Steady State Conduction ( Heat Transfer ) MCQs – Mechanical Engineering MCQs
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Mechanical Engineering MCQs – Steady State Conduction ( Heat Transfer ) MCQs
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Conduction Through a Plane Wall
1. In Cartesian coordinates the heat conduction equation is given by
a) d2t/dx2 + d2t/dy2 + d2t/dz2 + q g = (1/α) (d t/d T)
b) 2d2t/dx2 + d2t/dy2 + d2t/dz2 + 34q g = (d t/d T)
c) d2t/dx2 + 3d2t/dy2 + d2t/dz2 = (1/α) (d t/d T)
d) 4d2t/dx2 + d2t/dy2 + d2t/dz2 + 1/2q g = (1/α) (d t/d T)
Answer: a
Explanation: This is one dimensional heat conduction through a homogenous, isotropic wall with constant thermal conductivity.
2. The temperature distribution in a large thin plate with uniform surface temperature will be
(Assume steady state condition)
a) Logarithmic
b) Hyperbolic
c) Parabolic
d) Linear
Answer: d
Explanation: The temperature increases with increasing value of x. Temperature gradient will be positive i.e. linear.
3. Let us assume two walls of same thickness and cross-sectional area having thermal conductivities in the ratio 1/2. Let us say there is same temperature difference across the wall faces, the ratio of heat flow will be
a) 1
b) 1/2
c) 2
d) 4
Answer: b
Explanation: Q1 = k1 A1 d t1/δ1 and Q2 = k2A2 d t2/δ 2
Now, δ1 = δ2 and A1 = A2 and d t1 = d t2
So, Q1/Q2 = ½.
4. The interior of an oven is maintained at a temperature of 850 degree Celsius by means of a suitable control apparatus. The oven walls are 500 mm thick and are fabricated from a material of thermal conductivity 0.3 W/m degree. For an outside wall temperature of 250 degree Celsius, workout the resistance to heat flow
a) 0.667 degree/W
b) 1.667 degree/W
c) 2.667 degree/W
d) 3.667 degree/W
Answer: b
Explanation: R t = 0.5/0.3 = 1.667 degree/W.
5. A plane slab of thickness 60 cm is made of a material of thermal conductivity k = 17.45 W/m K. Let us assume that one side of the slab absorbs a net amount of radiant energy at the rate q = 530.5 watt/m2. If the other face of the slab is at a constant temperature t2 = 38 degree Celsius. Comment on the temperature with respect to the slab?
a) 87.5 degree Celsius
b) 32 degree Celsius
c) 47.08 degree Celsius
d) 32.87 degree Celsius
Answer: c
Explanation: Heat flux, q = k (t s – t f) / Thickness. So, t s = 56.17 degree Celsius. Now, t = t s + (t f – t s) x/Thickness.
6. The rate of heat transfer for a plane wall of homogenous material with constant thermal conductivity is given by
a) Q = kA (t1-t2)/δ
b) Q = 2kAx/ δ
c) Q = 2kAδx
d) Q = 2k/δ x
Answer: a
Explanation: Computations for heat flow can be made by substituting the value of temperature gradient into the general equation. The heat flow somehow doesn’t depend on x.
7. In case of homogeneous plane wall, there is a linear temperature distribution given by
a) t = t1 + (t2-t1) δ/x
b) t = t2 – (t2-t1) x/ δ
c) t = t1 + (t2-t1) x
d) t = t1 + (t2-t1) x/ δ
Answer: d
Explanation: The expression for steady state temperature distribution can be set up by integrating the Fourier rate equation.
8. The rate of convective heat transfer between a solid boundary and adjacent fluid is given by
a) Q = h A (t s – t f)
b) Q = h A
c) Q = (t s – t f)
d) Q = h (t s – t f)
Answer: a
Explanation: Here, h is heat transfer coefficient i.e. convective.
9. A homogeneous wall of area A and thickness δ has left and right hand surface temperatures of 0 degree Celsius and 40 degree Celsius. Determine the temperature at the center of the wall
a) 10 degree Celsius
b) 20 degree Celsius
c) 30 degree Celsius
d) 40 degree Celsius
Answer: b
Explanation: At the midpoint x = δ/2. So, temperature = 40 + (0 – 40)/2 = 20 degree Celsius.
