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Natural Convection Heat Transfer and Home Ventilation System

Task 1:

1. What do you know about convection heat transfer? What limits must be met in order for a

process of heat transfer by convection can be said to occur naturally?Natural convection heat transfer occurs due to temperature differences which affect the density

and thus relative buoyancy of the fluid. Heavier components will fall and less dense material will

rise, leading to bulk fluid movement.

Natural convection can only occurs in a gravitational field. Natural convection will be more rapid

with a greater variation in density between the two fluid, a larger acceleration due to gravity

that drives the convection, and or a larger distance through the convecting medium. Natural

convection will be less rapid with diffusion and results to a more sticky fluid. The beginning of a

natural convection can be determined by using Rayleigh's number (Ra).

The limitation on how a natural convection can occur is based on the density of the fluid.2. Explain what you know about the buoyancy force and the body force?

Magnitude of buoyancy force is the weight of the fluid displaced by the body.

=

The net force is:

= −

= −

Buoyancy is proportional to density difference, thus the larger the temperature differencebetween the fluid and the body, the larger the buoyancy force. The buoyancy force causing free-

convection currents are called body force. Body force is a force that acts throughout the volume

of the body. Gravity and electromagnetic forces are examples of body forces. Centrifugal force

may also be considered as body force.

3. How these forces can affect the movement of fluid on natural convection heat transfer?

The temperature close to the hot object is higher, thus density is lower. The heated air rises and

natural convection current occurs. In gravitational field, buoyance force causes light fluid placed

in heavier fluid to move upward. Buoyancy force in free-convection heat transfer results in the

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Prandtl number =

The general equation for average free-convection heat transfer is:

= ( ) (7.25)

where the constants C and m are specified for specific cases. f indicates that the properties is

evaluated at the film temperature.

=∞

+ 2

Rayleigh number: =

The characteristic dimension to be

used in Nusselt and Grashof numbers

are the height of the plate, L, for

vertical flat plate, and diameter, d, for

horizontal cylinder .

The general equation below may beused to calculate the local free-

convection heat transfer coefficient:

= 0.10( )1/3

For wider range of Rayleigh number,

the equations for average free-

convection heat transfer coefficient

are:

= 0.68 +0.6701/4

[1 + (0.492/Pr)9/16]4/9 < 109

1/2 = 0.852 +0.3871/6

1 + (0.492/Pr)9/168/27 10−1 <

< 1012

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Constant Heat flux Surfaces

In the case of constant heat flux surfaces, we use modified Garshof number:

∗=

=

4

2

Equation for local heat transfer coefficient for laminar range:

= = 0.60( ∗ )1/5 105 < ∗ < 1011 ; =

For the turbulent flow region, the local heat transfer coefficients:

= 0.17( ∗ Pr)1/4 2 × 1 013 < ∗ < 1016; =

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FREE CONVECTION FROM HORIZONTAL PLATES

Isothermal Surfaces

The average heat transfer coefficient for horizontal plate is also calculated using equation 7-25

with the constants listed in table 7-1. The characteristic dimension is the length of a side for a

square or mean of the two dimensions for rectangle surface. Alternatively, the characteristic

dimension may also be calculated by:

=

where A is the area and P is the perimeter of the surface. This equation is also applicable to

unsymmetrical platforms.

Constant Heat Flux

In the case of constant heat flux for horizontal plate with heated surface facing upward:

= 0.13(Pr)

1/3

< 2 × 1 0

8

and

= 0.16(Pr)1/3 2 × 108 < < 1011

For heated surface facing downward:

= 0.58(Pr)1/5 106 < < 1011

All properties are measured at temperature Te (except beta): = − 0.25( − ∞

)

Where T w is the average wall temperature related to heat flux by:

= −

∞

Nusselt number is:

=

=

−∞

Irregular Solid

There’s no general equation to be applied for irregular solid. For vertical cylinder with height

equal to diameter, equation 7-25 can be applied with C = 0.775 and m = 0.208. Nusselt and

Grashof are evaluated using diameter as characteristic length. For unlisted geometric shape,

Lienhard take the characteristic length as the distance a fluid travels in the boundary layer and

uses values of C = 0.52 and m = ¼ in equation 7-25 in the laminar range.

