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# Multiple choice question for engineering

## Set 1

1. The convective heat transfer coefficient in laminar flow over a flat plate
a) Increases with distance
b) Increases if a higher viscosity fluid is used
c) Increases if a denser fluid is used
d) Decreases with increase in free stream velocity

Answer: c [Reason:] It mostly increasers if a denser fluid is used.

2. For laminar flow over a flat plate, the average value of Nusselt number is prescribed by the relation
Nu = 0.664 (Re) 0.5 (Pr) 0.33
Which of the following is then a false statement?
a) Density has to be increased four times
b) Plate length has to be decreased four times
c) Specific heat has to be increased four times
d) Dynamic viscosity has to be decreased sixteen times

Answer: d [Reason:] The dynamic viscosity has an inverse relation to 1/6 power. To double the convective heat transfer coefficient, the dynamic viscosity has to be decreased 64 times.

3. For turbulent flow over a flat plate, the average value of Nusselt number is prescribed by the relation
Nu = 0.664 (Re) 0.5 (Pr) 0.33
Which of the following is then a false statement?
The average heat transfer coefficient increases as
a) 1/5 power of plate length
b) 2/3 power of thermal conductivity
c) 1/3 power of specific heat
d) 4/5 power of a free stream velocity

Answer: a [Reason:] The average heat transfer coefficient reduces with length as 1/5th power of the length.

4. A nuclear reactor with its core constructed of parallel vertical plates 2.25 m high and 1.5 m wide has been designed on free convection heating of liquid bismuth. Metallurgical considerations limit the maximum surface temperature of the plate to 975 degree Celsius and the lowest allowable temperature of bismuth is 325 degree Celsius. Estimate the maximum possible heat dissipation from both sides of each plate. The appropriate correlation for the convection coefficient is
Nu = 0.13 (Gr Pr) 1/3
a) 143 MW
b) 153 MW
c) 163 MW
d) 173 MV

Answer: b [Reason:] Q = 2 h A d t = 153 MW.

5. Consider the above problem, find the value of Grashoff number
a) 101.3 * 10 12
b) 102.3 * 10 12
c) 103.3 * 10 12
d) 104.3 * 10 12

Answer: d [Reason:] Grashof number = l 3 p 2 β g d t/µ 2.

6. A thin walled duct of 0.5 m diameter has been laid in an atmosphere of quiescent air at 15 degree Celsius and conveys a particular gas at 205 degree Celsius. Base your calculations on one meter length of the duct, estimate the convective coefficient of heat transfer
a) 5.086 W/m2 K
b) 6.086 W/m2 K
c) 7.086 W/m2 K
d) 8.086 W/m2 K

Answer: a [Reason:] h = 1.37 (d t/l) 0.25 = 5.086 W/m2 K.

7. Free correction modulus is given by
a) p 2 β g c P
b) p 2 β g c P/k
c) p 2 β g c P/µ k
d) p 2 β g c P

Answer: c [Reason:] It contains only fluid properties and is called the free convection modulus.

8. The free convection coefficient is given by
h = C 1 d t m/l 1 – 3m
The value of exponent for laminar flow is
a) 0.5
b) 0.6
c) 0.7
d) 0.8

Answer: a [Reason:] For laminar flow h = C 1 (d t/l) 0.25.

9. For inclined plates we multiply Grashoff number with
a) Cos 2 α
b) Sin 2 α
c) Sin α
d) Cos α

Answer: d [Reason:] It should be multiplied with cos α, as α is angle with the horizontal.

10. The free convection coefficient is given by
h = C 1 d t m/l 1 – 3m
The value of exponent for turbulent flow is
a) 0.43
b) 0.33
c) 0.23
d) 0.13

Answer: b [Reason:] For turbulent flow h = C (d t).

## Set 2

1. The following factors need consideration for the optimum design of fins
(i) Cost
(ii) Space considerations
(iii) Weight considerations
Choose the correct option
a) 1 only
b) 1 and 2 only
c) 1, 2 and 3
d) 2 only

Answer: c [Reason:] The design will be considered optimum when the fins require minimum cost of manufacture, are light in weight.

