

1

+

t₁

t₂

t

₃

Δx_b

k

_b

t₃

t

₄

Δx_c

k

_c

t₄

Analogy to Ohm’s law for electric conduction

A k

x A

k x A

k x

t q t

c c b

b a

a + Δ + Δ

Δ

= ¹ − ⁴

Driving force (Temp Difference)

Thermal resistances Heat

flow

3 2

1

4 1

R R

R

V I V

+ +

= − Electric flow

(Current)

Electric resistances

Driving force (Voltage Difference)

t₁ t₂ t₃ t₄

Distance, x

R₁ R₂ R₃

(a) (b) (c)

Δx_a

Temperature

k

_a

Δx_b

k

_b

Δx_c

k

_c

t

₁

t

₂

t

₃

t

₄

Supercritical Fluid Process Lab

A k

x A

k x A

k x

t q t

c c b

b a

a

+ Δ + Δ

Δ

=

¹

−

⁴

Overall resistance Individual resistance

Overall Thermal resistances

Because q/A is the same for all layers, it

follows that k(Δt/Δx) is the same for all layers;

thus Δt/Δx is inversely proportional to the thermal conductivity (열전도도가 크면 온도구 배가 작다).

c c

b b

a

a

x

t k t

x t k t

x t k t

A q

Δ

= − Δ

= − Δ

=

¹

−

^{2 2 3 3 4}

t₁ t₂ t₃ t₄

Distance, x

R₁ R₂ R₃

(a) (b) (c)

Δx_a

Temperature

k

_a

Δx_b

k

_b

Δx_c

k

_c

t

₁

t

₂

t

₃

t

₄

Δt/Δx

Example 18-1

A cold-storage room has walls constructed of a 4-in layer of corkboard contained between double wooden walls, each 1/2 in thick. (1) Find the rate of heat loss in But/(h)(ft²) if the wall surface temperature is 10°F inside room and 70 °F outside the room. In addition, (2) find the temperature at the interface between the outer wall and the corkboard.

Although thermal conductivity is a function of temperature, it is often assumed to be constant at the arithmetic average temperature of the layer involved. The conductivities of many materials have not been measured over a temperature range; in addition, other factors such as the density (in the case of corkboard) and the presence of impurities (e.g., moisture) can have an effect on the conductivity. Data limitations such as these often have a direct effect on the accuracy of the solution and guide the engineer in determining the degree of simplification he can use in his calculations.

Supercritical Fluid Process Lab

Example 18-1

Thermal conductivity of corkboard ?

Cork board ρ=10 lb/ft³,k=0.024 Btu/(h)(ft²)(^oF/ft) @30^oC Thermal conductivity of wood ?

Wood Balsa (벽오동) ρ=7-8 lb/ft³, k=0.025 Btu/(h)(ft²)(^oF/ft) @30^oC Wood Oak (참나무) ρ=51.5 lb/ft³, k=0.12 Btu/(h)(ft²)(^oF/ft) @15^oC Wood Maple (단풍나무)ρ=44.7 lb/ft³, k=0.11 Btu/(h)(ft²)(^oF/ft) @50^oC Wood Pine (소나무) ρ=34.0 lb/ft³, k=0.087 Btu/(h)(ft²)(^oF/ft) @15^oC Wood Teak (티크나무) ρ=40.0 lb/ft³, k=0.10 Btu/(h)(ft²)(^oF/ft) @15^oC White fir (전나무) ρ=28.1 lb/ft³, k=0.062 Btu/(h)(ft²)(^oF/ft) @60^oC (전나무의 가격이 싸고 벽의 건축자재로 널리 사용됨)

Table A-14 Thermal conductivities of some building materials (Page 805)

Btu ft F h

Btu ft F h

) )(

)(

9( . ) 13 024 . 0 (

12 ) 1 4 ( k

board x cork

resistance Thermal

) )(

)(

67( . ) 0

062 . 0 (

12 1 2 1 k

layer x wood

resistance Thermal

2

b b

2

a a

o o

=

⎟⎠

⎜ ⎞

⎝

⎛ Δ =

=

=

⎟⎠

⎜ ⎞

⎝

⎟⎛

⎠

⎜ ⎞

⎝

⎛ Δ =

=

a a b

b a

a

k x k

x k

x

t A t

q Δ + Δ + Δ

=

¹

−

⁴

/

) )(

/(

9 . 67 3

. 0 9 . 13 67

. 0

10

/ A 70 Btu h ft²

q =

+ +

= −

(1) the rate of heat loss:

Cold storage

4“Cork board , k=0.024 Btu/(h)(ft)(F)

½” Wood layer, k=0.062 Btu/(h)(ft)(F) t₁ t₂ t₃ t₄

Δt/Δx

Supercritical Fluid Process Lab 4“Cork board , k=0.024 Btu/(h)(ft)(F)

½” Wood layer, k=0.062 Btu/(h)(ft)(F)

The temperature at the interface between the outer wooden wall and the corkboard can be determined by rearranging the equation for the individual temperature drops which led to Eq. (18-5).

t₁ t₂ t₃ t₄

) )(

/(

9 . 3

/ ^{1 4 1 2 2 3 3 4} Btu h ft²

k x

t t k

x t t k

x t t k

x k

x k

x

t A t

q

a a b

b a

a a

a b

b a

a

Δ =

= − Δ

= − Δ

= − + Δ

+ Δ Δ

= −

67 . 0

10 67

. 0 9 70 . 3

/ ^{1 4 2 3} −

− =

= Δ =

Δ + Δ +

= − t t

k x k

x k

x

t A t

q

a a b

b a

a

t₂=67.4^oF t₃=12.6^oF

70^oF

10^oF

Heat Conduction in the walls of a Hollow cylinder

rL A

dr kA dt q

π

= 2

−

=

L

t₂ r₂ r₁ t₁

The Fourier Equation in cylindrical Coordinates

The area normal to heat flow

r A A

t A k A

r r A A

t t A k A

r r r r

t t r r q Lk

r r

t t q Lk

dt r Lk

q dr ^t

t r

r

Δ Δ

= −

−

−

= −

−

−

= −

= −

−

=

∫

∫

) / ln(

) (

) )(

/ ln(

) )(

( )

)(

/ ln(

) )(

( 2

) / ln(

) (

2

2

1 2

1 2

1 2 1 2

2 1 1 2

1 2 1 2

2 1 1 2

1 2

2 1

2

1 2

1

π π

π (18-7)

(18-8)

(18-6)

Integration Eq (18-6) yields

(18-10)

Supercritical Fluid Process Lab

Heat Conduction in the walls of a Hollow cylinder

L

t₂ r₂ r₁

t₁

r

kA t r

A A

t A

k A

q

_lm

Δ

= Δ Δ Δ

= −

) / ln(

) (

1 2

1 2

1 2

2 1

1 2

1 2

) / ln(

) (

r r r

t t t

A A

A A_lm A

−

= Δ

−

= Δ

= − The log mean area

(18-10)

x t t q kA

Δ

= ( ₁ − ₂)

Similar to

in flat wall

In most engineering applications (e.g., pipe), r₂/r₁<<2. In these

circumstance the arithmetic mean area may be used in Eq. (18-10), with a consequent error in q less than 4%.

Heat Conduction in Multi-layer cylinder

Consider the case of three concentric hollow cylinders, e.g., a pipe with two layers of insulation around it. The thickness the three layers will be designated Δr_a, Δr_b, and Δr_c, and the temperature drops over the individual layer Δt_a, Δt_b, and Δt_c .

The total heat-transfer rate, which will be the same for all the cylinders, can be written

Δt_a

Δt_b

Δt_c

c lm

b lm

a

lm

r

kA t r

kA t r

kA t

q ⎟

⎠

⎜ ⎞

⎝

⎛

Δ

= Δ

⎟ ⎠

⎜ ⎞

⎝

⎛

Δ

= Δ

⎟ ⎠

⎜ ⎞

⎝

⎛

Δ

= Δ

(18-11)

pipe

Insulator 1 Insulator 2

Supercritical Fluid Process Lab

Heat Conduction in Multi-layer cylinder

The individual temperature drops may be found by rearranging (18-11).

Δt_a

Δt_b

Δt_c ^{c lm} _c

lm b b

lm a a

kA q r t

kA q r t

kA q r t

⎟⎟⎠

⎜⎜ ⎞

⎝

= ⎛ Δ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝

= ⎛ Δ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝

= ⎛ Δ Δ

(18-12)

Adding and rearranging these three equations gives

lm c lm b

lm a

overall

kA r kA

r kA

r q t

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Δ

= Δ

Individual resistance

+

A 6-in, schedule-80 steel pipe is covered with a 0.1-m layer of 85 percent magnesia insulation. The temperature of the inner surface of the pipe is 250 ° C, and the temperature of the outer surface of the insulation is 40 ° C. (1) Calculate the rate of heat loss per meter of pipe and (2) the temperature at the interface between the pipe and the insulation.

