gate 20204. refrigeration & air-conditioning 4.1 - 4.000 1. vapour refrigeration 4.1 - 4.0 2....

GATE 2020Mechanical Engineering

(Volume - I)

TOPIC WISE GATE SOLUTIONS

1987 - 2019

TM

GATE A PCADEMY UBLICATIONS®

S. No. Topics Page No.

1. Fluid Mechanics 1.1 - 1.000

1. Properties of Fluid 1.1 - 1.00

2. Pressure & Its Measurement 1.00 - 1.00

3. Hydrostatic Forces 1.00 - 1.00

4. Buoyancy & Floatation 1.00 - 1.00

5. Kinematics of Fluid 1.00 - 1.00

6. Dynamics of Fluid 1.00 - 1.00

7. Laminar & Turbulent Flow, Viscous Flow, Flow Through Pipes 1.00 - 1.00

8. Boundary Layer Theory 1.00 - 1.00

9. Hydraulic Machines 1.00 - 1.00

2. Thermodynamics 2.1 - 2.000

1. Thermodynamics System & Process 2.1 - 2.00

2. First Law of Thermodynamics 2.00 - 2.00

3. Second Law of Thermodynamics 2.00 - 2.00

4. Entropy 2.00 - 2.00

5. Availability & Irreversibility 2.00 - 2.00

6. Pure Substances 2.00 - 2.00

7. Steam Power Cycles 2.00 - 2.00

8. Gas Power Cycles 2.00 - 2.00

9. Internal Combustion Engine 2.00 - 2.00

3. Heat Transfer 3.1 - 3.000

1. Conduction 3.1 - 3.0

2. Fins & Unsteady Heat Transfer 3.0 - 3.00

3. Free & Forced Convection 3.00 - 3.00

4. Heat Exchanger 3.00 - 3.00

5. Radiation 3.00 - 3.00

CONTENTS

4. Refrigeration & Air-Conditioning 4.1 - 4.000

1. Vapour Refrigeration 4.1 - 4.0

2. Properties of Moist Air & Psychrometric Processes 4.0 - 4.00

3. Heat Pumps & Cycles 4.00- 4.00

5. Mechanics of Solids 5.1 - 5.000

1. Stress & Strain 5.1 - 5.0

2. Complex Stress 5.0 - 5.00

3. Elastic Constants & Theory of Failure 5.00 - 5.00

4. Thin Cylinder 5.00 - 5.00

5. Shear Force & Bending Moment Diagrams 5.00 - 5.00

6. Bending of Beam 5.00 - 5.00

7. Torsion of Shaft 5.00 - 5.00

8. Springs 5.00 - 5.00

9. Euler's Theory of Column 5.00 - 5.00

10. Strain Energy & Thermal Stresses 5.00 - 5.00

6. Machine Design 6.1 - 6.000

1. Design Against Static Load (Theory of Failure) 6.1 - 6.0

2. Design Against Dynamic Load (Fatigue Strength & SN Diagram) 6.0 - 6.00

3. Gears 6.00 - 6.00

4. Bearings, Shaft & Keys 6.00 - 6.00

5. Clutches, Ropes & Belts 6.00 - 6.00

6. Brakes 6.00 - 6.00

7. Joints (Bolted, Riveted & Welded) 6.00 - 6.00

8. Power Screws & Springs 6.00 - 6.00

7. General Aptitude 7.1 - 7.000

1. Numerical Ability 7.1 – 7.0

2. Logical Reasoning 7.0 - 7.00

3. Verbal Ability 7.00 - 7.00

© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176

B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401

1.1 Thermal conductivity is lower for

[1 Mark] (A) Wood (B) Air (C) Water at 0100 C (D) Steam at 1 bar

1.2 Match the property with their units

[2 Marks] Property Units

A. Bulk modulus 1. W/s B. Thermal

conductivity 2. 2N/m

C. Heat transfer coefficient

3. 3N/m

D. Heat flow rate 4. W 5. W/mK

6. 2W/m K

1.3 For a current carrying wire of 20 mm

diameter exposed to air 2( 25W/m K)h , maximum heat distribution occurs when the thickness of insulation is ( 0.5W/mK)k [2 Marks]

(A) 20 mm (B) 10 mm (C) 25 mm (D) 0 mm

1.4 Two insulating materials of thermal

conductivity k and 2k are available for lagging a pipe carrying a hot fluid. If the radial thickness of each material is same.

[1 Mark] (A) Material with higher thermal

conductivity should be used for inner layer and one with lower thermal conductivity for the outer.

(B) Material with lower thermal conductivity should be used for inner layer and one with higher thermal conductivity for the outer.

(C) It is immaterial in which sequence the insulating material are used.

(D) It is not possible to judge unless numerical values of dimensions are given.

1.5 For a given heat flow and for the same

thickness, the temperature drop across the material will be maximum for

[1 Mark] (A) Copper. (B) Steel. (C) Glass wool. (D) Refractory brick.

1 Conduction

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3.2 Topic Wise GATE Solutions [ME] GATE ACADEMY®



1.6 The temperature variation under steady

heat conduction across a composite slab of two materials with thermal conductivities 1k and 2k is shown in figure then, which one of the following statements holds? [2 Marks]

(A) 1 2k k (B) 1 2k k

(C) 1 0k (D) 1 2k k

1.7 It is proposed to coat a 1 mm diameter

wire with enamel paint ( 0.1W/mK)k to increase heat transfer with air. If the air side heat transfer co-efficient is

2100 W/m K , the optimum thickness of enamel paint should be [2 Marks]

(A) 0.25 mm (B) 0.5 mm

(C) 1 mm (D) 2 mm

1.8 In descending order of magnitude, the

thermal conductivity of (a) pure iron, (b) liquid water, (c) Saturated water vapour, (d) Pure aluminum can be arranged as

[1 Mark]

(A) a b c d (B) b c a d

(C) d a b c (D) d c b a

Heat is being transferred by convection from water at 048 C to a glass plate whose surface that is exposed to the water is at 040 C . The thermal conductivity of water is 0.6 W/mK and the thermal conductivity of glass is 1.2 W/mK. The spatial gradient of temperature in the water at the water glass interface is

41 10dT

dy K/m.

