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Lesson

Review of

fundamentals: Heat

and Mass transfer

The objective of this lesson is to review fundamentals of heat

and mass transfer and discuss:

1. Conduction heat transfer with governing equations for heat conduction, concept of thermal conductivity with typical values, introduce the concept of heat transfer resistance to conduction
2. Radiation heat transfer and present Planck’s law, Stefan-Boltzmann equation, expression for radiative exchange between surfaces and the concept of radiative heat transfer resistance
3. Convection heat transfer, concept of hydrodynamic and thermal boundary layers, Newton’s law of cooling, convective heat transfer coefficient with typical values, correlations for heat transfer in forced convection, free convection and phase change, introduce various non-dimensional numbers
4. Basics of mass transfer – Fick’s law and convective mass transfer
5. Analogy between heat, momentum and mass transfer
6. Multi-mode heat transfer, multi-layered walls, heat transfer networks, overall heat transfer coefficients
7. Fundamentals of heat exchangers

At the end of the lesson the student should be able to:

1. Write basic equations for heat conduction and derive equations for simpler cases
2. Write basic equations for radiation heat transfer, estimate radiative exchange between surfaces
3. Write convection heat transfer equations, indicate typical convective heat transfer coefficients. Use correlations for estimating heat transfer in forced convection, free convection and phase change
4. Express conductive, convective and radiative heat transfer rates in terms of potential and resistance.
5. Write Fick’s law and convective mass transfer equation
6. State analogy between heat, momentum and mass transfer
7. Evaluate heat transfer during multi-mode heat transfer, through multi-layered walls etc. using heat transfer networks and the concept of overall heat transfer coefficient
8. Perform basic calculation on heat exchangers

7.1. Introduction

Heat transfer is defined as energy-in-transit due to temperature difference. Heat transfer takes place whenever there is a temperature gradient within a system or whenever two systems at different temperatures are brought into thermal contact. Heat, which is energy-in-transit cannot be measured or observed directly, but the effects produced by it can be observed and measured. Since heat transfer involves transfer and/or conversion of energy, all heat transfer processes must obey the first and second laws of thermodynamics. However unlike thermodynamics, heat transfer

thermal conductivity materials are required for insulating refrigerant pipelines, refrigerated cabinets, building walls etc.

Table 7.1. Thermal conductivity values for various materials at 300 K

Material Thermal conductivity (W/m K) Copper (pure) 399 Gold (pure) 317 Aluminum (pure) 237 Iron (pure) 80. Carbon steel (1 %) 43 Stainless Steel (18/8) 15. Glass 0. Plastics 0.2 – 0. Wood (shredded/cemented) 0. Cork 0. Water (liquid) 0. Ethylene glycol (liquid) 0. Hydrogen (gas) 0. Benzene (liquid) 0. Air 0.

General heat conduction equation:

Fourier’s law of heat conduction shows that to estimate the heat transfer through a given medium of known thermal conductivity and cross-sectional area, one needs the spatial variation of temperature. In addition the temperature at any point in the medium may vary with time also. The spatial and temporal variations are obtained by solving the heat conduction equation. The heat conduction equation is obtained by applying first law of thermodynamics and Fourier’s law to an elemental control volume of the conducting medium. In rectangular coordinates, the general heat conduction equation for a conducting media with constant thermo-physical properties is given by:

k

q z

T

y

T

x

T

τ

1 T g 2

2 2

2 2

2 ⎥^ + ⎦

α

In the above equation, c p

k ρ

α = is a property of the media and is called as thermal

diffusivity, qg is the rate of heat generation per unit volume inside the control volume and τ is the time.

The general heat conduction equation given above can be written in a compact form using the Laplacian operator, ∇^2 as:

k

q T τ

1 T 2 g =∇ + ∂

α

If there is no heat generation inside the control volume, then the conduction equation becomes:

T τ

1 T 2

α

If the heat transfer is steady and temperature does not vary with time, then the equation becomes:

∇ 2 T^ = 0 (7.5)

The above equation is known as Laplace equation.