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FREE CONVECTION FOR SPHERES

The empirical equation for spheres free convection:

= = 2 + 0.392 1/4

1 < < 105

This equation may be modified to:

= 2 + 0.43( )1/4

As the Grashof-Prandtl number approaches 0, Nusselt number approaches 2. This is the value

for pure conduction through an infinite stagnant fluid surrounding the sphere.

For higher ranges Rayleigh number:

= 2 + 0.50 14 3 × 105 < < 8 × 1 08

For wider range of Rayleigh number:

= 2 +0.5891/4

[1 + 0.469Pr

9/16

]4/9

< 1011 Pr < 0.5

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Table 7-2

Empirical Equation for Horizontal Cylinder

For cylinders of sufficient length and negligible end effects, Churchill and Chu has the following

correlation for

Empirical Equation for Vertical Cylinder

For laminar flows in the range of , the following equation can be further improved.

For cylinders with their axes vertical, the expressions for plane surfaces can be used provided

the curvature effect is not too significant. This represents the limit where boundary layer

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thickness is small relative to cylinder diameter D. The correlations for vertical plane walls can be

used when

where GrL is the Grashof number.

Emperical Equation for Enclosed System

The free-convection flow pehenomena inside an enclosed space are examples of very complex

fluid system that yied to analytical, numerical and empirical solution.

In a vertical enclosure,the fluid adjacent to the hotter surface rises and

the fluid adjacent to the cooler one falls, setting off a rotationary motion

within the enclosure that enhances heat transfer through the enclosure.

The figure below shows where a fluid is contained between two vertical

plates separated by the distance δ.

As a temperature difference (delta) Tw = T1 − T2 is impressed on the fluid,a heat transfer will be experienced with the approximate flow regions

shown in Figure below, according to MacGregor and Emery.

In this figure, the Grashof

number is calculated by

At very low Grashof

numbers, there are very

minute free-convection

currents and the heat

transfer occurs mainly by

conduction across the

fluid layer. As the Grashof

number is increased,

different flow regimes are

encountered, as shown,

with a progressively increasing heat transfer as expressed through the Nusselt number

Heat transfer through a horizontal enclosed spaes involves two distinct situation. If the upper

part of the plate is maintained at at higher temperature that the lower pate, the lower-density

fluid is above the higher-density fluid, and there will be no convection current. In this case the

ehat transfer across the space will only be based on the conduction alone and Nu = 1.0, where

is still the separation distance between the plates .

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The second condition is when the lower plate has a higher tempeautre thn the upper plate. For

values of Gr below bout 1700, pure conduction is still observed ang Nu = 1.0. As convetion

starts we can see a pattern of hexagonal cell. The pattern are called Bernard Cell. Turbulence

begins at about Gr = 50000 and destroys the cellular pattern.

In enclosed vertical or horizontal cylinders enclosure can be calculated with

NuF = 0.55(GrFPrF)1/4

Where now the gap spacing is = ro-ri. The effective thermal conductivity given by the equation

a is to be used with the convecntional raltion for the steady state conduction for spherical shell.

a) = 0.228 ()0.226 b) =

4∆−

Equation a is valid for 0.25/ri1.5 and;

1.2x102GrPr1.1x10

9 0.7Pr4150

3. Explain the mechanics and heat transfer calculations involving the steady state conduction,

natural convection, and radiation mechanisms simultaneously.

For steady-state conduction, first we need the thermal conductivity of a material and the

convection coefficient. By using =∆∙ we can calculate the heat transfer through steady-state

conduction. For natural convection the equation depends on the shape of the system. By

calculating the Nusselt number, Prandtl number, and Rayleigh number we can find the heat

transfer coefficient through convection. By using

=

∆ we can calculate the heat transfer

through natural convection. For radiation, we can just use = 14 − 24 and substitute allthe values to their variables. The overall heat transfer can be calculated by adding all the q(s).

Problem Calculations

1. Temperature at surface of the vertical wall 4ft x 10ft, kept constant at 530oF, while the air

temperature 70oF and pressure 1 atm.

a. Calculate the heat lost from the surface by natural convection from wall to air.