1.9 The value of the temperature gradient in

the glass at the water glass interface in K/m is [2 Marks]

(A) 42 10 (B) 0.0

(C) 40.5 10 (D) 42 10 1.10 The heat transfer coefficient h in

2W/m K is [2 Marks] (A) 0.0 (B) 4.8 (C) 6 (D) 750

1.11 One dimensional unsteady state heat

transfer equation for a sphere with heat generation at the rate of ‘q’ can be written [1 Mark]

(A) 1 1T q Tr

r r r k t

(B) 22

1 1T q Tr

r r r k t

1k

1T2k

3T

2T

040 C

048 C

y

WaterGlass

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Common Data for Questions 1.9 & 1.10

2004 IIT Delhi

3.3GATE ACADEMY® Heat Transfer : Conduction



(C) 2

2

1T q T

r k t

(D) 2

2

1( ) q TrT

r k t

1.12 A stainless steel tube ( 19 W/mK)sk of 2 cm ID and 5 cm OD in insulated with 3 cm thick asbestos ( 0.2 W/mK)ak . If the temperature difference between the inner most and outermost surface is

0600 C , the heat transfer rate per unit length is [2 Marks]

(A) 0.94 W/m (B) 9.44 W/m

(C) 944.72 W/m (D) 9447.21 W/m

1.13 A well machined steel plate of thickness

L is kept such that the wall temperatures are hT and cT as shown in the figure below. A smooth copper plate of the same thickness L is now attached to the steel plate without any gap as indicated in the figure below. The temperature at the interface is iT . The temperatures of

the outer walls are still the same at hT

and cT . The heat transfer rates are 1q and

2q per unit area in the two cases respectively in the direction shown.

Which of the following statements is correct? [1 Mark]

(A) h i cT T T and 1 2q q

(B) h i cT T T and 1 2q q

(C) ( ) / 2h i cT T T and 1 2q q

(D) ( ) / 2i h cT T T and 1 2q q

1.14 In a case of one dimensional heat conduction in a medium with constant properties, T is the temperature at

position x, at time t. Then T

t

is

proportional to [1 Mark]

(A) Tx

(B) T

t

(C) 2T

x t

(D) 2

2

T

x

1.15 Heat flows through a composite slab, as shown below. The depth of the slab is 1 m. The k values are in W/mK. The overall thermal resistance in K/W is

[2 Marks]

(A) 17.2 (B) 21.9

(C) 28.6 (D) 39.2

1q1q

hT

L

CT

STEEL

Ti

2q 2q

cThT

L

CopperSteel

L

1 0.02k �

0.5m

2 0.10k �

3 0.04k �

0.25m

3

2

1

0.5

m

1 mq

2005 IIT Bombay




1.16 In a composite slab, the temperature at

the interface inter( )T between two materials is equal to the average of the temperature at the two ends. Assuming steady one dimensional heat conduction, which of the following statements is true about the respective thermal conducti-vities? [1 Mark]

(A) 1 22k k (B) 1 2k k

(C) 1 22 3k k (D) 1 22k k 1.17 With an increase in the thickness of

insulation around a circular pipe, heat loss to surrounding due to [2 Marks]

(A) Convection increases while heat due to conduction decreases.

(B) Convection decreases, while that due to conduction increases.

(C) Convection and conduction decreases.

(D) Convection and conduction increases.

1.18 A long glass cylinder of inner diameter

0.03 m and outer diameter 0.05 m carries hot fluid inside. If the thermal conductivity of glass 1.05W/mKk , the thermal resistance (K/W) per unit length of the cylinder is [2 Marks]

(A) 0.031 (B) 0.077 (C) 0.17 (D) 0.34

1.19 Heat is being transferred convectively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 075 Cunder steady state condition, the rate of heat generation within the fuel element is 6 350 10 W/m and the convective heat

transfer coefficient is 21 kW/m K , the outer surface temperature of the fuel element would be [2 Marks]

(A) 0700 C (B) 0625 C (C) 0500 C (D) 0400 C

Consider steady one dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 380MW/m . The left and right

faces are kept at constant temperature of 0160 Cand 0120 C respectively. The plate has a constant thermal conductivity of 200 W/mK. 1.20 The location of maximum temperature

within the plate from its face is [2 Marks]

(A) 15 mm (B) 10 mm (C) 5 mm (D) 0 mm 1.21 The maximum temperature within the

plate in 0 C is [2 Marks] (A) 160 (B) 165 (C) 200 (D) 250

1.22 Steady two dimensional heat conduction

takes place in the body shown in the figure below. The normal temperature gradients at surface P and Q can be considered to be uniform. The

temperature gradients T

x

over a surface

bb

Ti

1k

1T

2T

2k

2 b2 b

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Common Data for Questions 1.20 & 1.21

2008 IISc Bangalore




Q is equal to 10 K/m. Surfaces P and Q are maintained at constant temperature as shown in the figure, while the remaining part of the boundary is insulated.

The body has a constant thermal conductivity of 0.1 W/mK. The value of

T

y

and T

x

at surface P are [2 Marks]

(A) 20K/m, 0K/mT T

x y

(B) 0K/m, 10K/mT T

x y

(C) 10K/m, 10K/mT T

x y

(D) 0K/m, 20K/mT T

x y

1.23 Consider one-dimensional steady state

heat conduction, without heat generation, in a plane wall, with boundary conditions as shown in the figure below. The conductivity of the wall is given by 0k k bT , where 0k

and b are positive constants, and T is temperature.

As x increases, the temperature gradient dT

dx

will [1 Mark]

(A) Remain constant. (B) Be zero. (C) Increase. (D) Decrease. 1.24 Consider one-dimensional steady state

heat conduction along x-axis ( 0 x L ), through a plane with the boundary surfaces ( 0x and x L ) maintained at temperatures of 00 C and 0100 C . Heat is generated uniformly throughout the wall. Choose the correct statements.

[1 Mark] (A) The direction of heat transfer will be

from the surface at 0100 C to the surface at 00 C .

(B) The maximum temperature inside the wall must be greater than 0100 C.

(C) The temperature distribution is linear within the wall.

