Category: Objective Questions

Energy Engineering Multiple Choice Questions on “Mechanical Dust Collectors”.

1. Gravitational separator works on the principle of __________
A. Size of particulate
B. Shape of particulate
C. Weight of particulate
D. Color of the particulate
View Answer

Answer: C
clarification: Gravitational separation is an industrial method of separating two components, either a suspension, or dry granular mixture where separating the components with gravity is sufficiently practical, i.e. the components of the mixture have different specific weights.

2. What is the use of baffles in the gravitational separators?
A. To separate types of dust
B. To settle the dust by letting them to strike
C. To control the flow of dust particles
D. To blow the dust

Answer: B
clarification: Baffles are placed in the direction of flow of flue gases. The dust particle strike down and settle. Heavier dust particles settle at the bottom by striking on to the baffles and lighter dust particles are exhaust into atmosphere.

3. What is the most effective advantage of gravitational separators?
A. They consume no power
B. They just need small amount of space for operation
C. They are cost effective
D. Time taken for operation is very less

Answer: C
clarification: The most notable advantage of the gravitational separators is their cost effectiveness. Gravity separation is an attractive unit operation as it generally has low capital and operating costs, uses few if any chemicals that might cause environmental concerns.

4. Which principle does cyclone separator use?
A. Gravitational force
B. Vortex velocity
C. Inertia
D. Temperatures of air
View Answer

Answer: C
clarification: Cyclone separators or simply cyclones work on the principle of inertia to move particulate matter from flue gases. In these high velocity dusts carrying gas is made to pass through conical separator in tangential direction. This results in centrifugal action, throwing away the heavier dust particles to the sides of conical chamber.

5. Cyclone separators are also known as __________
A. Twist cleaners
B. Squall
C. Pre-cleaners
D. Zephyr cleaners

Answer: C
clarification: Cyclone separators are a part of a group of air pollution control devices known as pre-cleaners as they generally used to roughly remove large pieces of particulate matter. This prevents finer filtration methods from having to deal with large, more abrasive particles later on.

6. What is called when several cyclone separators are operated parallely?
A. Octa-cyclone
B. Multi-cyclone
C. Center-cyclone
D. Para-cyclone

Answer: B
clarification: When several cyclone operators are operated parallely, when this system is setup it is called multi-cyclone. It is important to note that cyclones can vary drastically in their size. The size of the cyclone depends largely on how much flue gas must be filtered, and thus larger operations tend to need larger cyclones.

7. What is the range of particulate removing efficiency of cyclone separators?
A. 50-99%
B. 20-80%
C. 70-90%
D. 70-95%
View Answer

Answer: A
clarification: Cyclone separators are generally able to remove somewhere in the range of 50-99% of all particulate matter. How well the cyclone separators are actually able to remove this matter depends largely on particle size. If there is large amount of lighter particulate matter, less of these particles are able to be separated out. Because of this, cyclone separators work best on flue gases that contain large amounts of big particulate matter.

8. What is Baghouse?
A. Filters arranged in parallel form
B. Filters arranged throughout the system
C. Filters arranged randomly
D. Filters arranged alternatively

Answer: A
clarification: A fabric filter element made up of a long hollow cylindrical tube, that provides a large surface per unit of gas volumetric flow rate is called as baghouse filter and a system consisting of large number of such filter elements arranged in parallel rows is called a baghouse.

9. What is the work of the baghouse filter?
A. To remove the hot air from furnace
B. To separate the solid particles from dust produced
C. To remove dust particles from flue gas
D. To wash away the contamination of dust on the walls of furnace

Answer: C
clarification: Baghouse filters are used to remove dust particles from the flue gas produced from the combustion in boiler. These are also called as fabric filters or fabric dust collectors, designed to use fabric filter tubes, envelopes or cartridges to capture or separate dust and particulate matter when the gas is made to pass through the voids.

10. What is the use of wet scrubber in the dust collection?
A. Remove flue gas
B. Remove Scales on the furnace surface
C. Remove the dust that has the moisture content
D. Remove pollutants
View Answer

Answer: D
clarification: Wet scrubber is used remove the pollutants. The polluted gas stream is brought into contact with the scrubbing liquid, by spraying it with the liquid, by a pool of liquid, or by some other contact method, so as to remove the pollutants.

Energy Engineering online test focuses on “Engine Starting Methods – 2”.