10 ft = 3.048m

4 ft = 1.219 m

Tw = 530oF = 549.82K

T∞ = 70oF = 294.26 K

P = 1 atm = 101325 Pa

= = ∆

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calculating h for a vertical plate and isothermal surface

Film Temperature (Tf )

=∞

+ 2

=549.82 + 294.26

2= 422.04

β, =1

= 2.37 × 10−3

Kinematic viscosity (v)

Interpolated at reference 2 by using the film temperature

= 29 × 10−6

Prandtl Number (Pr)

Interpolated at reference 2 by using the film temperature

= 0.648

GrPr

= − ∞32

=9.8 × 2.37 × 10−3 × 255.56 × 3.053 × 0.64829×10−62

= 1.37 × 1011

1/2 = 0.825 +0.378 × 1.37 × 1011 × 0.6841/6

1 +

0.4920.648

916

8/27= 475.24

= ∙

=475.24 × 35.3 × 10−3

3.048= 5.50/2

Heat Transfer (q)

=

∆= 5.50 ×

3.048 × 1.219

× 255.56 = 5222.45

b. If the wall is insulated with 2 inches thick insulation material and thermal conductivity =

0.121 Btu/hr.ft2.oF, calculate the heat loss by conduction and natural convection when the

temperature at the surface of the insulator 250oF.

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=294.26 + 394.26

2= 344.26

All properties are measured at film temperature and by calculation and interpolation:

=1

344.26 = 2.90 × 10−3

= 2.034 × 10−5

= 0.701

= 29.58 × 10−3

=( −

∞)3

2 =

9.82.90 × 10−3394.26 − 294.263.0483

2.034 × 10−520.701

= 1.364 × 1011

Based on this value of Rayleigh number, the Nusselt equation is:

1/2 = 0.825 +0.3871/6

[1 + 0.492/Pr9/16]8/27= 0.825 +

0.387(1.364 × 1011)1/6

[1 + (0.492/0.701)9/16]8/27= 24.08

Thus,

= 579.85

Therefore, h is obtained by:

= =(579.85)(29.58 × 10−3)

3.048= 5.63 2. = 0.1533 . 2.℉

Heat loss by conduction and natural convection:

Since k = 0.121 Btu/hr.ft2.oF,

= −

∞∆

+ 1

=

(530℉− 70℉)

0.0833(0.121)(40)

+ 1

(0.1533)(40)

= 2541.83/

2. A gas flows through a pipe that has a surface temperature of 800oF planted in the ground. The

distance between the axis of the pipe to the ground 6 ft. outside diameter of pipe 2.38 in (pipe

thickness ignored), length of pipe = 50ft. Outside air temperature is 80oF, soil heat

conductivity = 1.4 Btu/hr.ft.oF.

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a. Calculate the heat lost from the surface of the pipe into the air.

If we assume that, soil temperature = air temperature, therefore, heat lost from natural

convection is:

= − ∞ = 80℉− 80℉ = 0 = 0

Therefore, heat lost by natural convection is 0. Therefore, the heat lost from surface of the pipe

to air become:

=( −

∞)∆x (1

2 ) =(800℉− 80℉)

5.900851.4(1

2 0.198350) = 2660.5 /

Where:

Twp = 800oF ∆x = 5.90085 ft

T∞ = 80oF d = 0.1983 ft

k = 1.4 Btu/hr.ft.oF L = 50 ft

b. If the pipe was placed in a room where the temperature is 80oF, calculate the heat lost by

natural convection and radiation from the pipe surface to air (ε pipe = 0.8).

=800℉ + 80℉

2= 440℉

All properties are calculated at Tf , therefore:

β = 1/Tf = 1/440 = 2.273 x 10-3

v =4.0786 x 10-4

Pr = 0.69566

k = 0.02284

= ( − ∞)32= 32.172.273 × 10−3800℉− 80℉0.198334.0786 × 10−42 0.69566

= 1,716,824.642

From table 7.1, for this value of Rayleigh number, the appropriate constants are, C = 0.53 and m

= ¼:

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= 0.531,716,824.64214 = 19.18

Therefore:

=

= 19.18 0.02284

0.1983 = 2.21

.

2.

℉

Then,

= − ∞ = −

∞ = 2.210.198350800℉− 80℉

= 49557.09 /

Radiation

= (14 − 2

4)

Where:

σ = 0.1714 x 10-8

Btu/hr.ft2.oR

4

A = πdL = π(0.1983)(50) = 31.149 ft

T1 = 800oF

T2 = 80oF

ε = 0.8

= 14 − 2

4 = 0.1714 × 10−831.1490.88004 − 804 = 17492.88 /

Therefore, the total heat lost by both radiation and natural convection is:

= + = 49557.09 + 17492.88 = 67049 /

c. Same with (b) but the pipe surface painted with paint from aluminum so ε pipe = 0.3.