10. A rod of 3 cm diameter and 20 cm length is maintained at 100 degree Celsius at one end and 10 degree Celsius at the other end. These temperature conditions are attained when there is heat flow rate of 6 W. If cylindrical surface of the rod is completely insulated, determine the thermal conductivity of the rod material
a) 21.87 W/m degree
b) 20.87 W/m degree
c) 19.87 W/m degree
d) 18.87 W/m degree
Answer: d
Explanation: Q = k A C (t 1 – t 2)/δ = 0.318 k.
Conduction Through a Composite Wall
1. A composite wall generally consists of
a) One homogenous layer
b) Multiple heterogeneous layers
c) One heterogeneous layer
d) Multiple homogenous layers
Answer: b
Explanation: Walls of houses where bricks are given a layer of plaster on either side.
2. Three metal walls of the same thickness and cross sectional area have thermal conductivities k, 2k and 3k respectively. The temperature drop across the walls (for same heat transfer) will be in the ratio
a) 3:2:1
b) 1:1:1
c) 1:2:3
d) Given data is insufficient
Answer: a
Explanation: As, δ1 = δ2 = δ3 and cross sectional areas are same i.e. temperature drop varies inversely with thermal conductivity.
3. A composite wall is made of two layers of thickness δ1 and δ2 having thermal conductivities k and 2k and equal surface area normal to the direction of heat flow. The outer surface of composite wall are at 100 degree Celsius and 200 degree Celsius. The minimum surface temperature at the junction is 150 degree Celsius. What will be the ratio of wall thickness?
a) 1:1
b) 2:1
c) 1:2
d) 2:3
Answer: c
Explanation: Q = k 1 A 1 d t 1 / δ1 = k 2 A 2 d t 2 / δ2 Also areas are same.
4. Let us say thermal conductivity of a wall is governed by the relation k = k0 (1
+ α t). In that case the temperature at the mid-plane of the heat conducting wall would be
a) Av. of the temperature at the wall faces
b) More than average of the temperature at the wall faces
c) Less than average of the temperature at the wall faces
d) Depends upon the temperature difference between the wall faces
Answer: b
Explanation: k0 is thermal conductivity at 0 degree Celsius. Here β is positive so it is more than average of the temperature at the wall faces.
5. Heat is transferred from a hot fluid to a cold one through a plane wall of thickness (δ), surface area (A) and thermal conductivity (k). The thermal resistance is
a) 1/A (1/h1 + δ/k + 1/h2)
b) A (1/h1 + δ/k + 1/h2)
c) 1/A (h1 + δ/k + h2)
d) A (h1 + δ/k + 1/h2)
Answer: a
Explanation: Net thermal resistance will be summation of resistance through plane wall and from left side and right side of the wall.
6. Find the heat flow rate through the composite wall as shown in figure. Assume one dimensional flow and take
k 1 = 150 W/m degree
k 2 = 30 W/m degree
k 3 = 65 W/m degree
k 4 = 50 W/m degree
AB = 3 cm, BC = 8 cm and CD = 5 cm. The distance between middle horizontal line from the top is 3 cm and from the bottom is 7 cm
a) 1173.88 W
b) 1273.88 W
c) 1373.88 W
d) 1473.88 W
Answer: b
Explanation: Q = d t/ R T. R T = R 1 + R e q + R 2 = 0.02 + 0.01469 + 0.1 = 0.2669 degree/W.
7. A pipe carrying steam at 215.75 degree Celsius enters a room and some heat is gained by surrounding at 27.95 degree Celsius. The major effect of heat loss to surroundings will be due to
a) Conduction
b) Convection
c) Radiation
d) Both conduction and convection
Answer: c
Explanation: As there is temperature difference so radiation suits well.
8. “Radiation cannot be affected through vacuum or space devoid of any matter”. True or false
a) True
b) False
Answer: b
Explanation: It can be affected only by air between molecules and vacuum of any matter.