1. What is a relay?
A. A device that resists the flow of Current
B. A device that resists the flow of direction in two ways
C. Device that allows small electrical current to control large electrical current
D. A device that allows the operator to control the current flow in the required direction

Answer: C
clarification: A relay is a device that allows a small amount of electrical current to control a large amount of current. An automobile starter uses a large amount of current to start an engine. If we allow much current to go through the ignition switch, we would need a very large switch.

2. Years ago diesel engines were started with small engines called ________
A. mini engines
B. runty engines
C. pup engines
D. puny engines

Answer: C
clarification: Years ago diesel engines were started with the small engines called pup engines. And another way used to start a diesel engine was by gasoline and switch it over to diesel fuel. A 12v electrical system became very popular and later on they were used.

3. How is minimum cranking speed achieved?
A. By providing the hp
B. By providing speed
C. By providing torque
D. By proving sufficient amount of fuel

Answer: C
clarification: Minimum cranking speed is achieved by providing the necessary torque, which is actually a purpose of starter motor. As the starter motor rotates the flywheel, the crankshaft is turned, which then starts piston movement.

4. Engine cranking is difficult in which temperatures?
A. Moderate
B. Hot
C. Humid
D. Cold

Answer: D
clarification: The engine cranking is difficult in colder temperatures especially if engine is directly driving other machine equipments such as torque converter or hydraulic pumps. Cold engine oil adds to the load on starter, and this load may increase by 3 to 4 times what it would normally be in warmer weather.

5. Which is the recent difficulty added to the starting systems?
A. Fuel flow
B. Gear transmissions
C. Electronic controls
D. Fuel consumption
View Answer

Answer: C
clarification: The recent difficulty added to the stating system is electronic controls in engines. Some electronic control machines may need some minimum number of revolutions at a minimum speed before energizing the fuel system.

6. If the starter is used properly, for how many starts will it last?
A. About 8000
B. Over 10000
C. Upto 40000
D. At least 25000
View Answer

Answer: B
clarification: If the starter is used properly, it will last for over 10000 starts. The biggest factor affecting the life of a electric starter is overheating and over-cranking. Never run the starter switch for more than 30 seconds continuously.

7. What is CRDi?
A. Common rail direct infusion
B. Common rail direct injection
C. Common rail diesel injection
D. Common rail diesel infusion

Answer: B
clarification: CRDi id common rail direct injection, which says direct injection of fuel into cylinders via a common single rail called single line which is connected to all the fuel injectors. This system is commonly equipped in diesel engines.

8. What is CDI?
A. Common diesel injection
B. Common direct injection
C. Capacitor discharge ignition
D. Capacitor direct ignition

Answer: C
clarification: CDI stands for capacitor discharge ignition or thyristor ignition is a type of automotive electronic ignition system which is widely used in the outboard motors, motorcycles, chainsaws, lawn mowers and some cars.

9. What does TDI stand for?
A. Turbocharged diesel injection
B. Turbocharged direct injection
C. Turbocharged discharged infusion
D. Turbocharged Direction ignition
View Answer

Answer: B
clarification: TDI stands for Turbocharged direct injection is a design of turbo diesel engines. These are used in vehicles, also used in the marine engines. In this fuel injector sprays atomized fuel into main combustion chamber of each cylinder.

10. What is GDI engine?
A. Gasoline direct injection
B. Gasoline diesel injection
C. Gasoline discharge injection
D. Gasoline direct infusion

Answer: A
clarification: In non-diesel IC engines, GDI (Gasoline direct injection), also known as petrol direct injection, Fuel satisfied ignition etc. Emission levels can be easily and accurately controlled with diesel systems.

Energy Engineering Multiple Choice Questions on “Solar Constant”.

1. The amount of energy received in unit time on a unit area perpendicular to the sun’s direction at the mean distance of the earth from the sun is called ________
A. Solar radiation
B. Solar constant
C. Intensity of solar radiation
D. Air Mass
View Answer

Answer: B
clarification: The amount of energy received in unit time on a unit area perpendicular to the sun’s direction at the mean distance of the earth from the sun is called solar radiation. It is defined as the solar energy receiving at the top of the atmosphere, denoted by I_sc.

2. What is ‘n’ in the following solar intensity formula?
I = I_sc {1 + 0.033cos (360n/365)}
A. Day of the year
B. Month of the year
C. The year
D. Week of the year

Answer: A
clarification: The following ‘n’ denotes day of the year in the formula. Since the distance between sun and earth varies, extra-terrestrial flux also varies. Earth is closest to the sun in the summer and farthest away in the winter.