=800℉ + 80℉

2= 440℉

All properties are calculated at Tf , therefore:

β = 1/Tf = 1/440 = 2.273 x 10-3

v =4.0786 x 10-4

Pr = 0.69566

k = 0.02284

=( −

∞)3

2=32.172.273 × 10−3800℉− 80℉0.19833

4.0786 × 10−420.69566

= 1,716,824.642

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From table 7.1, for this value of Rayleigh number, the appropriate constants are, C = 0.53 and m

= ¼:

=

0.53

1,716,824.642

14 = 19.18

Therefore:

= = 19.18 0.02284

0.1983 = 2.21 . 2.℉

Then,

= − ∞ = −

∞ = 2.210.198350800℉− 80℉

= 49,557.09/

Radiation

=

(

1

4

− 2

4)

Where:

σ = 0.1714 x 10-8 Btu/hr.ft2.oR4

A = πdL = π(0.1983)(50) = 31.149 ft

T1 = 800oF

T2 = 80oF

ε = 0.3

=

14

−24

=

0.1714 × 10−8

31.149

0.3

8004

−804

= 6,559.8

/

Therefore, the total heat lost by both radiation and natural convection is:

= + = 49,557.09 + 6559.8 = 56116.89/

d. If the pipes are insulated with 2 in. thick asbestos insulation (k = 0.04 Btu/hr.ft.oF) and outside

is isolated by 1 in. thick diatomic soil (k = 0.06 Btu/hr.ft.oF), calculate the heat lost to the air

per unit length of pipe.

Again, if we assume that, soil temperature = air temperature, therefore, heat lost from natural

convection is:

= 2 − ∞ = 80℉− 80℉ = 0 = 0

.

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Therefore, heat lost by natural convection is 0. Then, the heat lost from surface of the pipe to air

become: (by assuming that the pipe is very thin and therefore conduction heat transfer through

the pipe is negligeable)

=2( −

∞)

ln21

+

32

=2(800℉− 80℉)

0.26560.09915

0.04 +

0.34870.2656

0.06 = 159.8 .

If we assume that the insulators are in rectangular shape and only cover half of the circular

surface of the pipe:

=

12( −

∞)∆ + ∆ =

1/2(0.1983)(800℉− 80℉)

0.16640.04 + 0.08319

0.06 = 40.435 .

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Forced Convection Heat Transfer and Heat Exchanger

Task 1:

1. What do you know abut the heat exchanger and how the principle of Heat Exchanger?

General function of heat exchanger is to transfer heat from one fluid to another. There are 3

heat transfer operations that occur in a heat exchanger:

1. Convective heat transfer from fluid to the inner wall of the tube,

2. Conductive heat transfer to the tube wall,

3. Convective heat transfer from the outer tube wall to the outside fluid.

Heat exchangers are classified according to the flow arrangement

1. Parallel flow : the two fluids enter the exchanger from the same end

2.

Counter-flow : the fluids enter the exchanger at opposite end. This type of flow is mosteffective because there’s greater temperature difference along any unit length

3. Cross-flow : the fluids travel perpendicular to each other.

Heat exchangers are designed to maximize surface area of the wall between the two fluids and

minimizing the resistance of fluid flow. The performance is also affected by the addition of fins

which increase surface area and induce turbulence.

2. Describe the type of heat exchanger based on the complexity of the instrument.

Double pipe arrangement

Counterflow or parallel flow may be used in this type and either the hot or cold fluid occupies

the annular space or inner pipe. In its simplest form, this type of heat exchanger only consists of

pipes inside another larger pipe. The wall of the inner pipe is the heat transfer surface. One of

the most advantageous functions of double is that it can be operated in completely

countercurrent flow which is the most efficient mechanism to transfer heat. It will give the

highest overall heat transfer coefficient for the double pipe heat exchanger.

Parallel flow Counter flowCross-flow

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Small double pipe heat exchanger may also be arranged in coiled form.

Shell-and-tube arrangement

This type of heat exchanger is the most widely used in chemical-process industries. It consists of

a series of tubes arranged in a set called tube bundles and are made up of several types of

tubes: plain, longitudinal.