The solution of heat conduction equation along with suitable initial and boundary conditions gives temperature as a function of space and time, from which the temperature gradient and heat transfer rate can be obtained. For example for a simple case of one-dimensional, steady heat conduction with no heat generation (Fig. 7.1), the governing equation is given by:

dx

d T 2

2 = (7.6)

The solution to the above equation with the specified boundary conditions is given by:

L

x T = T 1 + ( T (^) 2 −T 1 ) (7.7)

Tx=0 = T 1 Tx=L = T 2

q x q x

Fig. 7.1. Steady 1-D heat conduction

x

and the heat transfer rate, Qx is given by:

cond

1 2 x R

T

L

T T

kA x

T

Q kA d

d (7.8)

where ΔT = T 1 -T 2 and resistance to conduction heat transfer, Rcond = (L/kA)

Similarly for one-dimensional, steady heat conduction heat transfer through a cylindrical wall the temperature profile and heat transfer rate are given by:

ε = Emissivity of the surface σ = Stefan-Boltzmann’s constant, 5.669 X 10-8^ W/m^2 .K^4 A = Surface area, m^2 Ts = Surface Temperature, K

The emissivity is a property of the radiating surface and is defined as the emissive power (energy radiated by the body per unit area per unit time over all the wavelengths) of the surface to that of an ideal radiating surface. The ideal radiator is called as a “black body”, whose emissivity is 1. A black body is a hypothetical body that absorbs all the incident (all wave lengths) radiation. The term ‘black’ has nothing to do with black colour. A white coloured body can also absorb infrared radiation as much as a black coloured surface. A hollow enclosure with a small hole is an approximation to black body. Any radiation that enters through the hole is absorbed by multiple reflections within the cavity. The hole being small very small quantity of it escapes through the hole.

The radiation heat exchange between any two surfaces 1 and 2 at different temperatures T 1 and T 2 is given by: 4 4 Q1-2 =σ.A.F F (T -T )ε A 1 2 (7.14)

where Q1-2 = Radiation heat transfer between 1 and 2, W Fε = Surface optical property factor FA = Geometric shape factor T 1 ,T 2 = Surface temperatures of 1 and 2, K

Calculation of radiation heat transfer with known surface temperatures involves evaluation of factors Fε and FA.

Analogous to Ohm’s law for conduction, one can introduce the concept of thermal resistance in radiation heat transfer problem by linearizing the above equation:

rad

1 2 (^1 2) R

(T T )

Q

where the radiative heat transfer resistance Rrad is given by:

1 2 rad (^4 ) ε A 1 2

T -T

R =

σAF F (T -T )

7.2.3. Convection Heat Transfer:

Convection heat transfer takes place between a surface and a moving fluid, when they are at different temperatures. In a strict sense, convection is not a basic mode of heat transfer as the heat transfer from the surface to the fluid consists of two mechanisms operating simultaneously. The first one is energy transfer due to molecular motion (conduction) through a fluid layer adjacent to the surface, which remains stationary with respect to the solid surface due to no-slip condition. Superimposed upon this conductive mode is energy transfer by the macroscopic motion of fluid particles by virtue of an external force, which could be generated by a pump or fan (forced convection) or generated due to buoyancy, caused by density gradients.

When fluid flows over a surface, its velocity and temperature adjacent to the surface are same as that of the surface due to the no-slip condition. The velocity and temperature far away from the surface may remain unaffected. The region in which the velocity and temperature vary from that of the surface to that of the free stream are called as hydrodynamic and thermal boundary layers, respectively. Figure 7.2 show that fluid with free stream velocity U∞ flows over a flat plate. In the vicinity of the surface as shown in Figure 7.2, the velocity tends to vary from zero (when the surface is stationary) to its free stream value U∞. This happens in a narrow region whose thickness is of the order of ReL-0.5^ (ReL = U∞L/ν) where there is a sharp velocity gradient. This narrow region is called hydrodynamic boundary layer. In the hydrodynamic boundary layer region the inertial terms are of same order magnitude as the viscous terms. Similarly to the velocity gradient, there is a sharp temperature gradient in this vicinity of the surface if the temperature of the surface of the plate is different from that of the flow stream. This region is called thermal boundary layer, δt whose thickness is of the order of (ReLPr)-0.5, where Pr is the Prandtl number, given by:

f

f f

p,f f k

c Pr α

ν

μ = (7.17)

In the expression for Prandtl number, all the properties refer to the flowing fluid.