(D) The temperature distribution is symmetric about the mid-plane of the wall.

1.25 Consider a long cylindrical tube of inner

and outer radii, ir and 0r respectively, length L and thermal conductivity k. its inner and outer surfaces are maintained at 1T and 0T , respectively 1 0( )T T . Assuming one-dimensional steady state heat conduction in the radial direction, the thermal resistance in the wall of the tube is [1 Mark]

(A) 0

1 ln2

ir

kL r

(B) 2 i

L

rk

(C) 01 ln2 i

r

kL r

(D) 01 ln4 i

r

k L r

Surface

Surface

0,0 CQ

0,100 CP

2 m

1m

y

x

T1 T2 Where

x

2 1T T�

2013 IIT Bombay

2014 IIT Kharagpur




1.26 As the temperature increases, the thermal conductivity of a gas [1 Mark]

(A) Increases (B) Decreases (C) Remain constant (D) Increases up to a certain temperature

and then decreases 1.27 Consider one dimensional steady state

heat conduction across a wall (as shown in figure below) of thickness 30 mm and thermal conductivity 15 W/mK. At

0x a constant heat flux, 5 2'' 1 10 W/mq is applied. On the

other side of the wall, heat is removed from the wall by convection with a fluid at 025 C and heat transfer coefficient of

2250 W/m K. The temperature (in 0 C ), at 0x is ________. [2 Marks]

1.28 A material P of thickness 1 mm is

sandwiched between two steel slabs, as shown in the figure below. A heat flux

210kW/m is supplied to one of the steel slabs as shown. The boundary temperatures of the slabs are indicated in the figure. Assume thermal conductivity of this steel is 10 W/mK. considering one-dimensional steady state heat conduction for the configuration, the thermal conductivity (k, in W/mK) of material P is ________. [2 Marks]

1.29 Heat transfer through a composite wall

is shown in figure. Both the sections of the wall have equal thickness ( l ). The conductivity of one section is k and that of the other is 2k. The left face of the wall is at 600 K and the right face is at 300 K. [2 Marks]

The interface temperature (in K)iT of

the composite wall is ______. 1.30 A plane wall has a thermal conductivity

of 1.15 W/mK. If the inner surface is at 01100 Cand the outer surface is at

0350 C , then the design thickness (in meter) of the wall to maintain a steady heat flux of 22500 W/m should be ____.

[2 Marks]

1.31 If a foam insulation is added to a 4 cm

outer diameter pipe as shown in the figure, the critical radius of insulation (in cm) is _______. [1 Mark]

0�x

x

025 C� �T

2T

1T

5 2" 1 10 W/m� �q

20 mm 20 mm1 mm

2 360K�T2" 10 kW/m�q

1 500 K�T

?�k

P

Steelslab

Steelslab

l l

300 K600 K iT

k 2k

Heat flow

2015 IIT Kanpur




1.32 A 10 mm diameter electrical conductor

is covered by an insulation of 2 mm thickness. The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 210W/m K . Addition of further insulation of the same material will

[2 Marks]

(A) Increases heat loss continuously

(B) Decreases heat loss continuously

(C) Increases heat loss to a maximum and then decreases heat loss

(D) Decreases heat loss to a minimum and then increases heat loss

1.33 A brick wall ( 0.9 W/mK)k of thickness 0.18 m separates the warm air in a room from the cold ambient air. On a particular winter day, the outside air temperature is 05 C and the room needs

to be maintained at 027 C . The heat transfer coefficient associated with outside air is 220 W/m K . Neglecting the convective resistance of the air inside the room, the heat loss, in 2W/m , is [2 Marks]

(A) 88 (B) 110

(C) 128 (D) 160

1.34 A cylindrical uranium fuel rod of radius 5 mm in a nuclear reactor is generating heat at the rate of 7 34 10 W/m . The rod is cooled by a liquid (convective heat transfer coefficient 21000 W/m K ) at

025 C . At steady state, the surface temperature (in K) of the rod is

[2 Marks] (A) 308 (B) 398 (C) 418 (D) 448

1.35 A plastic sleeve of outer radius

0 1 mmr covers a wire (radius 0.5 mmr ) carrying electric current.

Thermal conductivity of the plastic is 0.15 W / m K . The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 2W / m K . Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will [1 Mark]

(A) Increase (B) Remain the same (C) Decrease (D) Be zero 1.36 A hollow cylinder has length ,L inner

radius 1,r outer radius 2 ,r and thermal conductivity k . The thermal resistance of the cylinder for radial conduction is

[1 Mark]

(A)

2

1

ln

2

rr

kL

(B)

1

2

ln

2

rr

kL

(C) 2

1

2

ln

kL

rr

(D) 1

2

2

ln

kL

rr

2

0 2 W/m Kh �

0.1W/mKfoamk �

15 W/m-K�pipek

Pipe

FoamFoam

2016 IISc Bangalore




1.37 Steady one-dimensional heat conduction takes place across the faces 1 and 3 of a composite slab consisting of slabs A and B in perfect contact as shown in the figure, where ,A Bk k denote the respective thermal conductivities. Using the data as given in the figure, the interface temperature 2T (in 0 C ) is ______. [1 Mark]

1.38 Heat is generated uniformly in a long

solid cylindrical rod (diameter = 10 mm) at the rate of 7 34 10 W/m . The thermal conductivity of the rod material is 25 W/m K. Under steady state conditions, the temperature difference between the center and the surface of the rod is ________ 0 C [2 Marks]

1.39 A plane slab of thickness L and thermal

conductivity k is heated with a fluid on one side (P) and other side (Q) is maintained at a constant temperature QT

of 025 C , as shown in the figure. The fluid is at 045 C and the surface heat transfer coefficient, h is 210 W/m K . The steady state temperature pT (in 0 C ) of the side which is exposed to the fluid is ______ (correct to two decimal places). [2 Marks]

1.40 A slender rod of length L and diameter

( )d L d and thermal conductivity 1k is joined with another rod of identical dimensions, but of thermal conductivity

2k , to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is

[1 Mark]