One fluid flows on the inside of the tubes, and the other fluid flows through the shells and in

between the tubes. Baffles are placed in the shell to ensure that the fluid in the shell flow across

the tube and induce higher heat transfer. One or more tube passes may be used depending on

the head arrangement of the heat exchanger.

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The miniature form of this type of heat exchanger is commonly used in biotechnology field.

Cross flow exchanger

Commonly used in cooling and heating of gas or air. In this

type of exchanger, gas is forced to flow across the tube

bundles, and another fluid is flown inside the tubes to allow

heating or cooling process. The gas flowing across the tube is

said to be mixed because it can move freely in the exchanger.

The other fluid flowing in the tube is said to be unmixed

because the fluid is confined in separate tubular channels and

cannot mix with itself during the heating or cooling process.

The above picture is a different type of cross flow

heat exchanger where the gas is unmixed becausethe flow is confined in between the fins as it passes

through the exchanger. The fluid is unmixed so that

there is a temperature gradient both parallel and

normal to the flow direction. While, when the fluid

is mixed, there’s a tendency that the fluid

temperature will equalize.

This is a picture of the temperature gradient that occurs in cross flow heat exchanger.

Compact heat exchanger

Commonly used in gas-flow system. The overall heat transfer coefficients are low and are

desirable to achieve large surface area in small volume. Compact heat exchanger has area

density greater than 700m2/m

3 for gases or greater than 300m

2/m

3 when operating in liquid or

two-phase streams.

Plate heat exchanger

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This exchanger is made up of multiple, thin, and separated plates with very large surface area

and fluid flow passage.

Plate and shell heat exchanger

Use the combination of plate heat exchanger with shell and tube heat

exchanger. The center of the heat exchange contains circular plate pack

and assembled into an outer shell that creates second flow path for the

fluid.

Plate fin heat exchanger

Made up of passages containing fins. The design include cross

flow and counter flow with various fin configurations such as

straight fins, offset fins, and wavy fins.

Pillow plate heat exchanger

This type of heat exchanger is commonly used in dairy industry for cooling milk. Constructed

using a thin sheet of metal welded to another thicker sheet of metal. The thin metal is welded

with regular pattern of dots. After welding, the enclosed space is pressurized so that the thin

metal bulge out around the welds and provide space for heat exchanger liquids to flow.

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3. How fouling phenomena and pressure drop can degrade the performance of the heat

exchanger?

Fouling is the accumulation of unwanted material on solid surfaces to the detriment of

function. The fouling material can consist of either living organisms (biofouling) or a non-

living substance (inorganic or organic).

Fouling effect on a heat exchanger

reduces thermal efficiency

decreases heat flux

increases temperature on the hot side

decreases temperature on the cold side

induces under-deposit corrosion

Increases use of cooling water

Pressure drop effect on a heat exchanger

Pressure drops have nearly no effect on the effectiveness. It is contact area, flow regime,

fluid conductivities, Reynolds numbers, pipe thickness and fowling which determine

effectiveness.

Pressure drops are nearly negligible in heat exchanger design, although they do exist in

practice. Pressure drops are only worth considering such that we can account for

pumping/fan/compressor power required to operate the system. The pressure drops affect

the entire system efficiency, not the effectiveness of just the heat exchanger.

Task 2:

1. In the forced convection heat transfer is defined a temperature called Bulk Temperature. Give

an explanation of Bulk Temperature.

Bulk temperature represents energy average or “mixing cup” conditions. Thus, for the tube flow

depicted in the figure below the total energy added can be expressed in terms of a bulk-temperature difference by

http://en.wikipedia.org/wiki/Biofouling

http://en.wikipedia.org/wiki/Biofouling

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In some differential length dx the heat added dq can be expressed either in terms of a bulk-

temperature difference or in terms of the heat-transfer coefficient. here Tw and Tb are the wall

and bulk temperatures at the particular x location.

The total heat transfer can also be expressed as

2. Describe energy balance that occurs in the heat exchanger.

Heat exchangers usually operate for long periods of time with ni change n their operating

conditions, therefore, they can be modeled as steady-flow devices. The mass flow rate of each

fluid remains constant and the fluid properties remain the same. Kinetic and potential energy

changer are negligible as the stream’s velocities and elevations experience little or no change at

all. The specific heat of a fluid, in genera changes with temperature. Axial heat conduction along

the tube can be neglected. The outer surface of the heat exchanger is assumed to be perfectly

insulated. From these assumptions, the first law of thermodynamics require heat transfer from

the hot fluid be equal to the rate of heat transfer to the cold one,

=

= ( − )

= ( − )

3. How does the mechanism of forced convection heat transfer in the flow parallel to the object?

Consider the parallel flow of a fluid over a flat plate of length L in the flow direction, as shown

in Figure below The x -coordinate is measured along the plate surface from the leading edge in

the direction of the flow.