Fourier Equation And Thermal Conductivity MCQs
9. A composite slab has two layers having thermal conductivities in the ratio of 1:2. If the thickness is the same for each layer then the equivalent thermal conductivity of the slab would be
a) 1/3
b) 2/3
c) 2
d) 4/3
Answer: d
Explanation: 2(1) (2)/1+2 = 4/3.
10. A composite wall of a furnace has two layers of equal thickness having thermal conductivities in the ratio 2:3. What is the ratio of the temperature drop across the two layers?
a) 2:3
b) 3:2
c) 1:2
d) log e 2 : log e 3
Answer: b
Explanation: We know that temperature is inversely proportional to thermal conductivity, so the ratio is 2:3.
Conduction Through a Cylindrical Wall
1. Typical examples of heat conduction through cylindrical tubes are not found in
a) Power plants
b) Oil refineries
c) Most process industries
d) Aircrafts
Answer: d
Explanation: Boilers have tubes in them, the condenser consist of a bank of tubes.
2. The rate of heat conduction through a cylindrical tube is usually expressed as
a) Per unit length
b) Per unit area
c) Only length
d) Only area
Answer: a
Explanation: It is expressed as per unit length rather than per unit area as done for plane walls.
3. A steel pipe of 20 mm inner diameter and 2 mm thickness is covered with 20 mm thick of fiber glass insulation (k = 0.05 W/m degree). If the inside and outside convective coefficients are 10 W/m2 degree and 5 W/m2 degree, calculate the overall heat transfer coefficient based on inside diameter of pipe. In the diagram, the diameter of small circle is 20 mm
a) 1.789 W/m2 degree
b) 2.789 W/m2 degree
c) 3.789 W/m2 degree
d) 4.789 W/m2 degree
Answer: b
Explanation: Q = 2 π l (t i – t 0)/ [(1/h i r i) + log e (r 3/r 2) (1/k 2) + (1/h 0 r 3)].
4. Logarithmic mean area of the cylindrical tube is given as
a) 2πr m
b) πr ml
c) 2πr ml
d) 2r ml
Answer: c
Explanation: It is known as equivalent area and r m = r2-r1/log e (r2/r1).
5. A hot fluid is being conveyed through a long pipe of 4 cm outer diameter and covered with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the surroundings to one-third of the present rate by further covering with same insulation. Calculate the additional thickness of insulation
a) 11 cm
b) 12 cm
c) 13 cm
d) 14 cm
Answer: b
Explanation: Heat loss with existing insulation = 2 π k l (t 1 – t 2)/log e (r 2/r 1) and heat loss with additional insulation = 2 π k l (t 1 – t 2)/log e (r 2 + x/r 1).
6. The heat flow equation through a cylinder of inner radius r1 and outer radius r2 is desired to be written in the same form as that for heat flow through a plane wall. For wall thickness (r 2-r 1) the area will be
a) A1 + A2/2
b) A1 + A2
c) A2 – A1/ log e (A2/A1)
d) A1 + A2/2 log e (A2/A1)
Answer: a
Explanation: Here A 1 and A 2 are the inner and outer surface areas of tubes. The net area is A M.
7. A cylinder of radius r and made of material of thermal conductivity k 1 is surrounded by a cylindrical shell of inner radius r and outer radius 2r. This outer shell is made of a material of thermal conductivity k 2. Net conductivity would be
a) k 1 + 3 k 2/4
b) k 1 + k 2/4
c) k 1 + 3k 2
d) k 1 + k 2
Answer: a
Explanation: Heat flowing per second is given by = k1 (πr2) (t1-t2) δ. Shell heat is k2 π [(2r)2 – r2 ] (t1 – t2)/ δ.
8. For steady state and a constant value of thermal conductivity, the temperature distribution associated with radial convection through a cylinder is
a) Linear
b) Parabolic
c) Logarithmic
d) Exponential
Answer: c
Explanation: As thermal conductivity is constant so we get a profile that is logarithmic in nature.