The thermal conductivity of the steel pipe can be taken as 44.8W/m·K (Perry, p. 3-220), and that of the 85 percent magnesia as 0.066 (Perry, p. 3-221). The OD of the pipe is 0.1683 m, and the ID is 0.1463 m.

Example 18-2

Supercritical Fluid Process Lab

r₂ r₁ t₂

t₁ =250^oC T₃=40^oC

r₃

85% magnesia k=0.066 W/mK Δr₃₂=r₃-r₂=0.1m

6” schedule 80 steel pipe k=44.8 W/mK

ID=5.761”=0.1463m OD=0.1683m

Δr₂₁=r₂-r₁=0.011m

W q

W kA K

R r

W kA K

R r

m A

m A

lm insulation

lm pipe

lm lm

89 111 . 1 0005 .

0

40 250

/ 89 . ) 1 80 . 0 )(

066 . 0 (

1 . 0

/ 00050 .

) 0 49 . 0 )(

80 . 44 (

011 . 0

80 . ) 0

1683 .

0 / 3683 .

0 ln(

) 1683 .

0 3683 .

0 (

49 . ) 0

1463 .

0 / 1683 .

0 ln(

) 1463 .

0 1683 .

0 (

32 21

2 2

+ =

= −

= Δ =

=

= Δ =

=

− =

=

− =

= π For the pipe π

For the insulation

For the pipe

For the insulation

Rate of heat loss:

lm c lm b

lm a

overall

kA r kA

r kA

r q t

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Δ

= Δ

q

0.8023 m²

1.8885 K/W

111.17 W

r₂ r₁ t₂

t₁ =250^oC T₃=40^oC

r₃

85% magnesia k=0.066 W/mK Δr₃₂=r₃-r₂=0.1m

6” schedule 80 steel pipe k=44.8 W/mK

ID=5.761”=0.1463m OD=0.1683m

Δr₂₁=r₂-r₁=0.011m

W

q 111

89 . 1 0005 .

0

40

250 =

+

= − Rate of heat loss:

Thermal resistance of the insulation is so large

c lm

b lm

a lm

lm c lm b

lm a

overall

r kA t r

kA t r

kA t kA

r kA

r kA

r

q t ⎟

⎠

⎜ ⎞

⎝

⎛

Δ

= Δ

⎟⎠

⎜ ⎞

⎝

⎛

Δ

= Δ

⎟⎠

⎜ ⎞

⎝

⎛

Δ

= Δ

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Δ

= Δ

Interface temperature (t₂) = = 249.94(210) ^oC 00050

. 0 89 . 1

00050 .

250 0 ⎟

⎠

⎜ ⎞

⎝

⎛

− +

Supercritical Fluid Process Lab

Heat Conduction to the walls of a Hollow Sphere

4 r2

A

dr kA dt q

π

=

−

= The Fourier Equation

in spherical Coordinates

The area normal to heat flow

r kA t

r r

r t r

t k q

t t r k

q r

dt r k

q dr

gm t

t r

r

Δ

= Δ

− −

=

−

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛ −

−

=

∫

∫

) ) (

( 4

) (

1 4 1

4

1 2

2 1 2

1

2 1 2

1 2

2

1 2

1

π

π π

(18-6)

Integration Eq (18-6) yields

(18-13)

(18-14)

(18-15, 16)

r₂ r₁ t₂

t₁

Heat Conduction to the walls of a Hollow Sphere

r kA t

r r

r t r

t k

q _gm

Δ

= Δ

− −

= 4 ( ) ( )

1 2

2 1 2

π

1

r₂ r₁ t₂

t₁

2 1

A A

A

_gm

=

: Geometric mean area

Integration of Eq. (18-13) to give the temperature t at any radial position r inside the wall of the sphere shows that

“the temperature is a linear function of 1/r.”

