Heat Transfer(BSAU) UnitTest-02 2014 JJ (B.E.-6th Sem)

8/12/2019 Heat Transfer(BSAU) UnitTest-02 2014 JJ (B.E.-6th Sem)

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Prepared by Asst. Professor Mohammad (AERO)

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B.S. ABDUR RAHMAN UNIVERSITY

Heat Transfer (AE309) B.E.-3(AERO)

Continuous AssessmentII (05.03.2014)

Answer all the questions and dont mix-up Part-A with other answers.

Only non-programmable calculator is allowed

PartA (Marks: 5 x 2 = 10)

A1.Write the total thermal resistance of a cylindrical conducting layer followed by a convection

layer. Draw appropriate diagram and use appropriate symbols.

A2.Define Reynolds number and write its expression.

A3.Explain briefly what does transient mean.

A4.An ideal gas is heated from 500C to 800C (a) at constant volume and (b) at constant pressure.

For which case do you think the energy required will be greater? Why?

A5.Draw a typical velocity boundary layer over a flat plate. Also mention the various regions.

PartB (Marks: 2 x 20 = 40)

B1 (i) Define viscosity and also explain the effect of temperature on the viscosity of

gases and liquids.

(8)

(ii) Explain the concept of thermal boundary layer with an analogy of velocity

boundary layer. Also explain the significance of Prandtl number in heat

transfer.

(12)

(OR)

B2 (i) Explain the differences between boundary condition and initial condition.

Write the boundary conditions at the radial distance and in the givenfigure-1.

(8)

(ii) Write short notes on heat and moment transfer in turbulent flow. Also derive

the momentum equation for the fluid flow inside a velocity boundary layer by

using appropriate diagrams.

(12)

B3 (i) 10.9 kg of milk at 800C is added with cool water of 5 kg at 400C. Find the

final temperature and also the final specific heats of the mixture. Specificheats of milk and water is given to be 3.93 kJ/kg and 4.186 kJ/kg respectively.

Assume constant heat capacities within the range of operating temperature.

(4)

(ii) A spherical container of inner radius 2, outer radius 2.1, andthermal conductivity 30 W/mC is filled with iced water at 00C. thecontainer is gaining heat by convection from the surrounding air at250C with a heat transfer coefficient of 18W/m2K. Assuming the inner

(16)


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surface temperature of the container to be 0C, (a) express the differential

equation and the boundary conditions for the steady one-dimensional heat

conduction through the container, (b) obtain a relation for the variation of

temperature in the container by solving the differential equation, (c) evaluate

the rate of heat gain to the iced water, and (d) amount of ice turned into water

in 150 seconds in steady condition (latent heat of water is 333.7 kJ/kg).(Marks: 4+6+4+2)

B4 (i) In a thermal circuit, draw the isotherms and adiabatic surfaces. Take at least

two layers in parallel and two in series

A pipe is insulated to reduce the heat loss from it. However, measurementindicates that the rate of heat loss has increased instead of decreasing. Can the

measurement bet right? Justify your answer.

(4)

(ii) A 2.2-mm-diameter and 10-m-long copper 401W/m0C) electric wireis tightly wrapped with a 1-mm-thick plastic cover whose thermal

conductivity is 0.15W/m0C. Electric current or 13A passes through thewire with a voltage drop of 8V across it. If the insulated wire is exposed tothe medium of 300C with a heat transfer coefficient of 24W/m20C, determine (a) the temperature at the interface of the wire and the

plastic cover in steady operation. Also determine (b) Critical Radius of

Insulation and (c) Maximum temperature inside the electrical conductor. Also

explain by doubling the thickness will increase or decrease the interface

temperature. (Ref. Figure-2)

(16)

Figure-1

Figure-2


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Answers (Part-A)

A1. The total thermal resistance for the given situation is as

follows:

+

1

2

A2. Reynolds number is defined as the ratio of the inertial force

to the viscous force.

A3. Transient means the unsteady behaviour but it is in the process of being steady. In due course

of time the transient will subside and steady nature will prevail. It is the large fluctuation about

the slow buildup of the average value of the quantity.

A4. The energy required to heat the gas at constant pressure will be higher because in this way two

job will be done. 1. To heat the gas from 500C to 800Cand 2. To do the job by expanding

the gas at constant pressure.

A5.

A typical velocity boundary layer is drawn below with its various regions indicated.

Answers (Part-B)

B1.(i) Viscosity is the property of the material by the virtue of

which, it opposes or resists the relative motion in the adjacent

layers of fluid. When there is a fluid flowing over any solid

surface, the fluid particle at the adjacent to the surface comes

to rest relative to the surface movement. This also results a

gradual increment in the velocity in the layers away from the

solid surface and after a small distance velocity of the layer

becomes equal to that of the free stream. This happens due to

the inherent property of the fluid called viscosity.