3. When the sun is directly on the top of head, it as referred to _________
A. Zenith
B. Azimuth
C. Declination
D. Hour angle

Answer: A
clarification: When the sun is directly on the top of the head, it is referred to as sun at zenith. The zenith is an imaginary point directly “above” a particular location, on the imaginary celestial sphere. The zenith is the “highest” point on the celestial sphere.

4. Path length of radiation through the atmosphere to the length of path when the sun is at zenith is called ___________
A. Declination
B. Air mass
C. Azimuth
D. Solar Constant

Answer: B
clarification: Path length of radiation through the atmosphere to the length of path when the sun is at zenith is called Air mass.
Air mass, m = Cos (altitude angle), except for very low solar altitude angles.
m = 1; When the sun is at the zenith
m = 2; When zenith angle is 60^o
m = 3; sec (θ_z) for m>3.

5. Radiation intensity ‘I’ normal to the surface is given by __________
A. ICosθ
B. Itanθ
C. ICotθ
D. Isinθ
View Answer

Answer: A
clarification: I = ICosθ, Let θ = Angle between an incident beam radiation I and the normal to the plane surface. And the θ is referred to as incident angle. And further by this formula latitude angle, declination, hour angle, zenith angle and solar azimuth angles are found out.

6. Angle made by radial line joining the location to the centre of the earth with the projection of the line on the equatorial plane is called _________
A. Latitude
B. Zenith angle
C. Hour angle
D. Declination
View Answer

Answer: A
clarification: Angle made by radial line joining the location to the centre of the earth with the projection of the line on the equatorial plane is called latitude. And it is denoted by Φ_l. It is also given by the angular distance north or south of the equator measured from the centre of the earth.

7. Angular distance of sun’s rays north or south of the equator is called _______
A. Declination
B. Hour angle
C. Latitude
D. Air mass

Answer: A
clarification: Declination is the angular distance of sun’s rays north or south of the equator. It is the angle between the line extending from the centre of the sun to the centre of the earth and the projection of this line upon earth’s equatorial plane.

8. By which of the following symbol is solar Declination denoted by ____________
A. δ
B. ρ
C. Δ
D. γ

Answer: A
clarification: Solar declination is denoted by Greek letter δ(DELTA). Solar declination is the angle between the earth-sun line and the equatorial plane. Solar declination varies throughout the year. And is given by cooper equation,
δ = 23.45sin [(360/365) (284+n)] ∵n is day of the year.

9. The angle through which the earth must turn to bring the meridian of a point directly in sun’s rays is called __________
A. Hour angle
B. Declination
C. Latitude
D. Air mass

Answer: A
clarification: The angle through which the earth must turn to bring the meridian of a point directly in sun’s rays is called Hour angle. And it is denoted by Greek letter ω (OMEGA). It is measured from noon based on the solar local time (LST).

10. Solar Altitude is also called as ________
A. Declination
B. Altitude angle
C. Zenith angle
D. Azimuth angle

Answer: B
clarification: The vertical angle between the projection of the sun’s rays on the horizontal plane and the direction of sun’s rays passing through the point s called solar altitude and is also referred to altitude angle and is denoted by Greek letter α (ALPHA).

11. The angle between the sun’s rays and a line perpendicular to the horizontal plane through angle the beam of the sun and vertical is called __________
A. Solar Azimuth angle
B. Zenith angle
C. Altitude angle
D. Declination

Answer: B
clarification: The angle between the sun’s rays and a line perpendicular to the horizontal plane through angle measured from the north to the horizontal projection of rays is called zenith angle. And it is denoted by θ_z.

12. The solar angle in degrees along the horizon east or west of north or it is the horizontal angle measured from north to the horizontal projection of sun’s rays is called ___________
A. Solar azimuth angle
B. Zenith angle
C. Altitude angle
D. Declination

Answer: A
clarification: The solar angle in degrees along the horizon east or west of north or it is the horizontal angle measured from north to the horizontal projection of sun’s rays is called solar azimuth angle. And it is denoted by Greek letter γ_s (GAMMA).

Tough Network Theory Questions and Answers on “Calculation of I(0+), di/dt(0+), d2I/dt2(0+) in Circuits Involving both Capacitor and Inductor”.

1. A coil of inductance 10 H, resistance 40 Ω is connected as shown in the figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. If, at t=0⁺, the voltage across the coil is 120 V, the value of the resistance R is __________

A. Zero
B. 20 Ω
C. 40 Ω
D. 60 Ω

Answer: A
Clarification: I_L at 0^– = (frac{120}{60}) = 2A
120 = 2(R + 40 + 20)
∴ R = 0.