(A) 1 2

1 2

2k k

k k (B) 1 2

1 2

k k

k k

(C) 1 2k k (D) 1 2k k

1.41 One-dimensional steady state heat conduction takes place through a solid whose cross section area varies linearly in the direction of heat transfer. Assume there is no heat generation in solid and thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is [1 Mark]

(A) Linearly (B) Quadratic (C) Logarithmic (D) Exponential 1.42 Three slabs are joined together as shown

in the figure. There is no thermal contact resistance at the interfaces. The center slab experiences a non-uniform internal heat generation with an average value equal to 310000 Wm , while the left and right slabs have no internal heat

A B

1 2 3

0

1 130 CT � 0

3 30 CT �20

W/mK

Ak � 100 W/mKBk �

0.1 m 0.3 m

L = 20 cm

TQ = 25 C0h = 10 W/m K

2

0= 45 CT

TP

k = 2.5 W/mK

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generation. All slabs have thickness equal to 1 m and thermal conductivity of each slab is equal to 1 15Wm K . The two extreme faces are exposed to fluid with heat transfer coefficient

2 1100 Wm K and bulk temperature 030 C as shown.

The heat transfer in the slabs is assumed to be one dimensional and steady, and all properties are constant. If the left extreme face temperature 1T is measured

to be 0100 C, the right extreme face temperature 2T is _____ 0 C . [2 Marks]

1 m 1 m 1 mLeft extremeface = 100 CT1

0

100 W/m K2

30 C0

100 W/m K2

30 C0

T2




1.1 B 1.2 A-2, B-5, C-6, D-4

1.3 B 1.4 B 1.5 C

1.6 A 1.7 B 1.8 C 1.9 C 1.10 D

1.11 B 1.12 C 1.13 D 1.14 D 1.15 C

1.16 D 1.17 A 1.18 B 1.19 A 1.20 C

1.21 B 1.22 D 1.23 D 1.24 B 1.25 C

1.26 A 1.27 625 1.28 0.1 1.29 400 1.30 0.345

1.31 5 1.32 C 1.33 C 1.34 B 1.35 A

1.36 A 1.37 67.5 1.38 10 1.39 33.88 1.40 A

1.41 C 1.42 60

Thermal conductivity is lower for air. Hence, the correct option is (B). Key Point (i) In case of solids thermal conductivity of

material is due to flow of free electrons and lattice vibration.

(ii) In case of liquids and gases, thermal conductivity is due to molecular wave.

(iii) Pure metals will have highest thermal conductivity from its alloys.

'puremetal it s alloyk k , iron steelk k ,

copper brassk k , ice water water vapour airk k k k

Maximum thermal conductivity is of diamond due to crystalline lattice. 2300 W/mKdiamondk

Key Point (i) The ratio of the change in pressure to the

volume compression is the Bulk Modulus.

22

3 3

N/m N/m/ m /mp

KV V

(ii) Thermal conductivity is ability of material to conduct heat through it. S.I. unit W/m K.

2

Wm K/m

Qk

dTA

dx

W/m K

(iii) Heat transfer coefficient (h) is used as a proportionality constant between heat flow and driving force for heat flow, i.e., temperature difference.

22

W W/m Km K

Qh

A T

Given : Wire diameter (d) = 20 mm Heat transfer coefficient 225 W/m Kh Thermal conductivity 0.5 W/mKk Critical radius of insulation,

0.5 0.020m25c

kr

h = 20 mm

or 20 10mm2 2d

r

Answers Conduction

Explanations Conduction

1.1 (B)

1.2 A-2, B-5, C-6, D-4

1.3 (B)




Hence, thickness of insulation

c ct r r 20 10 10mm

Hence, the correct option is (B).

Key Point In case of plane wall the area

perpendicular to the direction of heat flow adding more insulation to a wall always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate. This is due to the fact the outer surface have always the same area.

But in cylindrical and spherical coordinates, the addition of insulation also increases the outer surface, which decreases the convection resistance at the outer surface. Moreover, in some cases, a decrease in the convection resistance due to the increase in surface area can be more important than an increase in conduction resistance due to thicker insulation. As a result the total resistance may actually decrease resulting in increased heat flow.

The thickness upto which heat flow increases and after which heat flow decreases is termed as critical thickness. In the case of cylinders and spheres it is called critical radius. It can be derived the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.

Two insulating materials of thermal conductivity k and 2k are available for lagging a pipe carrying a hot fluid here the critical radius for insulation must be

(i) For first material 1

( )c

kr

h

(ii) For second material 2

2( )c

kr

h

It is apparent that 2 1

( ) 2c cr r

Hence, the first material is to be used inside and second material is to be used for outside as it has bigger critical radius.


For heat flow over the same thickness, the temperature drop across the material will be maximum for that material which is having more resistance to heat flow i.e., value of thermal conductivity will be less, from Fourier equation we can infer as follows :

Fourier equation,

dTQ kA

dx

QdxkdT

A

kdT Constant or 1dT

k

Hence, the correct option is (C).

Key Point

glass wool refractory bricks steel copperk k k k

Whichever the material is having lowest thermal conductivity the corresponding material has highest temperature drop.

r

rc

tc

1.4 (B)

1.5 (C)




From Fourier law, we can infer as follows :

dTQ k

dx as Q constant.

1dT

dx k

1 1 2 2

Constant Constant,dT dT

dx k dx k

1 21 2

ordT dTk k

dx dx

Hence, the correct option is (A).

Key Point (i) Here, k is called thermal conductivity of

Material that measures the ability of material to conduct heat. It is a property that depends on temperature but not on too extent. For all gases, k increase with temperature due to collision as result vibration increase.

(ii) On the other side we can infer conductivity as opposite of resistivity, the more resistance, temperature drop will be more.

Given : Wire diameter ( ) 1 mmd Thermal conductivity of enamel point ( ) 0.1W/mKk

Heat transfer coefficient 2( ) 100 W/m Kh

Critical radius,

0.1 0.001 m 1mm100c

kr

h

Critical thickness of enamel point

11 0.5mm2c ct r r


Out of the given substances pure aluminum has high conductivity and steam has low conductivity. Aluminum - 250 W/mK, Iron - 80 W/mK, Liquid water - 0.7 W/mK, Water vapour - 0.016 W/mK Hence, the correct option is (C).