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Viscosity is basically a mutual interaction of the two types of molecular forces called adhesive

and cohesive forces in liquids and a process of momentum transfer in the gases. Due to the

different mechanism of viscosity being active in the liquid and gases, they behave differently with

the rise in temperature. In liquids viscosity decreases as the temperature rises, because with the

rise in temperature, the intermolecular forces decrease so the resistance based upon these forces

to decrease. On the other hand, in gases, with the rise in temperature, molecules start moving more

vigorously, hence causing larger momentum transfer. So among gases, viscosity increases with

the rise in temperature.

(ii) When flow takes place over any flat plate, there exists a thin region in which velocity of

the flow varies from zero velocity relative to the plate to the full velocity of the free stream. This

thin region is called boundary layer and specifically velocity boundary layer because velocity acts

as the characteristic quantity in this phenomenon. In the similar manner there exists a thin region

in the same flow field where temperature of the fluid varies from the temperature of the surface to

the ambient temperature of the fluid. This region is called thermal boundary layer because the

behaviour of this layer is characterised by the temperature variation in the region.

In the case of velocity boundary layer there exists non slip condition and the velocity of the fluid

at the surface is equal to velocity of the surface itself

similarly the temperature of the fluid in the thermal

boundary layer will be equal to the temperature of the

surface itself. Then there will be a variation in the

characteristic quantity with the distance away from

the surface.

Analogous to the velocity boundary layer thickness,

thermal boundary layer thickness is also defined as

the normal distance from the surface where temperature gain of the fluid becomes almost 99% of

the total temperature difference between the fluid and surface.

The existence of these boundary layers is caused by the difference in their respective characteristic

quantities between the fluid and surface. In the velocity boundary layer there seems to be a loss of

momentum at the surface which is slowly gained as we go away from it. In the similar manner in

the thermal boundary layer there exist a loss of temperature at the surface in the fluid which is

gained as we go away from the surface. These two boundary layers always exist as long as the


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difference in their characteristic quantities between the surface and fluid exists. The mutual

existence and their grown is characterised by a non-dimensional called Prandtl number, which is

defined as below:

Molecular diffusivity of momentumMolecular diffusivity of heat A material of higher Prandtl number will have high momentum diffusion but low thermal

diffusion, so the conduction of velocity will be high compared to conduction of heat. So a thin

velocity boundary layer and a thick thermal boundary layer is expected in such material. On the

other hand if a material has low Prandtl number then, there will be low momentum diffusion but

high thermal diffusion and hence a thick velocity boundary layer and a thin thermal boundary

layer is expected.

B2.(i) Boundary conditions and initial conditions are the required spatial and temporal values

to solve the problem related to heat transfer. By the name itself it is clear that boundary conditions

will exist at the edge of the control volume i.e. 0, . On the other hand initial conditionswill exist at the edge of the time or at the beginning of the time i.e. 0.In the given problem there exists a convection in the inner region of the cylinder, so the heat

transfer rate by convection will be same as that of conduction at the inner surface of the cylindrical

conductor. On the other hand, at the outer there region there is an insulation so there will be no

heat transfer at the outer surface of the cylindrical conductor. This will give us two boundary

conditions at both the boundaries as below:

|=

|= 0 |= 0

(ii) Turbulent flows are characterised by random and rapid fluctuations of swirling region

of fluid. These fluctuating regions are called eddies. So apart from the mean velocity there exists

an additional fluctuating velocity for the particles in the turbulent flows i.e. + . In thelaminar flow the mass, momentum and energy are transferred across the streamlines using

molecular diffusion only as the molecules remain. But in the case of turbulent flow, the fluctuating

part of the velocity provides an additional means to transfer the mass, momentum and energy.


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Thus turbulent flow is associated with much higher values of friction, heat transfer and mass

transfer coefficients.

Sometimes even the average flow fluctuates and then these transport phenomena become much

more vigorous.

Laminar and turbulent flows are detected by the non-dimensional parameter called Reynolds

number which includes all the possible parameters affecting the flow characteristics. A high

Reynolds number flow will be prone to the turbulent flow and hence will make the higher values

of momentum and energy transfer. Also the in turbulent flow the velocity profile is very steep so

a very high value of shear resistance force (moment transfer) will exist near the wall.

Moment equation for the boundary layer:

While considering the forces acting on the elemental volume, we must consider all the possible

forces there. One type of the force is body force that acts throughout the entire body of the control

volume such as gravity force, electrostatic force, magnetic force etc. in which each and every

molecule of the mass is affected and such forces are proportional to the volume of the body.