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The flow in the velocity boundary layer starts out as laminar, but if the plate is sufficiently long,

the flow will become turbulent at a dis- y tance x cr from the leading edge where the Reynolds

number reaches its critical value for transition.

The transition from laminar to turbulent flow depends on the surface geometry, surface

roughness, upstream velocity, surface temperature, and the type of fluid, among other things,

and is best characterized by the Reynolds number

For flow over a flat plate, transition from laminar to turbulent is usually taken to occur at the

critical Reynolds number of

4. What do you know about overall dirty fouling factor? How to determine the Log Mean

Temperature Difference (LMTD) in the Heat Exchanger?

The log mean temperature difference (also known by its initialism LMTD) is used to determine

the temperature driving force for heat transfer in flow systems, most notably inheat exchangers.

The LMTD is a logarithmic average of the temperature difference between the hot and cold

streams at each end of the exchanger. The larger the LMTD, the more heat is transferred. Theuse of the LMTD arises straightforwardly from the analysis of a heat exchanger with constant

flow rate and fluid thermal properties.

We assume that a generic heat exchanger has two ends (which we call "A" and "B") at which the

hot and cold streams enter or exit on either side; then, the LMTD is defined by thelogarithmic

mean as follows:

http://en.wikipedia.org/wiki/Acronym_and_initialism

http://en.wikipedia.org/wiki/Heat_transfer

http://en.wikipedia.org/wiki/Heat_exchanger

http://en.wikipedia.org/wiki/Logarithmic_mean





http://en.wikipedia.org/wiki/Heat_exchanger

http://en.wikipedia.org/wiki/Heat_transfer

http://en.wikipedia.org/wiki/Acronym_and_initialism

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where ΔTA is the temperature difference between the two streams at end A, andΔTB is the

temperature difference between the two streams at end B. With this definition, the LMTD can

be used to find the exchanged heat in a heat exchanger:

Where Q is the exchanged heat duty (in watts), U is the heat transfer coefficient (in watts

per kelvin per square meter) and Ar is the exchange area. Note that estimating the heat transfer

coefficient may be quite complicated.

After a period of time, heat exchanger may experience reduction in performance. This reduced

performance may be triggered by the accumulation of materials or pollutants from the flow

system that coat the surface area of heat exchanger. The heat exchange may also undergo

corrosion as a result of the interaction with the fluids or materials used. This corrosion or coat

give additional resistance to the heat flow. The overall effect is represented by fouling factor , Rf .Fouling factor is obtained experimentally and is defined as:

Rf = 1/Udirty – 1/Uclean

Problem Calculation:

1. Helium flow rate 5 g/s in the pipe of countercurrent flow double pipe heat exchanger. Helium

entry at a temperature of 300 K and exit at 84 K, and pipe diameter = 2 cm. Nitrogen gas flow

in the annulus at flow rate 35 g/s. Annulus equivalent with 8 cm pipe diameter. Temperature

N2 = 78 K. Physical properties of N2 and He as follows:

He : Cp = 1.25 cal/g. C; = 0.018 cp; k = 0.082 BTU/hr.ft. F

N2 : Cp = 0.25 cal/g. C; = 0.0165 cp; k = 0.014 BTU/hr.ft. F

Determine:

a. Heat transfer between He and N2

We assume that the system is adiabatic, where Back principle is applicable

= 2

= ( − )

= 5 ∙ 1.25 ∙ 300 − 84 = 1350 b. Temperature N2 out of the Heat exchanger

= 2

1350 = 2( − )

1350 = 35 ∙ 0.25 ∙ ( + 195)

= −40.71 = 232.44

http://en.wikipedia.org/wiki/Watt

http://en.wikipedia.org/wiki/Heat_transfer_coefficient

http://en.wikipedia.org/wiki/Kelvin

http://en.wikipedia.org/wiki/Kelvin

http://en.wikipedia.org/wiki/Heat_transfer_coefficient

http://en.wikipedia.org/wiki/Watt

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c. LMTD

300

232.44

84

78

=300−232.44−(84−78)