Fig. 7.2. Velocity distribution of flow over a flat plate

In the thermal boundary layer region, the conduction terms are of same order of magnitude as the convection terms.

The momentum transfer is related to kinematic viscosity ν while the diffusion of heat is related to thermal diffusivity α hence the ratio of thermal boundary layer to viscous boundary layer is related to the ratio ν/α, Prandtl number. From the expressions for boundary layer thickness it can be seen that the ratio of thermal boundary layer thickness to the viscous boundary layer thickness depends upon Prandtl number. For large Prandtl numbers δt < δ and for small Prandtl numbers, δt > δ. It can also be seen that as the Reynolds number increases, the boundary layers become narrow, the temperature gradient becomes large and the heat transfer rate increases.

the problem. The convective heat transfer coefficient can vary widely depending upon the type of fluid and flow field and temperature difference. Table 7.2 shows typical order-of-magnitude values of convective heat transfer coefficients for different conditions.

Convective heat transfer resistance:

Similar to conduction and radiation, convective heat transfer rate can be written in terms of a potential and resistance, i.e.,

conv

w c w R

(T T )

Q h A(T T∞) ∞

where the convective heat transfer resistance, Rconv = 1/(hcA)

Determination of convective heat transfer coefficient:

Evaluation of convective heat transfer coefficient is difficult as the physical phenomenon is quite complex. Analytically, it can be determined by solving the mass, momentum and energy equations. However, analytical solutions are available only for very simple situations, hence most of the convection heat transfer data is obtained through careful experiments, and the equations suggested for convective heat transfer coefficients are mostly empirical. Since the equations are of empirical nature, each equation is applicable to specific cases. Generalization has been made possible to some extent by using several non-dimensional numbers such as Reynolds number, Prandtl number, Nusselt number, Grashoff number, Rayleigh number etc. Some of the most important and commonly used correlations are given below:

Heat transfer coefficient inside tubes, ducts etc.:

When a fluid flows through a conduit such as a tube, the fluid flow and heat transfer characteristics at the entrance region will be different from the rest of the tube. Flow in the entrance region is called as developing flow as the boundary layers form and develop in this region. The length of the entrance region depends upon the type of flow, type of surface, type of fluid etc. The region beyond this entrance region is known as fully developed region as the boundary layers fill the entire conduit and the velocity and temperature profiles remains essentially unchanged. In general, the entrance effects are important only in short tubes and ducts. Correlations are available in literature for both entrance as well as fully developed regions. In most of the practical applications the flow will be generally fully developed as the lengths used are large. The following are some important correlations applicable to fully developed flows:

a) Fully developed laminar flow inside tubes (internal diameter D):

Constant wall temperature condition:

3. 66

h D Nusselt number,NuD c ⎟⎟= ⎠

k (^) f

Constant wall heat flux condition:

4. 364

h D Nusselt number,NuD c ⎟⎟= ⎠

k (^) f

b) Fully developed turbulent flow inside tubes (internal diameter D):

Dittus-Boelter Equation:

1. (^8) n D

c D 0.^023 Re Pr

h D Nusselt number,Nu ⎟⎟= ⎠

k (^) f

where n = 0.4 for heating (Tw > Tf) and n = 0.3 for cooling (Tw < Tf).

The Dittus-Boelter equation is valid for smooth tubes of length L, with 0.7 < Pr < 160, ReD > 10000 and (L/D) > 60.