9. A cylindrical cement tube of radii 0.05 cm and 1.0 cm has a wire embedded into it along its axis. To maintain a steady temperature difference of 120 degree Celsius between the inner and outer surfaces, a current of 5 ampere is made to flow in the wire. Find the amount of heat generated per meter length. Take resistance of wire equal to 0.1 ohm per cm of length
a) 150 W/m length
b) 250 W/m length
c) 350 W/m length
d) 450 W/m length
Answer: b
Explanation: Resistance of wire = 10 ohm per m length. Heat generated = (5) 2 10 = 250 W/m length.
10. A stainless steel tube with inner diameter 12 mm, thickness 0.2 mm and length 50n cm is heated electrically. The entire 15 k W of heat energy generated in the tube is transferred through its outer surface. Find the intensity of the current flow
a) 52 amps
b) 62 amps
c) 72 amps
d) 82 amps
Answer: a
Explanation: Power generated = 15 k W = 15000 W. Therefore, intensity of current flow = (15000/5.548) ½ = 52 amps.
Conduction Through a Sphere
1. The temperature distribution associated with radial conduction through a sphere is represented by
a) Parabola
b) Hyperbola
c) Linear
d) Ellipse
Answer: b
Explanation: As conduction is radial i.e. in outward direction, so it follows the hyperbola equation..
2. The thermal resistance for heat conduction through a spherical wall is
a) (r2-r1)/2πkr1r2
b) (r2-r1)/3πkr1r2
c) (r2-r1)/πkr1r2
d) (r2-r1)/4πkr1r2
Answer: d
Explanation: We get this on integrating the equation Q = -k A d t/ d r from limits r1 to r2 and T1 to T2.
3. The rate of conduction heat flow in case of a composite sphere is given by
a) Q = t1 – t2/ (r2 – r1)/4πk1r1r2 + (r3 – r2 )/4πk2r2r3
b) Q = t1 – t2/ (r2 – r1)/4πk1r1r2 + (r3 – r2 )/4πk2r2r3
c) Q = t1 – t2/ (r2 – r1)/4πk1r1r2 + (r3 – r2 )/4πk2r2r3
d) Q = t1 – t2/ (r2 – r1)/4πk1r1r2 + (r3 – r2 )/4πk2r2r3
Answer: c
Explanation: Here, convective film coefficient at the inner and outer surfaces are also considered.
4. The thermal resistance for heat conduction through a hollow sphere of inner radius r1 and outer radius r2 is
a) r 2 – r 1/4πk r 1r 2
b) r 2 /4πk r 1r 2
c) r 1/4πk r 1r 2
d) 4πk r 1r 2
Answer: a
Explanation: As Q = d t/ R T. Here R T is thermal resistance.
5. A spherical vessel of 0.5 m outside diameter is insulated with 0.2 m thickness of insulation of thermal conductivity 0.04 W/m degree. The surface temperature of the vessel is – 195 degree Celsius and outside air is at 10 degree Celsius. Determine heat flow per m2 based on inside area
a) – 63.79 W/m2
b) – 73.79 W/m2
c) – 83.79 W/m2
d) – 93.79 W/m2
Answer: b
Explanation: Heat flow based on inside area = Q/4 π r 2 = – 73.79 W/m2.
6. The quantity d t/Q for conduction of heat through a body i.e. spherical in shape is
a) ln (r2/r1)/2πLk
b) ln (r2/r1)/πLk
c) ln (r2/r1)/2Lk
d) ln (r2/r1)/2πk
Answer: a
Explanation: We get this on integrating the equation Q = -k A d t/ d r from limits r1 to r2 and T1 to T2.
7. A spherical vessel of 0.5 m outside diameter is insulated with 0.2 m thickness of insulation of thermal conductivity 0.04 W/m degree. The surface temperature of the vessel is – 195 degree Celsius and outside air is at 10 degree Celsius. Determine heat flow
a) – 47.93 W
b) – 57.93 W
c) – 67.93 W
d) – 77.93 W
Answer: b
Explanation: Q = 4 π k r 1 r 2 (t 1 – t 2)/r 2 – r 1 = -57.93 W.
8. If we increase the thickness of insulation of a circular rod, heat loss to surrounding due to
a) Convection and conduction increases
b) Convection and conduction decreases
c) Convection decreases while that due to conduction increases
d) Convection increases while that due to conduction decreases
Answer: d
Explanation: In convection energy is transferred between solid and fluid but in conduction from T 1 to T 2.