300−232.29

84−78

= 25.42

d. hN2

=4 =

4 × 3 5 × 8 × 1 . 6 5 × 1 0−4= 33777.26

= =

1.0467 × 0.0165

0.024= 0.68

= 0.023 ×

0.8 ×

0.3 = 0.023 × 33777.260.8 × 0.680.3 = 85.98

= ∙ =85.98 ∙ 0.024

8 × 1 0−2= 25.79 /

e. The length of the pipe if Ul = 400 BTU/hr.0F.ft2

1350 cal/s = 5652.18 W

400 Btu/hr ft2 oF = 117.23 W

= ∙ =

5652.18

117.23 ∙ 25.52= 1.892

= 2

=

1.89

2

10×10−2

= 3.00

2. A total of 96,00 lb/hr of liquid to be cooled from a temperature of 400 F to 200 F. Used as acoolant liquid B which will increase the temperature of 100 F to 200 F. The unit available for

this purpose is a shell and tube heat exchanger with ID shell = 29 in, the number of tubes in

shell = 338. OD pipe = 1 in (BWG = 14), 16 ft length, triangular pitch pipe with PT = 1.25 in, the

distance between two baffle two baffles 10 in. 4 passes pipe flow and 1 pass shell. Given the

properties of the fluid (constant with temperature) as follow:

A : Cp = 0.4 BTU/lb. F ; = 0.6 cp; k = 0.07 BTU/hr.ft. F

B : Cp = 0.6 BTU/lb. F ; = 0.8 cp; k = 0.08 BTU/hr.ft. F

Determine: a. The amount of Liquid B that can be heated

= − = 960000.4200 − 400 = −7680000 /

7680000 = ( − )

=7680000

(0.6)(200 − 100)= 128000

8/12/2019 Report PBL 2 (1)


b. The heat transfer coefficient in pipe

=4 =

(4)(96000)(0.8333)(1.4514)= 101,099

=

=

(0.4)(1.4514)

0.07= 8.293

= 0.023(101099)0.8

(8.293)0.3

= 297.6 =

= =

(297.6)(0.07)

0.833= 25.01

.℉

Or using Dittus-Boelter correlation = 0.023 0.8

( )0.3

= =

(0.8332

4 )= 521.16

= 0.023521.160.83311.4514 0.8

(0.41.45140.07)0.3 = 0.023(299.1)0.8(8.294)0.3

= 4.1496 .

c. The heat transfer coefficient in shell

= =

(29)(0.5)(10)

1.25= 116 2

=

(4)(1.252 − 124 )

(1)

= 0.989

= 0.0824

= =(0.082417)(13236.81)

1.9352= 563

= =

(0.6)(1.9352)

0.08= 14.514

= 0.36 0.08

0.082417 5630.5514.5140.33 1.9352

1.4514

= 0.360.970732.5672.4191.333 = 36.676 . 2.℉

d. Overall heat transfer coefficient of the HE in new condition (clean)

= ∆

= 2

∆ = 400 − 200 − (200 − 100)

ln (400 − 200200 − 100

)= 144.27℉

= 23380.083316 = 2830 2 = ∆ = 960000.4200 − 400 = −7680000

=2 − 11 − 1 =

200 − 400

100 − 400= 0.667

8/12/2019 Report PBL 2 (1)


=1 − 22 − 1 =

100 − 200

200 − 400= 0.5

From correction graph, F = 0.82

=

7680000

(2830)(0.82)(44.27)= 22.94

.

2.

℉

Using other equation to obtain Uclean:1 =

1 +0

1 =1

4.1496+

0.833

(36.676)(2.417)= 0.2504

=1

0.2504= 3.994

. 2.℉

e. Dirt factor (Rd)

8/12/2019 Report PBL 2 (1)


References

1. Holman, JP. (2010). “Heat Transfer”. 10th Edition. New York: The McGraw-Hill Companies, Inc.

2. Incropera. “Fundamentals of Heat and Mass Transfer”. 6th Edition.

3. http://www.sfu.ca/~mbahrami/ENSC%20388/Notes/Natural%20Convection.pdf

http://www.sfu.ca/~mbahrami/ENSC%20388/Notes/Natural%20Convection.pdf



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