Given : Temperature of water 0( ) 48 CwT

Temperature of glass plate 0( ) 40 CgT

Thermal conductivity of water( ) 0.6 W/mKwk

Thermal conductivity of glass ( ) 1.2 W/mKgk

Spatial gradient of temperature of water-glass

interface, 41 10 K/mw

dT

dy

Heat generation (flux),

w g

w g

dT dTq k k

dy dy

w

gg w

kdT dT

dy k dy

4 40.6 (1 10 ) 0.5 10 K/m1.2

g

dT

dy

Hence, the correct option is (C)

Key Point The temperature gradient in the glass interface must be lesser than water as here thermal conductivity is higher, that suggest the resistance will be lesser in glass.

r

rc

tc

1.6 (A)

1.7 (B)

1.8 (C)

1.9 (C)




At steady state condition, heat will pass through water-glass interface from either side i.e. from water to glass and vice versa.

At interface ( 0)y , following the no slip condition, water will not move. Hence at 0y , heat transfer will be due to conduction in water and then the same amount of heat will be convected into water.

0

( )water water glass

y

dTk A h A T T

dy

40.6 1 10 8h 2750 W/m Kh Hence, the correct option is (D).

One dimensional unsteady state heat transfer equation in spherical coordinate system is

22

Internal heat Unsteady conduction heatgeneration term termtransfer term

1 1T q Tr

r r r k t


Key Point The one dimensional time dependent heat conduction equation can be written more compactly as in a simple manner as,

1 nn

T q c Tr

r r r k k t

Where, n = 0, for rectangular section n = 1, for cylindrical section n = 2, for spherical section.

k

c

Where, is called thermal diffusivity and ability to transport heat through conduction.

Given : Temperature difference between innermost and outermost surface 0

1 2( ) 600 CT T Thermal conductivity of stainless steel ( ) 19 W/mKsk Thermal conductivity of asbestos ( ) 0.2 W/mKak

Internal diameter of tube ( ) 2cmid

Outer diameter of tube 0( ) 5cmd

Thickness of insulation ( ) 3cmit Length of tube (L) = 1 m

Insulation diameter of asbestos, 2 5 2 3 11 cmc o id d t

0 5ln ln2

2 2 19 1i

steels

d

dR

k L

0.00767 K/WsteelR

0

11ln ln5

2 2 0.2 1

c

asbestosa

d

dR

k L

0.6274asbestosR K/W

WaterGlass

(Stagnation Layer)048 C

040 C

y

ri

tc

r0

rc

L = 1 m

Insulation

1T aT

2T

steelR asbestosR

1.10 (D)

1.11 (B)

1.12 (C)




1 2 600 6000.00767 0.6274th s a

T TQ

R R R

944.77 944.72 W/mQ Hence, the correct option is (C).

1h c

s

T Tq

R

and 2

h c

s c

T Tq

R R

Because s c sR R R 1 2q q

1Resistance

Q

Resistance is inversely proportional to thermal conductivity. Hence it is fact that conductivity of steel is less than conductivity of copper. s cR R

h i i cT T T T

Q Q

h i i cT T T T 2h c iT T T

2

h ci

T TT

Hence, the correct option is (D).

For one dimensional heat conduction, we have

2

2

1T T

x t

Since, (thermal diffucsivity) is constant as it a property, we have

2

2

T T

t x


Given : Thermal conductivity of 1st slab

1( ) 0.02 W/mKk Thermal conductivity of 2nd slab

2( ) 0.10 W/mKk

Thermal conductivity of 3rd slab

3( ) 0.04 W/mKk

Length of 1st slab 1( ) 0.5mL

Length of 2nd and 3rd slab 2 3( ) ( ) 0.25mL L Assumption : (i) Heat transfer is steady since there is no

indication of change with time. (ii) Heat transfer can be approximated as

being one-dimensional since it is predominantly in the x-direction.

(iii) Thermal conductivities are constant. (iv) Heat transfer by radiation is negligible. This given composite slab can redrawn in the form of thermal circuit.

Assuming unit width of slab,

11

1

LR

k A

0.50.02 1 1

25K/W

22

2

LR

k A

0.250.1 0.5 1

5K/W

33

3

LR

k A

0.250.04 0.5 1

12.5K/W

Overall thermal resistance,

2 31

2 3th

R RR R

R R

5 12.525(5 12.5)

28.6K/WthR Hence, the correct option is (C).

Key Point There is no variation in the horizontal (x-direction). So, we consider a unit depth and unit height portion of the slab since it is representative of the entire wall. Assuming any cross- section of the slab normal to the x-direction to be isothermal.

2R

A Bq

1R

3R

1.13 (D)

1.14 (D)

1.15 (C)




Given : 1 2

2i

T TT

Thermal circuit,

At steady state, 1 2Q Q

1 2 1 21 2

1 2

2 22

T T T TT T

Qb b

k A k A

1 1 2 2 1 2( ) ( )2 1

k T T k T TQ

1 22k k


Heat loss to surrounding in a circular pipe,

0

0

2 ( )ln[ / ]1 i

L TQ

r r

hr k

The addition of insulation to a cylindrical piece or spherical shell increase the conduction resistance of the insulation layer but decrease the convection resistance of the surface because of the increase in the outer surface area for convection. The heat transfer from the pipe may increase or decrease, depending on which effect dominates. Hence, the correct option is (A).

Given : Inner diameter of glass cylinder ( ) 0.03mid

Outer diameter of glass cylinder 0( ) 0.05md

Thermal conductivity of glass ( ) 1.05 W/mKk

This is case of radial heat transfer,

Resistance network for above figure is,

0ln

2i

th

r

rR

kL

0.05ln0.03

2 (1.05) 1

(Per unit length 1mL )

0.077 K/WthR per unit length


Given : Diameter of nuclear reactor fuel rod

3( ) 50mm 50 10 md Convective heat transfer coefficient

2 3 2( ) 1kW/m K 1 10 W/m Kh Heat generation with in the fuel element

6 3( ) 50 10 W/mq

Ambient temperature 0( ) 75 CT

Under steady state conditions, by energy conversion. generated convection lossQ Q

1T 2T

1 2

2

T T�� 2

b

k A1

2b

k A

Q Q

ri

r0

Hot fluid

1.0 m

iT 0T

�0ln /

2

ir rR

Lk�

�

Fuel element

Convective losssT T��

q�

1.16 (D)

1.17 (A)

1.18 (B)

1.19 (A)




( ) ( )sq V hA T T

2 ( )4 sq d L hA T T

2 ( )( )4 sq d L h dL T T

( )4 s

dq h T T

6 33 050 10 50 10 1 10 ( 75 )

4 sT

0625 75 700 CsT Hence, the correct option is (A).