Petukhov equation: This equation is more accurate than Dittus-Boelter and is applicable to rough tubes also. It is given by:

1 / 2 2 / 3

n

w

D b D

f whereX 1. 07 12. 7 (Pr 1 )

f X

Re Pr Nu

μ

μ ⎟ ⎠

where n = 0.11 for heating with uniform wall temperature n = 0.25 for cooling with uniform wall temperature, and n = 0 for uniform wall heat flux or for gases

‘f’ in Petukhov equation is the friction factor, which needs to be obtained using suitable correlations for smooth or rough tubes. μb and μw are the dynamic viscosities of the fluid evaluated at bulk fluid temperature and wall temperatures respectively. Petukhov equation is valid for the following conditions:

104 < ReD < 5 X 10^6 0.5 < Pr < 200 with 5 percent error 0.5 < Pr < 2000 with 10 percent error

0.08 < (μb/μw) < 40

c) Laminar flow over a horizontal, flat plate (Rex < 5 X 10^5 ):

Constant wall temperature:

1. (^51) / 3 x

c x 0.^332 Re Pr

h x Local Nusseltnumber,Nu ⎟⎟= ⎠

k (^) f

The values of c and n are given in Table 7.3 for different orientations and flow regimes.

Table 7.3 Values of c and n

Orientation of plate Range of GrLPr c n Flow regime Hot surface facing up or cold 105 to 2 X 10^7 0.54 1/4 Laminar surface facing down, constant Tw 2 X 10^7 to 3 X 10^10 0.14 1/3 Turbulent Hot surface facing down or cold surface facing up, constant Tw

3 X 10^5 to 3 X 10^10 0.27 1/4 Laminar

Hot surface facing up, constant < 2 X 10^8 0.13 1/ qw 5 X 10^8 to 10^11 0.16 1/ Hot surface facing down, constant qw

106 to 10^11 0.58 1/

In the above free convection equations, the fluid properties have to be evaluated at a mean temperature defined as Tm = Tw−0.25(Tw-T∞).

g) Convection heat transfer with phase change:

Filmwise condensation over horizontal tubes of outer diameter Do:

The heat transfer coefficient for film-wise condensation is given by Nusselt’s theory that assumes the vapour to be still and at saturation temperature. The mean condensation heat transfer coefficient, hm is given by: 1 / 4

o f

fg

2 f

3 f m (^) ND μ ΔT

k ρ gh h 0. 725 ⎥

where, subscript f refers to saturated liquid state, N refers to number of tubes above each other in a column and ΔT = Tr – Two , Tr and Two being refrigerant and outside wall temperatures respectively.

Filmwise condensation over a vertical plate of length L:

The mean condensation heat transfer coefficient, hm is given by, 1 / 4

f

fg

2 f

3 f m (^) μ L T

k ρ g h h 0. 943 ⎥

Nucleate pool boiling of refrigerants inside a shell:

2 to 3 h (^) r = CΔT (7.35)

where ΔT is the temperature difference between surface and boiling fluid and C is a constant that depends on the nature of refrigerant etc.

The correlations for convective heat transfer coefficients given above are only few examples of some of the common situations. A large number of correlations are available for almost all commonly encountered convection problems. The reader should refer to standard text books on heat transfer for further details.

7.3. Fundamentals of Mass transfer

When a system contains two or more components whose concentration vary from point to point, there is a natural tendency for mass to be transferred, minimizing the concentration differences within the system. The transport of one constituent from a region of higher concentration to that of lower concentration is called mass transfer. A common example of mass transfer is drying of a wet surface exposed to unsaturated air. Refrigeration and air conditioning deal with processes that involve mass transfer. Some basic laws of mass transfer relevant to refrigeration and air conditioning are discussed below.

7.3.1. Fick’s Law of Diffusion:

This law deals with transfer of mass within a medium due to difference in concentration between various parts of it. This is very similar to Fourier’s law of heat conduction as the mass transport is also by molecular diffusion processes. According to this law, rate of diffusion of component A (kg/s) is proportional to the

concentration gradient and the area of mass transfer, i.e.

m A

x

c m (^) A DABA A d

d  (^) = − (7.36)

where, DAB is called diffusion coefficient for component A through component B, and it has the units of m^2 /s just like those of thermal diffusivity α and the kinematic viscosity of fluid ν for momentum transfer.