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛ −

+

=

2 2

1 1

4 k r r t q

t π

x t t q kA

Δ

= ( ₁ − ₂)

Similar to

in flat wall

Supercritical Fluid Process Lab

Heat Conduction of Three Concentric Spherical Layer

r₂ r₁ t₂

t₁ t₃

r₃

gm c gm b

gm a

overall

kA r kA

r kA

r q t

⎟ ⎟

⎠

⎞

⎜ ⎜

⎝ + ⎛ Δ

⎟ ⎟

⎠

⎞

⎜ ⎜

⎝ + ⎛ Δ

⎟ ⎟

⎠

⎞

⎜ ⎜

⎝

⎛ Δ

= Δ

Individual resistance

Heat Conduction

gm c gm b

gm a

overall

kA r kA

r kA

r q t

⎟⎟

⎠

⎞

⎜⎜

⎝ +⎛ Δ

⎟⎟

⎠

⎞

⎜⎜

⎝ +⎛ Δ

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ Δ

= Δ

lm c lm b

lm a

overall

kA r kA

r kA

r q t

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝ +⎛ Δ

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Δ

= Δ

c b

a

overall

kA x kA

x kA

x q t

⎟ ⎠

⎜ ⎞

⎝ + ⎛ Δ

⎟ ⎠

⎜ ⎞

⎝ + ⎛ Δ

⎟ ⎠

⎜ ⎞

⎝

⎛ Δ

= Δ Multi-layer

flat wall

Walls of a

Hollow Sphere Multi-layer cylinder

2 1

A A A

_gm

=

) / ln(

) (

1 2

1 2

A A

A A_lm = A −

2

2

1

A

A

_am

A +

=

Thermal Contact Resistance

(1) Solid to solid conduction

(2) Conduction/convection/radiation through entrapped gas in the void spaces

For solids whose thermal conductivities exceed that of the interfacial fluid, the contact resistance may be reduced by increasing the area of the contact spots. Such an increase may be effected by increasing the joint pressure and/or by reducing the roughness of the mating surfaces. The contact resistance may also reduced by selecting an interfacial fluid of large thermal conductivity. (e.g.; thermal grease such as silicon oil)

Supercritical Fluid Process Lab

Thermal Contact Resistance

Protuberance at the contact point (spot)

TC c

c b

b a

a R

A k

x A

k x A

k x

t q t

Δ + Δ +

Δ +

= ¹ − ⁴

Vacant at the contact point Convection + radiation

Thermal Contact Resistance

Pressure ^{102 kN/m2 104 kN/m2} R_TC 1~10 0.1~1

Two copper plates Δx_a

t

₁

Distance, x

Temperature

t

₂

t

₃

t

₄

(a) (b) (c)

Δx_b Δx_c

k

_a

k

_b

k

_c

Supercritical Fluid Process Lab

Thermal Contact Resistance

Contact pressure 10² kN/m² 10⁴ kN/m²

SUS 6~25 0.7~4.0

Thermal Contact Resistance R_TCx10⁴ (m²·K/W)

Copper 1~10 0.1~0.5

Magnesium 1.5~3.5 0.2~0.4 Aluminum 1.5~5.0 0.2~0.4

Al-Al interface (10 μm roughness) under 10⁵ kN/m²

Thermal Contact Resistance R_TCx10⁴ (m²·K/W) with different interfacial fluids

Air 2.75

He 1.05

H₂ 0.720

Silicone oil 0.525

Glycerine 0.265

Fried, E., “Thermal Conduction Contribution to Heat Transfer at Contacts”, in R.P. Tye, Ed., Thermal Conductivity, vol 2, Academic Press, London, 1969

Thermal Dissipation of the Chip

Epoxy joint ~0.02 mm R_TC=9x10^-5 m²K/W Silicon chip

8 mm Aluminum Substrate

k_aluminum=240 W/mK

10⁵ W/m²

t₂=25^oC

q

₂

q

₁

W

q 10 8 . 11

240 10 10 8

9

25

35

₄

3 5

2

× =

+ ×

×

= −

₋ ⁻

−

R_TC

k_aluminum Cross-sectional area A=10 cm²

t₁=35^oC

Supercritical Fluid Process Lab

A heating element is constructed from carbon in the shape of a bar 3” wide, ½” thick, and 3 ft long. When a potential of 12 V is applied to the ends of the bar, its surface reaches a uniform temperature of 1400 ° F, as indicated by an optical pyrometer.

What is the temperature at the center of the bar? The electrical resistivity of the bar is 1.30*10

^-4

( Ω )(ft), and its thermal conductivity is 2.9 Btu/(h)(ft)( ° F).