Given : Thickness of plate ( ) 20 mmt

Uniform heat generation 3( ) 80MW/mq

Temperature at the left face 01( ) 160 CT

Temperature at the right face 02( ) 120 CT

Thermal conductivity of the plate ( ) 200 W/mKk

..Method 1.. From steady state one dimensional heat conduction equation with heat generation

2

2 0d T q

dx k

2

2

d T q

dx k

Integrating both sides,

1dT q

x Cdx k

…(i)

Again integrating,

2

1 22qx

T C x Ck

…(ii)

From equation (ii), At, 0,x 0160 CT 2 160C

At, 20mm,x 0120 CT

6

21

80 10120 (0.02) 0.02 1602 200

C

1 2000C

For maximum temperature 0dT

dx

1qx

Ck

1C kx

q 6

2000 20080 10

5mm


..Method 2.. Location of maximum temperature for steady state condition with internal heat generation when walls are maintain at different temperature.

2 1

max 2g

g

q LT Tkx

q L k

6

max 6

200 120 160 80 10 0.0280 10 0.02 2 200

x

3max 5 10 mx or 5 mm (from left face)


..Method 3..

0160 C

0120 C

q�

x

1T

2T

x L�0x �

1T

2T

x L� � x L�0x �

2L

0.01mx � � 0.01mx � �

1.20 (C)




max 2 1( )2 g

kx T T

q L

max 6

200 (120 160)2 80 10 0.01

x

3max 5 10 mx

max 5mmx (from mid place )

i.e., max 5 10 5 5mmx L (from left face).

Key Point In case of heat generation maximum temperature will occur where the heat is being generated. (i) If heat is being generated within the wall

at any location.

(a) If 1 2T T , x L (b) If 1 2T T , x L 1 2gQ Q Q

(c) maxx is location of maxT

At maxx , 0dT

dx

(i.e., slope of temperature profile is zero which represent maximum temperature).

(ii) If heat is being generated one side of the wall. Temperature will be maximum on that side of wall.

Here heat is being generated on left hand

face, hence 1T is the maximum temperature.

gQ Q 1 2

2gq LT T

kAL k

.

According to question, by using equation (ii),

2

1 22qx

T C x Ck

max 5( )x mmT

6280 10 (0.005)

400

0.005 2000 160 0

max 5mm( ) 165 CxT


Given : For surface Q :

10T

x

K/m,

Thermal conductivity of body ( ) 0.1k W/mK For surface P :

0T

x

Heat is transferred only in y direction.

T

maxT

1T1Q 2Q

2T

x

L

Plane where heat isbeing generatedat a rate of Qg

1T

2T

L

gQQ

0160 C

0120 C

q�

x

0165 C

1.21 (B)

1.22 (D)




For steady state condition,

P Q

T TkA kA

y x

( 2 )Q PA A

2T T

y x

2 10T

y

20T

y

K/m


Key Point Features of Fourier law : Fourier law is valid for all matter solid,

liquid or gas. The vector expression indicating that

heat flow rate is normal to an isotherm and is in the direction of decreasing temperature.

It cannot be derived from first principle. It helps to define the transport property

‘k’.

Thermal conductivity, 0k k bT

Since, heat conduction is one dimensional

Heat flux, q dTk

A dx

As x increase, temperature in the wall increase and so the thermal conductivity increase, then

the value of dT

dx

must decreases to ensure

constant heat flux. Hence, the correct option is (D).

gq Heat generation per unit volume.

Steady state, one dimensional heat conduction equation,

2

2 0gqd T

dx k

2

2gqd T

dx k

Integrating both sides,

1gqdT

x Cdx k

Again integration both sides,

21 22

gqT x C x C

k …(i)

Now, applying the boundary condition on equation (i), At 0,x 0T then,

2 0C

At ,x L 0100 CT then,

2

11002gq L

C Lk

1100

2gq L

CL k

Surface

Surface

0,0 CQ

0,100 CP

y

x

0100 C

00 C

q

0x � x L�x

1.23 (D)

1.24 (B)




So that, equation (i) becomes

2 100

2 2g gq x q L

T xk L k

This equation shows that temperature distribution in the wall will not be linear but parabolic. For maximum temperature,

0dT

dx

2 100 02 2

g gq x q L

k L k

1002

g

g

q Lkx

q L k

1002g

k Lx

q L

So the curve shows that maximum temperature will be more than that of 0100 C, which will be close to x L and heat will flow, both sides of this maximum temperature plane. Hence, the correct option is (B).

Let, iT Inner surface temperature of

cylindrical tube 0T Outer surface temperature of

cylindrical tube Using electrical analogy,

0i

th th

T T TQ

R R

…(i)

From Fourier law for one dimensional (radial) heat transfer through pipe,

Consider an elementary cylinder of radius ‘r’ and thickness dr. Area, 2A rL (where, L is length of tube) From Fourier law, (under steady state condition),

2dT dTQ kA k rL

dr dr

0 0

2i i

r T

r T

Q drdT

kLr

0 0ln2 i i

r Tr T

Qr T

kL

0 0(ln ln ) ( )2 i i

Qr r T T

kL

0

01 ln2

i

i

T TQ

r

kL r

…(ii)

Comparing equation (i) and (ii),

01 ln2th

i

rR

kL r


3

v

A

nV ck

N

Where, k = Thermal conductivity V = Mean particle speed = Mean free path vc = Molar heat capacity