9. The following data pertains to a hollow cylinder and a hollow sphere made of same material and having the same temperature drop over the wall thickness
Inside radius = 0.1 m and outside surface area = 1 square meter
If the outside radius for both the geometrics is same, calculate the ratio of heat flow in the cylinder to that of sphere?
a) 0.056
b) 2.345
c) 1.756
d) 3.543
Answer: c
Explanation: For sphere r 2 = (1/4 π) 1/2 = 0.282 m, for cylinder, l = A 2/2 r 2 π = 0.565 m.
10. The oven of an electric store, of total outside surface area 2.9 m2 dissipates electric energy at the rate of 600 W. The surrounding room air is at 20 degree Celsius and the surface coefficient of heat transfer between the room air and the surface of the oven is estimated to be 11.35 W/m 2 degree. Determine the average steady state temperature of the outside surface of the store
a) 38.22 degree Celsius
b) 48.22 degree Celsius
c) 58.22 degree Celsius
d) 68.22 degree Celsius
Answer: a
Explanation: Q = h A (t 0 – t a).
Shape Factor
1. “All the factors relating to the geometry of the sections are grouped together into a multiple constant called the shape factor”.
a) True
b) False
Answer: b
Explanation: They are grouped together into a single constant instead of multiple one.
2. Shape factor for a plane wall is equal to
a) A/δ
b) 2A/δ
c) 3A/δ
d) 4A/δ
Answer: a
Explanation: It should be obtained by dividing area with respect to length.
3. For a prescribed temperature difference, bodies with the same shape factor will allow heat conduction proportional to
a) k/2
b) 2k
c) k
d) k/4
Answer: c
Explanation: It is proportional to material thermal conductivity irrespective of size and configuration.
4. Shape factor for cylinder is
a) 6 π l/log e (r 2/r 1)
b) 4 π l/log e (r 2/r 1)
c) π l/log e (r 2/r 1)
d) 2 π l/log e (r 2/r 1)
Answer: d
Explanation: It is two times the rest of the product.
5. The annealing furnace for continuous bar stock is open at the ends and has interior dimensions of 0.6 m * 0.6 m * 1.5 m long with a wall 0.3 m thick all around. Calculate the shape factor for the furnace?
a) 15.24 m
b) 16.34 m
c) 14.54 m
d) 13.76 m
Answer: a
Explanation: Shape factor for 4 walls = 4 (area of wall)/ (thickness) = 4 (1.5) (0.6)/0.3 = 12 m. Shape factor for 4 edges = 4 (0.54) (1.5) = 3.24 m. Total = 15.24 m.
6. Shape factor for sphere is
a) 4 π r 1 r 2
b) 4 π r 1 r 2/r 2 – r 1
c) 4 π /r 2 – r 1
d) r 1 r 2/r 2 – r 1
Answer: b
Explanation: In case of sphere it is given in option b. Area of sphere is 4 π r 2.
7. Which is true regarding a complete rectangular furnace?
a) 6 walls, 12 edges and 6 corners
b) 0 walls, 2 edges and 4 corners
c) 6 walls, 12 edges and 8 corners
d) 2 walls, 6 edges and 8 corners
Answer: c
Explanation: It has 6 walls, 12 edges and 8 corners.
8. The shape factor for complete rectangular furnace is
Where a, b and c are the inside dimensions and d x is the wall thickness
a) 2/ d x (a b + b c + c a) + 4 (0.64) (a + b +c) + 8 (0.45) d x
b) 2/ d x (a b + b c + c a) + 4 (0.44) (a + b +c) + 8 (0.35) d x
c) 2/ d x (a b + b c + c a) + 4 (0.34) (a + b +c) + 8 (0.25) d x
d) 2/ d x (a b + b c + c a) + 4 (0.54) (a + b +c) + 8 (0.15) d x
Answer: d
Explanation: S edge = 0.54 times length of edge. S corner = 0.15 d x.
9. For the same material and same temperature difference, the heat flow in terms of shape factor is given by
a) S k d t
b) k d t/S
c) 2S k/d t
d) 2S/3
Answer: a
Explanation: The heat flow is proportional to the shape factor.