2. A heating unit is made in the form of a vertical tube of 50 mm outside diameter and 1.2 m height. The tube is fitted with 20 steel fins of rectangular section with height 40 mm and thickness 2.5 mm. The temperature at the base of fin is 75 degree Celsius, the surrounding air temperature is 20 degree Celsius and the heat transfer coefficient between the fin as well as the tube surface and the surrounding air is 9.5 W/m2 K. If thermal conductivity of the fin material is 55 W/m K, find the amount of heat transferred from the tube with fin
a) 1234 .98 W
b) 1004.84 W
c) 6539.83 W
d) 3829.46 W

Answer: b [Reason:] Heat flow rate convicted from the base, Q b = h A b (t 0 – t INFINITY) and heat flow rate convicted from the fins, Q f = n k A C m (t 0– t a).

3. The fins would be effective for heat conduction if the ratio P k/h A C is
a) Greater than 5
b) Less than 5
c) Equal to 5
d) Varies between 2 to 9

Answer: a [Reason:] The ratio perimeter divided by area multiply by length must be greater than 5.

4. Consider the following statements pertaining to heat transfer through fins
(i) They must be arranged at right angles to the direction of flow of working fluid
(ii) The temperature along the fin is variable and accordingly heat transfer rate varies along the fin elements
(iii) Fins are equally effective irrespective whether they are on the hot side or cold side of the fluid
(iv) Fins are made of materials that have thermal conductivity higher than that of wall
Identify the correct statements
a) 3 and 4
b) 1 and 3
c) 2 and 3
d) 1 and 2

Answer: d [Reason:] The statements made at serial number 3 and 4 are wrong. Fins are located on the side where the convective coefficient has a low value.

5. An increase in fin effectiveness is caused by high value of
(i) Convective coefficient
(ii) Thermal conductivity
(iii) Circumference
(iv) Area
Identify the correct statement
a) 1 and 3
b) 3 and 4
c) 2 and 4
d) 2 and 3

Answer: c [Reason:] Refer to expression for fin effectiveness, an increase in fin effectiveness is caused by high value of circumference and thermal conductivity.

6. A steel strap is serving as a support for the steam pipe. The strap is welded to the pipe and bolted to the ceiling. The junction between the support strut and the ceiling is adiabatic, and the outside temperature of steam pipe is 105 degree Celsius. The strut AB is 60 cm high and AD = BC = 12.5 cm. It is 0.3 cm thick. Workout the rate at which heat is lost to the surrounding air by the support strut. It may be assumed that thermal conductivity for steel is 45 W/m degree, the total outside surface coefficient is 17 W/m2 degree and the surrounding air is at 32 degree Celsius a) 178 W
b) 168 W
c) 158 W
d) 148 W

Answer: a [Reason:] α x0 = t x – t a/t 0 – t a = cos m (l – x)/cos ml.

7. Choose the correct option regarding fin efficiency and fin effectiveness
a) 2 Fin effectiveness = A FIN/A B (Efficiency of fin)
b) 3 Fin effectiveness = A FIN/A B (Efficiency of fin)
c) Fin effectiveness = A FIN/A B (Efficiency of fin)
d) ½ Fin effectiveness = A FIN/A B (Efficiency of fin)

Answer: c [Reason:] On simplify the equations of fin efficiency and fin effectiveness we get the result.

8. The handle of a saucepan, 30 cm long and 2 cm in diameter, is subjected to 100 degree Celsius temperature during a certain cooking operation. The average unit surface conductance over the handle surface is 7.35 W/m2 degree in the kitchen air at 24 degree Celsius. The cook is likely to grasp the last 10 cm of the handle and hence the temperature in this region should not exceed 38 degree Celsius. What should be the thermal conductivity of the handle material to accomplish it? The handle may be treated as a fin insulated at the tip
a) 18.36 W/m degree
b) 17.36 W/m degree
c) 16.36 W/m degree
d) 15.36 W/m degree

Answer: b [Reason:] α x0 = t x – t a/t 0 – t a = cos m (l – x)/cos ml. Now, for a circular handle of diameter d, P/A = 4/d.