AN = Avogadro's number

n = Particles per unit volume Gases transfer heat by direct collisions between molecules. As the temperature increases, the thermal conductivity increases due to increase in speed, movement and collisions in the molecules. From the above expression, by increasing mean particle speed, the thermal conductivity increases. Hence, the correct option is (A).

iTQ

oTthR

0r

ir

r

dr

1.25 (C)

1.26 (A)




Given : Thickness of wall ( ) 30mmL Thermal conductivity of wall ( ) 15W/mKk

Heat flux at 0x 5 2( ") 1 10 W/mq Convective heat transfer coefficient

2( ) 250 W/m Kh

Temperature of surrounding 0( ) 25 CT

For the given problem, the network of thermal resistance involves two resistances in series as shown in figure,

Conductive thermal resistance,

cond

LR

kA

Convective thermal resistance,

1convR

hA

Rate of heat transfer,

th

TQ

R

1 1

1cond conv

T T T TQ

LR RkA hA

Heat flux,

1"1

T Tq

Lk h

5 1 251 10 0.03 115 250

T

5 1 251 100.006T

5 01 (1 10 0.006) 25 625 CT

Hence, the temperature at 0x is 6250C.

Given : Thickness of 1st steel slab 1( ) 20mml

Thickness of 2nd steel slab 2( ) 20mml

Thickness of P material 3( ) 1mml

Heat flux 2( ) 10kW/mq

Thermal conductivity of 1st and 2nd steel slab

1 2( ) ( ) 10 W/mKk k

Temperature of 1st steel slab 1( ) 500KT

Temperature of 2nd steel slab 3( ) 360KT

As per steady state heat conduction equation,

According to question applying steady state heat conduction equation. Heat flux is given by,

1 2

31 2

1 2 3

" T Tq

ll l

k k k

3 3

2

500 3601000020 10 0.001 20 10

10 10k

0�x

x

025 C� �T

2T

1T

5 2" 1 10 W/m� �q

T1Rconductive RconvectiveT2 T� = 25 C

0

0.02 m 0.02 mm

2 360K�T2" 10 kW/m�q

1 500 K�T

2k

PSteelslab

Steelslab

1k 3k

1.27 625

1.28 0.1




2

14010000 0.0010.002 0.002k

2 0.1W/mKk

Hence, the thermal conductivity of material P is 0.1 W/mK.

Given : Thermal conductivity of 1st section 1( )k k

Thermal conductivity of 2nd section 2( ) 2k k

Temperature of left face of 1st section

1( ) 600KT

Temperature of right face of 2nd section

2( ) 300KT

At steady state, heat transfer is given by, Fourier’s equation of heat conduction, Q = – kA(dT/dx)

For 1st section,

1 1dT

Q k Adx

…(i)

For 2nd section,

2 2dT

Q k Adx

…(ii)

Now, equating equations (i) and (ii) 1 2Q Q

1 2dT dT

k A k Adx dx

1 2 21

( ) ( )i iT T k A T Tk A

l l

1 2( ) 2 ( )i ikA T T kA T T

l l

1 22 2i iT T T T

1 22 2 i iT T T T

1 22 3 iT T T

1 22 600 2 300 400K3 3i

T TT

Hence, the interface temperature iT of the composite wall is 400 K.

Given : Thermal conductivity ( ) 1.15W/mKk

Inner surface temperature 0( ) 1100 CiT

Outer surface temperature 00( ) 350 CT

Heat flux 2( ) 2500 W/mq

Fourier’s equation of heat conduction, Q = – kA(dT/dx)

0ik T Tq

L

1.15 1100 3502500

L

Solving, we get 0.345mL Hence, the design thickness of the wall to maintain a steady heat flux of 22500 W/mshould be 0.345 m.

l l

300 K600 K iT

k 2k

Heat flow

Q2Q1

1 2

0

0 350 CT �

01100 CiT �

22500W/mq �

L

1.29 400

1.30 0.345




Given : Diameter of pipe ( ) 4 cmd Convective heat transfer coefficient inside pipe

2( ) 15 W/m Kih

Convective heat transfer coefficient outside foam insulation 2

0 2 W/m Kh

Thermal conductivity of pipe ( ) 15 W/mKpipek

Thermal conductivity of foam insulation ( ) 0.1 W/mKfoamk

According to question it is given that the pipe is insulated by foam from outer side. So, we find the critical radius of insulation.

0

0.1 0.05m2

foamc

kr

h

Critical radius of insulation 5cmcr .

Hence, the critical radius of insulation is 5 cm.

Key Point In finding critical insulation the thermal conductivity of the insulation material is taken and hence, here there is no use of the thermal conductivity of the pipe.

Given : Diameter of electrical conductor ( ) 10mmd ( ) 5 mmr Thickness of insulation ( ) 2mmt Thermal conductivity of insulation ( ) 0.08W/mKik

Convective coefficient of insulation surface 2( ) 10 W/m Kih

Critical radius, 0.08 0.008m10

ic

i

kr

h

8 mmcr

Thickness of critical insulation is, 8 5 3mmc ct r r Heat flow for various radius is shown in above figure. Heat loss will increase up to thickness 3 mm of insulation then it will decreases. Hence, the correct option is (C).

Given : Thermal conductivity of brick wall ( ) 0.9 W/mKk Thickness of brick wall ( ) 0.18mL Outside (surrounding) temperature of air

0( ) 5 CT

Temperature of room 0( ) 27 CiT Convective heat transfer coefficient

2( ) 20 W/m Kh

Total thermal resistance,

1 11

thLR

h k

1 1 41 0.1820 0.9

thR

Q

Increase

Decrease

2 mm

3 mm

Thicknessof insulation

220W/m Kh �027 CiT �

0.18mL �

05 CT� � �Wall

0.9W/mkk �

1.31 5

1.32 (C)

1.33 (C)




1 W/K4thR

Heat loss inside the room is,

227 ( 5) 128W/m14

th

TQ

R


Given : Radius of uranium fuel rod ( ) 5mm 0.005 mr Heat generation in nuclear reactor

7 3( ) 4 10 W/mgq

Convective heat transfer coefficient 2( ) 1000 W/m Kh

Surrounding temperature 0( ) 25 CT

For steady state conduction, gen convQ Q

gq V hA T

2 (2 )gq r L h rL T

2gq r h T

74 10 0.005 2 1000( )sT T

74 10 0.005( ) 100

2 1000sT T

100sT T

0100 25 125 CsT

But in this question temperature asked in Kelvin so, 125 273 398KsT


Given : Outer radius of plastic sleeve 0( ) 1mmr

Electric current carrying wire radius( ) 0.5mmr

Thermal conductivity of plastic ( ) 0.15W/mKk Convective heat transfer coefficient

2( ) 25W/m Kh According to question it is said that the wire is insulated by a plastic sleeve which shown in figure below,

Critical radius, 30.15 6 10 m25c

kr

h

6mmcr Here, we can see that the critical radius is greater than the radius of coating 0( )cr r . Therefore, addition of plastic sleeve will increase the heat transfer. Hence, the correct option is (A).