7.3.2. Convective mass transfer:

Mass transfer due to convection involves transfer of mass between a moving fluid and a surface or between two relatively immiscible moving fluids. Similar to convective heat transfer, this mode of mass transfer depends on the transport properties as well as the dynamic characteristics of the flow field. Similar to Newton’s law for convective heat transfer, he convective mass transfer equation can be written as:

m^ =hm AΔc A (7.37)

where hm is the convective mass transfer coefficient and ΔcA is the difference between the boundary surface concentration and the average concentration of fluid stream of the diffusing species A.

Similar to convective heat transfer, convective mass transfer coefficient depends on the type of flow, i.e., laminar or turbulent and forced or free. In general the mass transfer coefficient is a function of the system geometry, fluid and flow properties and

To account for values of Schmidt number different from one, following correlation is introduced,

Re 2

Sc^2 3 f .Sc

Sh (^) / = (7.43)

Comparing the equations relating heat and momentum transfer with heat and mass transfer, it can be shown that, 2 / 3

p m

c c h D

h ⎟ ⎠

⎛ α ⎟⎟= ⎠

ρ

This analogy is followed in most of the chemical engineering literature and α/D is referred to as Lewis number. In air-conditioning calculations, for convenience Lewis number is defined as: 2 / 3

D

Lewis number,Le ⎟ ⎠

⎛ α = (7.45)

The above analogies are very useful as by applying them it is possible to find heat transfer coefficient if friction factor is known and mass transfer coefficient can be calculated from the knowledge of heat transfer coefficient.

7.5. Multimode heat transfer

In most of the practical heat transfer problems heat transfer occurs due to more than one mechanism. Using the concept of thermal resistance developed earlier, it is possible to analyze steady state, multimode heat transfer problems in a simple manner, similar to electrical networks. An example of this is transfer of heat from outside to the interiors of an air conditioned space. Normally, the walls of the air conditioned rooms are made up of different layers having different heat transfer properties. Once again the concept of thermal resistance is useful in analyzing the heat transfer through multilayered walls. The example given below illustrates these principles.

Multimode heat transfer through a building wall: The schematic of a multimode heat transfer building wall is shown in Fig. 7.3. From the figure it can be seen that:

1 2 1- total

(T -T )

Q =

R

(7.46a)

( )

conv,2 rad,2 conv,1 rad, total w,3 w,2 w, conv,2 rad,2 conv,1 rad,

R R R R

R = + R +R +R +

R +R R +R

(7.46b)

R (^) total = (^) ( R (^2) ) + R( (^) w ) + R( 1 ) (7.46c)

Q 1-2 =UA(T -T ) 1 2 (7.46d)

total

where, overall heat transfer coefficient, U =

R A

Fig. 7.3. Schematic of a multimode heat transfer building wall

omposite cylinders:

he concept of resistance networks is also useful in solving problems involving

Room 2

T 2

Room 1

T 1

T 2

T 1

q rad q rad

q conv

q conv

T 1

T 2

Rconv,

Rconv,

Rrad,2 Rrad,

Rw,3 Rw,2 Rw,

R 2 Rw R 1

T 2

T 2

T 1

T 1

C

T

composite cylinders. A common example of this is steady state heat transfer through an insulated pipe with a fluid flowing inside. Since it is not possible to perfectly insulate the pipe, heat transfer takes place between the surroundings and the inner fluid when they are at different temperatures. For such cases the heat transfer rate is given by:

Q = Uo Ao(Ti−To ) (7.47)

If we assume that the overall heat transfer coefficient does not vary along the length, and specific heats of the fluids remain constant, then the heat transfer rate is given by:

ln( T / T )

T T

Q UA(LMTD) UA

also

ln( T / T )

T T

Q U A (LMTD) U A

1 2

1 2 i i i i

1 2

1 2 o o o o

(7.50)

the above equation is valid for both parallel flow (both the fluids flow in the same direction) or counterflow (fluids flow in opposite directions) type heat exchangers. For other types such as cross-flow, the equation is modified by including a multiplying factor. The design aspects of heat exchangers used in refrigeration and air conditioning will be discussed in later chapters.

