Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

# Multiple choice question for engineering

## Set 1

1. Condensation refers to a change from the
a) Solid to a liquid phase
b) Vapor to a liquid phase
c) Liquid to a solid phase
d) Liquid to a vapor phase

Answer: b [Reason:] Condensation is a convective heat transfer process that is associated with change in the phase of a fluid.

2. Condensation process is very common in
(i) Boilers
(ii) Condensers
(iii) Evaporators
Identify the correct statements
a) 1 and 2
b) 2 and 3
c) 1, 2 and 3
d) 1 and 3

Answer: c [Reason:] This process is very common in power plants and refrigeration systems.

3. The convective coefficients for condensation usually lie in the range
a) 30-300 W/m2 K
b) 60-3000 W/m2 K
c) 300-10000 W/m2 K
d) 2500-10000 W/m2 K

Answer: d [Reason:] The convective coefficient for condensation should be high because condensation refers to a change from the vapor to a liquid phase.

4. Drop wise condensation usually occurs on
a) Oily surface
b) Glazed surface
c) Smooth surface
d) Coated surface

Answer: a [Reason:] It generally occurs on oily surface. It is the convective heat transfer process that is associated with change in the phase of a fluid.

5. Consider the following statements
(i) If a condensing liquid does not wet a surface, then drop wise condensation will not take place on it
(ii) Drop wise condensation gives a higher transfer rate than film wise condensation
(iii) Reynolds number of condensing liquid is based on its mass flow rate
(iv) Suitable coating or vapor additive is used to promote film wise condensation
Identify the correct statement
a) 1 and 2
b) 2, 3 and 4
c) 4 only
d) 1, 2 and 3

Answer: d [Reason:] All are correct except the last one because suitable coating or vapor additive is not used to promote film wise condensation.

6. Depending upon the behavior of condensate up on the cooled surface, the condensation process are classified into how many distinct modes?
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] Film condensation and drop wise condensation.

7. A plate condenser was designed to be kept vertical. How would the condensation coefficient be effected if due to site constraints, it has to be kept at 60 degree to the horizontal?
a) 1.53% reduction in condensation coefficient
b) 2.53% reduction in condensation coefficient
c) 3.53% reduction in condensation coefficient
d) 4.53% reduction in condensation coefficient

Answer: c [Reason:] h VER = 0.943 [k 3 p 2 g h f g/δ l (t sat – t s)] 0.25, h INC = 0.943 [k 3 p 2 g sin α h f g/δ l (t sat – t s)] 0.25.

8. Saturated steam is allowed to condense over a vertical flat surface and the condensate film flows down the surface. The local coefficient of heat transfer for condensation
a) Remains constant at all heights of the surface
b) Decreases with increasing distance from the top of the surface
c) Increases with increasing thickness of film
d) Increases with increasing temperature differential between the surface and vapour

Answer: b [Reason:] It decreases with increasing thickness of condensate film.

9. In condensation over a vertical surface, the value of convection coefficient varies as
a) k 0.25
b) k 0.33
c) k 0.75
d) k -0.5

Answer: a [Reason:] h = 0.943 [k 3 p 2 g h f g/δ l (t sat – t s)] 0.25.

10. For film wise condensation on a vertical plane, the film thickness δ and heat transfer coefficient h vary with distance x from the leading edge as
a) δ decreases, h increases
b) Both δ and h increases
c) δ increases, h decreases
d) Both δ and h decreases

Answer: c [Reason:] Thickness increases and heat transfer coefficient decreases.

## Set 2

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 [Reason:] Walls of houses where bricks are given a layer of plaster on either side.

2. Three metal walls of 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 (1/h1 + δ/k + 1/h2)

Answer: a [Reason:] 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 [Reason:] 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
d) Both conduction and convection

Answer: c [Reason:] 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 [Reason:] It can be affected only by air between molecules and vacuum of any matter.

9. A composite slab has two layers having thermal conductivities in the ratio of 1:2. If thickness is 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 [Reason:] 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 temperature drop across the two layers?
a) 2:3
b) 3:2
c) 1:2
d) log e 2 : log e 3

Answer: b [Reason:] We know that temperature is inversely proportional to thermal conductivity, so ratio is 2:3.

## Set 3

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 [Reason:] Boilers have tubes in them, the condenser consist of 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 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 [Reason:] 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 [Reason:] 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 current flow
a) 52 amps
b) 62 amps
c) 72 amps
d) 82 amps

Answer: a [Reason:] Power generated = 15 k W = 15000 W. Therefore, intensity of current flow = (15000/5.548) ½ = 52 amps.

## Set 4

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 [Reason:] 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 [Reason:] 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 [Reason:] Q1 = k1 A1 d t11 and Q2 = k2A2 d t22 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] Q = k A C (t 1 – t 2)/δ = 0.318 k.

## Set 5

1. The temperature distribution associated with radial conduction through a sphere is represented by
a) Parabola
b) Hyperbola
c) Linear
d) Ellipse

Answer: b [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 