9. Let us assume a square section fin split longitudinally and used as two fins. This will result in
a) Increase or decrease in heat transfer depending on material of fin
b) Heat flow remains constant
c) Decrease in heat transfer
d) Increase in heat transfer

Answer: d [Reason:] Heat transfer will definitely increases because it split into two fins i.e. more surface area.

10. Mark the false statement regarding effectiveness of fin
a) A high value of film coefficient adversely affects the fin effectiveness
b) Fin effectiveness is improved if fin is made from a material of low conductivity
c) Fin effectiveness is improved by having thin fins
d) It can also be improved by having closely spaced fins

Answer: b [Reason:] It should be of high thermal conductivity.

## Set 3

1. Which one of the following materials are quickly heated by applying high frequency?
a) Textiles
b) Engines
c) Rubber
d) Coal

Answer: a [Reason:] They can be heated at high voltage alternating current to the plated of the condenser.

2. Generally heat generated depends on some parameters. It is directly proportional to
a) Time
b) Conductivity
c) Voltage
d) Distance between plates

Answer: c [Reason:] It generally depends on voltage as directly proportional.

3. Consider a 1.2 m thick slab of poured concrete (k = 1.148 W/m degree) with both of side surfaces maintained at a temperature of 20 degree Celsius. During its curing, chemical energy is released at the rate of 80 W/m3. Workout the maximum temperature of concrete
a) 30.73 degree celsius
b) 29.73 degree celsius
c) 28.73 degree celsius
d) 27.73 degree celsius

Answer: b [Reason:] t = q g (δ – x) x/2 k + t w = 29.73 degree celsius.

4. The insulating material used in dielectric heating is
a) Coal
b) Silver
c) Coin
d) Wool

Answer: d [Reason:] Wool is good for heat conduction and from dielectric heating point of view.

5. A composite slab consists of 5 cm thick layer of steel (k = 146 kJ/m hr degree) on the left side and a 6 cm thick layer of brass (k = 276 kJ/m hr degree) on the right hand side. The outer surfaces of the steel and brass are maintained at 100 degree Celsius and 50 degree Celsius. The contact between the two slabs is perfect and heat is generated at the rate of 4.2 * 10 5 k J/m2 hr at the plane of contact. The heat thus generated is dissipated from both sides of composite slab for steady state conditions. Calculate the temperature at the interface a) 115.26 degree celsius
b) 125.26 degree celsius
c) 135.26 degree celsius
d) 145.26 degree celsius

Answer: b [Reason:] Q 1 + Q 2 = Q g. Q 1 = k 1 A 1 t i – t 1)/δ 1 and Q 2 = k 2 A 2 t i – t 2)/δ 2.

6. Unit of specific resistance is
a) Ohm mm2/m
b) Ohm mm
c) Ohm/m
d) Ohm mm/m

Answer: a [Reason:] Specific resistance is resistance per unit length.

7. What maximum thickness of concrete can be poured without causing the temperature gradient to exceed 98.5 degree Celsius per meter anywhere in the slab? Consider a 1.2 m thick slab of poured concrete (k = 1.148 W/m degree) with both of side surfaces maintained at a temperature of 20 degree Celsius. During its curing, chemical energy is released at the rate of 80 W/m3. Workout the maximum temperature of concrete
a) 2.64 m
b) 3.64 m
c) 4.64 m
d) 5.64 m

Answer: b [Reason:] d t/d x = q g (δ – 2 x)/2 k. The temperature is largest at x = 0.

8. Dielectric heating apparatus consists of
a) 4 electrodes
b) Elemental strip
c) No Insulating material
d) 4 plates

Answer: b [Reason:] It consists of elemental strip in the middle of the system.

9. The given expression can be used to solve the electrode temperature t w1 and t w2
q g δ = h 1 α 1 + h 2 α 2
Where, α 1 = A (t w 2 – t a) and α 2 = (t w1 – t a)
This statement is true or false
a) True
b) False

Answer: b [Reason:] α 1 = A (t w 1 – t a) and α 2 = (t w2 – t a).