Only heat conduction normal to the largest forces of the bar will be considered, since heat leaves the bar principally through these faces. A differential equation is obtained by writing an energy balance on a differential segment dx of the bar, as shown in Fig. 18-4.

Example 18-3

3”

1/2”

3’

dx x

12V

q_in q_out element

x dx

Energy Balance on the Element

@steady state

Heat flow in - heat flow out + rate of heat generation = 0

( )

75 2

. 0

12 3 3

ft A

=

⎟⎠

⎜ ⎞

⎝

= ⎛

( )

_{= Ω}

⎟⎠

⎜ ⎞

⎝

⎟⎛

⎠

⎜ ⎞

⎝

⎛

⋅ Ω

= × ⁻ 0.0375

12 3 12

2 / 1

) 3 ( 10

30 . Resistance 1

4

ft ft

ft ft

) )(

/(

419000 ) 3830

12 (

q q bar entire in

generation heat

of Rate

3 2

2

out in

ft h Btu

V = = W =

=

=

=

Supercritical Fluid Process Lab

Heat flow into element =

h Btu dx dx

t d dx

kA dt

dx d dt dx

kA dt

h dx Btu

kA dt

/ /

2 2

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ +

−

⎟⎠

⎜ ⎞

⎝

⎛ +

−

− ⁽¹⁾

(2) (3) Heat flow out of element =

@steady state: Heat flow in - heat flow out + rate of heat generation = 0

Energy Balance on the Element

Rate of heat generation in the element = 419,000 Adx Btu/h

k dx

t d

Adx dx dx

t d dx

kA dt dx

kA dt

000 , 419

0 000

, 419

2 2

2 2

−

=

=

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ +

+

Energy Balance: −

which reduces to

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛ ⎟

⎠

⎜ ⎞

⎝

⎛

∂ + ∂

− dx

dx dt x

dx

kA dt

2 2

2

/ 000 , 9 145

. 2

000 , 419 000

,

419 F ft

k dx

t

d = − = − = − _o

Integration with BCs: dt/dx=0 @x=0 (the bar is symmetrical)

000

1

,

145 x C

dx

dt = − +

0

Second Integration with BCs: t=1400^oF @x=1/48 ft

1431 500

, 72

1431 )

48 / 1 )(

2 / 000 ,

145 ( 1400

2 000 ,

145

2

2 2

2 2

+

−

=

= +

=

− +

=

x t

C

x C t

1400^oF 1431^oF

0 1/48 ft

x

Supercritical Fluid Process Lab

Homework #1 PROBLEMS

18-1; 18-4; 18-5; 18-7; 18-8

Due date: Before quiz on September 21

Quiz: September 21 13:00-14:15 (302-509)

Closed book, Calculator

dydz u

dydz u

d

u

_x

ρ + (

_x

ρ )] −

_x

ρ

[

(7-1)

@steady state: Mass in - Mass out = Accumulate

Differential element for mass balance

dy

dx

dz )

, , (x y z

x

z

y

dydz u

d (

_x

ρ )

^(7-2)

x dx u u

d

_x ^x

∂ ρ

= ∂

ρ ( )

)

(

(7-3)

FIGURE 7-1

Differential element of mass balance

The second term is the input in the x-direction through the face of area dydz located at a distance x from the plane x=0; the first term is the output through the parallel face located at x+dx. The express (7-1) reduces to

It will be convenient to express the differential change in u_xρ over the distance dx by

As a result the output minus input in the x-direction is In the x direction, output less input by mass flow is given by

dxdydz x

u

_x

∂ ρ

∂ ( )

(7-3)

Supercritical Fluid Process Lab

Differential element for mass balance

(7-5)

(7-6)

(7-7)

Similar expression can be written for flow in the y and z directions.

The rate of accumulation in the element is

) 0 ) (

) (

( =

θ

∂ ρ + ∂

∂ ρ + ∂

∂ ρ + ∂

∂ ρ

∂ dzdxdy dxdydz

x dydxdz u

x dxdydz u

x

u

x y _z

(7-8)

= 0 θ

∂ ρ

∂ dxdydz

) 0 ) (

) (

( =

θ

∂ ρ + ∂

∂ ρ + ∂

∂ ρ + ∂

∂ ρ

∂

x u x

u x

u

x y _z

= 0 θ

∂ ρ + ∂

∂ ρ + ∂

∂ ρ + ∂

∂ ρ + ∂

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂ ρ ∂

u x u x

u x x

u x

u x

u

z y

x y z

x