Questions:

1. Obtain an analytical expression for temperature distribution for a plane wall having uniform surface temperatures of T 1 and T 2 at x 1 and x 2 respectively. It may be mentioned that the thermal conductivity k = k 0 (1+bT), where b is a constant. (Solution)
2. A cold storage room has walls made of 0.3 m of brick on outside followed by 0.1 m of plastic foam and a final layer of 5 cm of wood. The thermal conductivities of brick, foam and wood are 1, 0.02 and 0.2 W/mK respectively. The internal and external heat transfer coefficients are 40 and 20 W/m^2 K. The outside and inside temperatures are 400 C and -10^0 C. Determine the rate of cooling required to maintain the temperature of the room at -10^0 C and the temperature of the inside surface of the brick given that the total wall area is 100 m^2. (Solution)
3. A steel pipe of negligible thickness and having a diameter of 20 cm has hot air at 1000 C flowing through it. The pipe is covered with two layers of insulating materials each having a thickness of 10 cm and having thermal conductivities of 0.2 W/mK and 0.4 W/mK. The inside and outside heat transfer coefficients are 100 and 50 W/m^2 K respectively. The atmosphere is at 35^0 C. Calculate the rate of heat loss from a 100 m long pipe. (Solution)
4. Water flows inside a pipe having a diameter of 10 cm with a velocity of 1 m/s. the pipe is 5 m long. Calculate the heat transfer coefficient if the mean water temperature is at 40^0 C and the wall is isothermal at 80^0 C. (Solution)
5. A long rod having a diameter of 30 mm is to be heated from 400^0 C to 600^0 C. The material of the rod has a density of 8000 kg/m^3 and specific heat of 400 J/kgK. It is placed concentrically inside a long cylindrical furnace having an internal diameter of 150 mm. The inner side of the furnace is at a temperature of 1100^0 C and has an

emissivity of 0.7. If the surface of the rod has an emissivity of 0.5, find the time required to heat the rod. (Solution)

6. Air flows over a flat plate of length 0.3 m at a constant temperature. The velocity of air at a distance far off from the surface of the plate is 50 m/s. Calculate the average heat transfer coefficient from the surface considering separate laminar and turbulent sections and compare it with the result obtained by assuming fully turbulent flow. (Solution)

Note: The local Nusselt number for laminar and turbulent flows is given by: 1/2 (^) 1/ x x 0.8 (^) 1/ x x

laminar : Nu = 0.331Re Pr turbulent: Nu = 0.0288Re Pr

Transition occurs at. The forced convection boundary layer flow

begins as laminar and then becomes turbulent. Take the properties of air to be , , k = 0.03 W/mK and Pr = 0.7.

5 Rex.trans = 2 X 10

ρ = 1.1 kg/m 3 μ = 1.7 X 10-5 kg/m s

7. A vertical tube having a diameter of 80 mm and 1.5 m in length has a surface temperature of 80^0 C. Water flows inside the tube while saturated steam at 2 bar condenses outside. Calculate the heat transfer coefficient. (Solution)

Note: Properties of saturated steam at 2 bar: , ,

; For liquid phase at 100

0 Tsat = 120.2 C h (^) fg= 2202 kJ/kgK ρ = 1.129 kg/m^3 0 C: , , and Pr = 1.73.

3 ρL = 958 kg/m c (^) p= 4129 J/kgK

μ (^) L= 0.279X10 kg/m s

8. Air at 300 K and at atmospheric pressure flows at a mean velocity of 50 m/s over a flat plate 1 m long. Assuming the concentration of vapour in air to be negligible, calculate the mass transfer coefficient of water vapour from the plate into the air. The diffusion of water vapour into air is 0.5 X 10-4^ m^2 /s. The Colburn j-factor for heat transfer coefficient is given by jH=0.0296 Re -0.2. (Solution)
9. An oil cooler has to cool oil flowing at 20 kg/min from 100^0 C to 50^0 C. The specific heat of the oil is 2000 J/kg K. Water with similar flow rate at an ambient temperature of 35^0 C is used to cool the oil. Should we use a parallel flow or a counter flow heat exchanger? Calculate the surface area of the heat exchanger if the external heat transfer coefficient is 100 W/m^2 K. (Solution)