10. A slab of insulating material of thickness 6 cm and thermal conductivity 1.4kJ/m hr deg is placed between and is in contact with two parallel electrodes, and is then subjected to high frequency dielectric heating at a uniform rate of 140,000kJ/m3 hr. At steady state coefficients of combined radiation and convection are 42 and 48 kJ/m2 hr deg. If atmospheric temperature is 25 degree Celsius, find surface temperatures?
a) 144.10 degree Celsius and 134.47 degree Celsius
b) 123.50 degree Celsius and 154.34 degree Celsius
c) 121.60 degree Celsius and 115.45 degree Celsius
d) 165.40 degree Celsius and 165.45 degree Celsius

Answer: c [Reason:] α = -q g x2 /2k + h 1 α 1/k + α 1. At x =0.06 m and α = α 2, α 2 = -180 + 2.8 α 1. Also q g A δ = h 1 α 1 + h 2 α 2.

## Set 4

1. How is dimensional homogeneity related with fundamental units of measurements?
a) Independent
b) Dependent
c) Dependent but can vary
d) Twice

Answer: a [Reason:] This implies that length dimension can be added to oe subtract from only a length dimension.

2. What is the time period of oscillation of a simple pendulum of length L and mass m?
a) 2 π (L/g) 3/2
b) 2 π (L/g)
c) 2 π (L/g) 2
d) 2 π (L/g) 1/2

Answer: d [Reason:] The time period of a simple pendulum means the time it takes for one complete revolution.

3. The principle of dimensional homogeneity serves the following useful concepts
(i) It helps to check whether an equation of any physical phenomenon is dimensionally homogenous or not
(ii) It helps to determine the dimensions of a physical quantity
(iii) It helps to convert the units from one system to another
Identify the correct statements
a) 1 and 2
b) 2 and 3
c) 1, 2 and 3
d) 2 only

Answer: c [Reason:] It serves all the aspects.

4. The equation of friction loss in a pipe of length l and diameter d through which fluid flows with velocity v is
a) h i = 4 f V 2/d g
b) h i = 4 f V 2/d 2 g
c) h i = 4 f V 2/d 2 g
d) h i = 4 f V 2/2 g

Answer: b [Reason:] Here f is any constant with no dimensions whatsoever.

5. Bernoulli’s equation for fluid flow along a stream line is given as
a) p/w + V 2/2 g + y = 2
b) p/w + V/2 g + y = constant
c) p/w + V 2/2 g + y = 1
d) p/w + V 2/2 g + y = constant

Answer: d [Reason:] This is the Bernoulli’s equation and is dimensionally homogenous.

6. The convective film coefficient in k cal/m2 hr degree can be converted to J/m2 s degree by multiplying it with a factor
a) 1.1627
b) 1.1527
c) 1.1427
d) 1.1327

Answer: a [Reason:] Both k cal/m2 hr degree and J/m2 s are the units of convective film coefficient.

7. The pressure in kg/cm2 can be converted to N/m2 by multiplying it with a factor
a) 9.807 * 10 2
b) 9.807 * 10 3
c) 9.807 * 10 4
d) 9.807 * 10 5

Answer: c [Reason:] Both kg/cm2 and N/m2 are the units of pressure.

8. How many Newton’s are there in one kg?
a) 9.507
b) 9.607
c) 9.707
d) 9.807

Answer: d [Reason:] 1 kg = 9.807 N. Newton is the unit of force and kg is the unit of mass.

9. How many Joule are there in 1 k cal?
a) 3186
b) 4186
c) 5186
d) 6186

Answer: b [Reason:] 1 k cal = 4186 J. Joule is the unit of work done.

10. How many fundamental quantities are there?
a) 1
b) 2
c) 3
d) 4

Answer: c [Reason:] These are M, L and T. M is mass, L is length and T is time.

## Set 5

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 [Reason:] 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) [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 depends on temperature
b) k depends on temperature
c) k is directly proportional to t
d) Data is insufficient

Answer: c [Reason:] 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 [Reason:] 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