L Length, k Thermal conductivity

Thermal resistance given by 2 1ln( / )2r r

kL


Given : Temperature at face 1 0

1( ) 130 CT

Temperature at face 3 03( ) 30 CT

Thermal conductivity of slab A ( ) 20 W/mKAk

r

r0

Wire

Plastic

h = 25 W/m K2

k=

0.15 W/mK

k

1r

L

r2r2

1.34 (B)

1.35 (A)

1.36 (A)

1.37 67.5




Thermal conductivity of slab B ( ) 100 W/mKBk Length of slab A ( ) 0.1mAL Length of slab B ( ) 0.3mBL

The thermal resistance network for the representative section of slab becomes as shown in figure. Then the individual thermal resistance is,

1

0.120

A

A

LR

k and 2

0.3100

B

B

LR

k

For steady one-dimensional heat conduction,

1 3 1 2

1 2 1

T T T TQ

R R R

2130130 300.1 0.3 0.120 100 20

TQ

21301000.8 0.1100 20

T

2130 62.5T

02 130 62.5 67.5 CT

Hence, the interface temperature 2T is 67.5 0C .

Given : Diameter of solid cylinder rod ( ) 10 mmD Heat generated in rod 7 3( ) 4 10 W/mgq Thermal conductivity of the rod ( ) 25W/mKk

Temperature distribution in a cylindrical rod with uniform heat generation under steady state is given by

2 2( )

4g

r s

q R rT T

k

At center 0r and 0rT T

2

0 4g

s

q RT T

k

And radius of solid cylinder,

10 5 mm2 2D

R

27 3

00

4 10 5 10 10 C4 25sT T

Hence, the temperature difference between the centre and the surface of the rod is 010 C .

Key Point Under steady state one dimensional heat conduction in cylinder is derived below,

gq rTr

r r k

…(i)

Integrating above equation, we get

2

12gq rT

r cr k

1

2gq r cT

r k r

At 00,r T T and 0T

r

1 0c

2

gq rT

r k

…(ii)

Again integrating equation (ii),

2

24gq r

T ck

…(iii)

A B

1 2 3

0

1 130 CT � 0

3 30 CT �20

W/m.K

Ak �100 W/m.KBk �

Q Q Q

0.1 m 0.3 m

T1 = 130 C0 R1 R2T2 T3 = 30 C

0

Q

R

Ts

T0

1.38 10




At , sr R T T

2

24g

s

q RT c

k

2

2 4g

s

q Rc T

k

Putting the value of 2c in equation (iii),

2 2

4 4g g

s

q r q RT T

k k

2 2

4 4g g

s

q R q rT T

k k

2 2( )4

gs

qT T R r

k …(iv)

In case of solid cylinder value of r is zero then the equation (iv) becomes,

2

4g

s

q RT T

k

Given : Thickness of slab ( ) 20cmL

Thermal conductivity of slab ( ) 2.5W/mKk

Convective heat transfer coefficient 2( ) 10 W/m Kh

Ambient temperature 0( ) 45 CT

Surface temperature 0( ) 25 CQT

Considering steady state condition,

Thermal resistance circuit for the given problem as,

1 110convR

hA A

0.22.5cond

LR

kA A

For steady state one dimensional heat generation,

conv cond

T TQ

R R

1 0.210 2.5

P QPT TT T

Q

A A

45 251 0.2

10 2.5

P PT TQ

45 250.1 0.08

P PT TQ

0.08(45 ) 0.1( 25)P PT T

033.88 CPT

Hence, the steady state temperature pT of the

side which is exposed to the fluid is 033.88 C .

1 2th th thR R R

1 2

2

e

L L L

k k k

1 2

1 2 1 2

2 1 1

e

k k

k k k k k

1 2

1 2

2e

k kk

k k


L = 20 cm

TQ = 25 C0h = 10 W/m K

2

0= 45 CT�

TP

k = 2.5 W/mK

045 CT� � PT

convR condR

025 CQT �

k1 k2

L L

T�

dQ

ke

2L

T�

Q

1.39 33.88

1.40 (A)




Fourier law, assumptions are, 1. Steady state, one dimensioned. 2. No internal heat generation. 3. Thermal conductivity = Constant

dTQ kA

dx

Given, area is varying linearly. A a bx (a and b are constant) Assuming, 0x , 0T and x L , T T

So, ( )Q dT

K a bx dx

0 0( )L TQ

dx dTK a bx

ln( )Qa bL T

bK

So, temperature is varying logarithmic. Hence, the correct option is (C).

According to question,

10000 ( )gE L A

10000 1 1 10000 WgE

and, 1 11( ) conv condhA T T q q

1

100 (100 30) condq

1

7000 Wcondq

Again, 2 2( )convq hA T T 2100 1 ( 30)T

2 22100( 30)conv condq T q

At steady state, g outE E

1 2g cond condE q q

210000 7000 100( 30)T

02

3000 30 60 C100

T

Hence, the right extreme face temperature 2T is 600C.

T1 = 100 C0

h = 100 W/m K2

T2 = ?

T� = 30

qcond 1 qcond 2

T� = 30

h = 100 W/m K2

1 m 1 m 1 m

5 W/m Kk � 5 W/m Kk �

1.41 (C)

1.42 60

gate 20204. refrigeration & air-conditioning 4.1 - 4.000 1. vapour refrigeration 4.1 - 4.0 2....

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