10. For the same amount of fabrication material and same inside capacity, the heat loss is lowest in
a) Sphere
b) Cylinder
c) Rectangle
d) Cube
Answer: b
Explanation: For cylinder, it is 0.439 i.e. lowest.
Effect of Variable Conductivity
1. With variable thermal conductivity, Fourier law of heat conduction through a plane wall can be expressed as
a) Q = -k0 (1 + β t) A d t/d x
b) Q = k0 (1 + β t) A d t/d x
c) Q = – (1 + β t) A d t/d x
d) Q = (1 + β t) A d t/d x
Answer: a
Explanation: Here k0 is thermal conductivity at zero degree Celsius.
2. The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation
K = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg
Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
a) 1355.3 kJ/m2 hr
b) 2345.8 kJ/m2 hr
c) 1745.8 kJ/m2 hr
d) 7895.9 kJ/m2 hr
Answer: c
Explanation: Q = -k A d t/d x, Q d x = – k A d t = – (1.45 + 0.5 * 10-5 t2) A d t. Integrating over the wall thickness δ, we get Q = 436.45/0.25 = 1745.8 kJ/m2 hr.
3. A plane wall of thickness δ has its surfaces maintained at temperatures T1 and T2. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k0 T2. Find the expression to work out the steady state heat conduction through the wall?
a) Q = 2A k0 (T 1 3 – T 2 3)/3 δ
b) Q = A k0 (T 1 3 – T 2 3)/3 δ
c) Q = A k0 (T 1 2 – T 2 2)/3 δ
d) Q = A k0 (T 1 – T 2)/3 δ
Answer: b
Explanation: Q = -k A d t/d x = k0 T2 A d t/d x. Separating the variables and integrating within the prescribed boundary conditions, we get Q = A k0 (T 1 3 – T 2 3)/3 δ.
4. The mean thermal conductivity evaluated at the arithmetic mean temperature is represented by
a) km = k0 [1 + β (t1 – t2)/2].
b) km = k0 [1 + (t1 + t2)/2].
c) km = k0 [1 + β (t1 + t2)/3].
d) km = k0 [1 + β (t1 + t2)/2].
Answer: d
Explanation: At arithmetic mean temperatures i.e. (t1 + t2)/2.
5. With respect to the equation k = k0 (1 +β t) which is true if we put β = 0?
a) Slope of temperature curve is constant
b) Slope of temperature curve does not change
c) Slope of temperature curve increases
d) Slope of temperature curve is decreases
Answer: a
Explanation: As temperature profile is linear so it is constant.
6. The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material?
a) 0.81 W/m K
b) 0.71 W/m k
c) 0.61 W/m K
d) 0.51 W/m K
Answer: b
Explanation: Q = k A (t 1 – t 2)/δ. So, k = 0.71 W/m K.
7. If β is greater than zero, then choose the correct statement with respect to given relation
k = k0 (1 +β t)
a) k doesn’t depend on temperature
b) k depends on temperature
c) k is directly proportional to t
d) Data is insufficient
Answer: c
Explanation: k increases with increases temperature.
8. The unit of thermal conductivity doesn’t contain which parameter?
a) Watt
b) Pascal
c) Meter
d) Kelvin
Answer: b
Explanation: Its unit is W/m K.
9. The temperatures on the two sides of a plane wall are t1 and t2 and thermal conductivity of the wall material is prescribed by the relation
K = k0 e (-x/δ)
Where, k0 is constant and δ is the wall thickness. Find the relation for temperature distribution in the wall?
a) t 1 – t x / t 1 – t 2 = x
b) t 1 – t x / t 1 – t 2 = δ
c) t 1 – t x / t 1 – t 2 = δ/x
d) t 1 – t x / t 1 – t 2 = x/δ
Answer: d
Explanation: Q = -k A d t/d x = -k0 e (-x/δ) d t/d x. Separating the variables and upon integration, we get Q/k0 A = (t 1 – t 2)/ δ (e – 1). Therefore heat transfer through the wall, Q = k0 A (t 1 – t 2)/ δ (e – 1). At x = x and t = t x we get the answer.