Second type is surface force which act only at the control surface such as hydrostatic pressure

force and shear force. Now applying the Newtons second law of motion:

Mass (Acceleration in aspecified direction) (Net force body and surfaceacting in that direction )Now considering the same elemental volume again, mass of the fluid element will be given by:

1Now let us analyse the flow and forces acting in the-direcion first. From the knowledge of partialand total derivative and velocity being two-dimensional, we can write that:

, , , + ,

+

+ +


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We have assumed steady flow so only convective acceleration will be available. Local

acceleration will be zero. Same will be true in they-direction also. Now the surface forces acting on the element

will be pressure force and the shear stress force. Pressure force

will be normal to the surface while shear stress force will be

along the surface. Net force in the positive x-direction will be:

, {( + )} 1+ { ( + )} 1

(

) 1

1

We got above equation by putting the Newtons law of viscosity . We are analysingthe laminar boundary layer so it is a valid argument. Now on equating the product of mass and

acceleration with this force:

(+

)

Above is the momentum equation or momentum balance equation in x-direction. The same can be

written for the y-direction also with an introduction of only body force acting in that direction in

the normal cases. If there is any body force acting on the fluid element, the body force per unit

volume can be added on the right had side in that case.

( + )

+ g

B3.

(i) The given data are as follows:

10.9 kg, 80, 3.93 kJkg C 5 kg, 40, 4.186 kJkg C


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Let after mixing the final temperature of the mixture is. Then loss of heat from the milk will besame as the gain of heat from water.

10.93.938 0 54.186 40

42.8378 0 20.93 40 3426.96 42.837 20.93 837.2

3426.96 + 837.220.93 + 42.837 .The finalof the mixture will be given by:

+ +

10.93.93+54.18610.9+5 .

(ii) The given situation is as given below:

2.0 m, 2.1 m, 30 Wm C,

0, 25, 18 W

m C

It is clear that there will be no sensible heat transfer between

inner surface and the water in the centre. Both are at the same

temperature, but there can be phase change which happens at the constant temperature.

Boundary conditions are:

0

|

=

The general equation for the conduction inside the solid container is:

1

(

) 0

No heat generation and no transient is mentioned. On integrating the equation (iii) we get that:


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+ +

Now we have three unknowns (A, B, T2) but only two equations. So we generate the third equation

here.

+ |=

Above two equations can be put into equation (ii) to reduce the number of equations by 1. On

rewriting the equation (ii) using (vii) and (viii)

(+ )

(+

)

Replacing B from (vi):

{+

(+

)}

On solving above equation we get that:

+11

+

On putting this in (viii) we get that:

+ +

On putting the numerical values:

2 5 0 .+ . 62.23

0 + 2 5 02.0 .+ . 31.11


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Differential equation and boundary conditions are already given above as equation (i), (ii) and

(iii).

(b) The relation of variation of temperature is:

62.23

+31.11

(c) The heat transfer rate:

|= 30462.23 .

Negative sign shows that heat transfer is inward.

(d) Total amount of ice turned into water in 150 seconds:

23460150

23460150333.7 . B4.(i) The isothermal and adiabatic lines are as shown below:

When we want to pass some hot fluid through pipes, we expect that there

would be as less loss of heat as possible. For this we insulate the pipe

using some insulating material. While insulating this we also increase the

external surface area through which the convection is taking place. So

virtually the area or volume which was doing convection is not doing

conduction and depending upon the conductivity of the insulation andconvection heat transfer of the surrounding fluid, the heat transfer may increase or decrease. There

is a critical outer radius of the insulation which gives the maximum heat transfer and any other

radius will actually decrease the lower heat transfer rate. Now it is possible that sometimes we

want to check the heat transfer rate and for this we put the insulation and increase the radius, but

if it is well below the critical radius of insulation, instead of decreasing the heat transfer rate, it

may increase it. The critical radius of insulation for the cylindrical and spherical object is given

below.

, , ,

(i) The given situation is as below:


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2.2 mm, 10 m, 401 Wm C , 1 mm, 0.15W

m C , 13A, 8V, 30C, 24 Wm C,

The heat generated per unit time is:

13 8 104 WThis is also the total heat transfer rate by any means. At the central axis of the conductor there will

be symmetric boundary condition, so there will be no heat transfer there.

Total thermal resistance will be:

1 1

24 0.0042 10 0.3157

ln

2 ln2.1 1.1

2 0.15 10 0.0686

+ 0.3157 +0.0686 0.3843(a) Now using the thermal circuit equation:

104 300.3843 .(b) Critical radius of insulation:

0.1524 . (c) Maximum temperature will be felt at the axis of the electrical wire.

Heat generation per unit volume per unit time will be:

2 104

1.1 10 2.7358Wm

Now we must have two boundary conditions.

69.96 , |= 0 Governing equation will be:

1

(

) +

0

1 (

)

2.7358401 6.822 10


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2+

2 +

2 +

4 +

On putting the expression (ii) in (iii) we get that 0.Then on putting the value of temperature from (i) in (iv) we get that:

4 +

+ 4 69.96+ 2.7358401 1.12 69.9618

Temperature at the centre will be given by:

4+ 69.9618 69.9618

There is a very little difference between the temperature between the temperature at the conductor

surface and the temperature at the conductor core axis.

(d) On doubling the thickness of the plastic cover the outer radius will go to 3.1 mmwhich is less

than the critical radius of insulation 6.25 mm. So the heat transfer will increase instead of

decreasing.

Documents

Heat Transfer(BSAU) UnitTest-02 2014 JJ (B.E.-6th Sem)