10. “If β is less than zero, then with respect to the relation k = k0 (1 + β t), conductivity depends on surface area”.
a) True
b) False
Answer: b
Explanation: k decreases with increasing temperature.
Critical Thickness of Insulation
1. A cable of 10 mm outside is to be laid in an atmosphere of 25 degree Celsius (h = 12.5 W/m2 degree) and its surface temperature is likely to be 75 degree Celsius due to heat generated within it. How would the heat flow from the cable be affected if it is insulated with rubber having thermal conductivity k = 0.15 W/m degree?
a) 43.80 W per meter length
b) 53.80 W per meter length
c) 63.80 W per meter length
d) 73.80 W per meter length
Answer: b
Explanation: Q = 2 π d t/ (1/k) log e(r c/r 0) = 53.80 W per meter length.
2. Chose the correct one with respect to the critical radius of insulation
a) There is more heat loss i.e. conductive
b) There occurs a decrease in heat flux
c) Heat loss increases with addition of insulation
d) Heat loss decreases with addition of insulation
Answer: c
Explanation: For a pipe heat loss is more at the critical radius.
3. A heat exchanger shell of outside radius 15 cm is to be insulated with glass wool of thermal conductivity 0.0825 W/m degree. The temperature at the surface is 280 degree Celsius and it can be assumed to remain constant after the layer of insulation has been applied to the shell. The convective film coefficient between the outside surface of glass wool and the surrounding air is estimated to be 8 W/m2 degree. What is the value of a critical radius?
a) 9.31 mm
b) 10.31 mm
c) 11.31 mm
d) 12.31 mm
Answer: b
Explanation: Critical radius of insulation = k/h = 0.0825/8 = 0.01031 m = 10.31 mm.
4. For an object i.e. spherical the value of critical radius would be
a) 2k/3h
b) 3k/h
c) 2k/h
d) k/h
Answer: c
Explanation: It depends on the variation of angle with layers of insulation.
5. Maximum value of critical radius is
a) 0.01 m
b) 0.04 m
c) 0.06 m
d) 0.0001 m
Answer: a
Explanation: K for common insulating material is 0.05 W/ m degree.
6. An electric cable of aluminum (k = 240 W/ m degree) is to be insulated with rubber (k = 6 W/ square meter degree). If the cable is in air (h = 6 W/square meter degree). Find the critical radius?
a) 80 mm
b) 160 mm
c) 40 mm
d) 25 mm
Answer: d
Explanation: Critical radius = 0.15/6 = 0.025 m = 25 mm.
7. The value of critical radius in case of a cylindrical hollow object is
a) 2k/h
b) 2h/k
c) k/h
d) h/k
Answer: c
Explanation: Unit is meter.
8. A wire of radius 3 mm and 1.25 m length is to be maintained at 60 degree Celsius by insulating it by a material of thermal conductivity 0.175 W/m K. The temperature of surrounding is 20 degree Celsius with heat transfer coefficient 8.5 W/ m2 K. Find percentage increase in heat loss due to insulation?
a) 134.46 %
b) 124.23 %
c) 100.00 %
d) 12.55 %
Answer: a
Explanation: Q = 8.5 (2 π 0.003 1.25) (60 – 20) = 8.01 W. % increase = (18.78 – 8.01/8.01) (100) = 134.46 %.
9. A pipe of outside diameter 20 mm is to be insulated with asbestos which has a mean thermal conductivity of 0.1 W/m degree. The local coefficient of convective heat to the surroundings is 5 W/square meter degree. Find the critical radius of insulation for optimum heat transfer from a pipe?
a) 10 mm
b) 20 mm
c) 30 mm
d) 40 mm
Answer: b
Explanation: Critical radius of insulation = k/h0 = 0.1/5 = 0.02 m = 20 mm.
10. For insulation to be properly effective in restricting heat transmission, the pipe radius r0 will be
a) Greater than critical radius
b) Less than critical radius
c) Equal to critical radius
d) Greater than or equal to critical radius
Answer: d
Explanation: Addition of insulating material doesn’t always decrease in the heat transfer rate.