2. A coil of inductance 10 H, resistance 40 Ω is connected as shown in figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. For the value of R obtained, the time taken for 95 % of the stored energy dissipated is ________

A. 0.10 s
B. 0.15 s
C. 0.50 s
D. 1.0 s

Answer: C
Clarification: For source free circuit,
I (t) = I_o(e^{-frac{R}{T} t})
∴ I (t) = 0.05 = 2 × (e^{-frac{60}{10} t})
Or, t = 0.61 ≈ 0.5 s.

3. In the circuit shown below steady state is obtained before the switch closes at t = 0. The switch is closed for 1.5 s and is then opened. At t = 1 s, V (t) will be?

A. – 3.24 V
B. 1.97 V
C. 5.03 V
D. 13.24 V

Answer: C
Clarification: V (0^–) = 10 V = V (0⁺)
For 0<t≤1.5 s, τ = 4 × 0.05 = 0.2 s
V_OC = 5 V
V (t) = 5 + (10 – 5)(e^{frac{-t}{0.2}}) = 5 + 5e^-5t
V (1s) = 5.03 V.

4. The circuit shown below is at steady state before the switch closes at t=0. The switch is closed for 1.5 s and is then opened. At t = 2 s, V (t) will be?

A. 5.12 V
B. 6.43 V
C. 8.57 V
D. 9.88 V

Answer: C
Clarification: V (1.5s) = 5.002 V
For t > 1.5 s, τ = 8 × 0.05 = 0.4
V (t) = 10+ (5-10)(e^{frac{-(t-1.5)}{0.4}}) = 10 – 5e^-2.5(t-1.5)
For t≥1.5 s, V (2 s) = 8.57 V.

5. In the circuit given below, the voltage across capacitor when switch is closed at t = ∞ is ____________

A. 50 V
B. 20 V
C. 30 V
D. 7.5 V

Answer: A
Clarification: From the figure, we can infer that,
Voltage across capacitor = voltage across 10 Ω resistance.
Now, voltage across the 10 Ω resistance = (frac{100}{10+10}) X 10
= (frac{100}{20}).10
= 50 V.

6. In the circuit given below, the current source is 1 A, voltage source is 5 V, R₁ = R₂ = R₃ = 1 Ω, L₁ = L₂ = L₃ = 1 H, C₁ = C₂ = 1 F. The current through R₃ is _________

A. 1 A
B. 5 A
C. 6 A
D. 8 A

7. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________

A. 14.7 A
B. 18.5 A
C. 40.0 A
D. 50.0 A

Answer: A
Clarification: Using KVL, 100 = R(frac{dq}{dt} + frac{q}{C})
100 C = RC(frac{dq}{dt}) + q
Or, (int_{q_o}^q frac{dq}{100C-q} = frac{1}{RC} int_0^t ,dt)
100C – q = (100C – q_o)e^-t/RC
I = (frac{dq}{dt} = frac{(100C – q_o)}{RC} e^{-1/1})
∴ e^-t/RC = 40e^-1 = 14.7 A.

8. In the circuit given below, the current source is 1 A, voltage source is 5 V, R₁ = R₂ = R₃ = 1 Ω, L₁ = L₂ = L₃ = 1 H, C₁ = C₂ = 1 F. The current through the voltage source V is _________

A. 1 A
B. 3 A
C. 2 A
D. 4 A

9. In the circuit given below, the voltage across R by mesh analysis is _________

A. 1.59 V
B. 1 V
C. 2.54 V
D. 1.54 V

Answer: D
Clarification: In loop 1, by KVL, (10 + 2) I₁ + (-2) I₂ + 0I₃ = 5
Or, 12 I₁ – 2I₂ = 5
In loop 2, -2I₁ + (10+2+20+2) I₂ + (-2) I₃ = 0
Or, -2I₁ + 34 I₂ – 2I₃ = 0
In loop 3, 0I₁ + – 2I₂ + (2+10) I₃ = 10
Or, 0I₁ – 2I₂ + 12I₃ = 10
Now, the voltage across R is V_R = (I₂ – I₃) R
Now, I₂ = (frac{360}{4800} = frac{3}{40})
Now, I₃ = (frac{4060}{4800} = frac{203}{240}) A
Therefore, V_R = (I₂ – I₃) R = (Big[frac{3}{40} – frac{203}{240}Big] 2) = 1.54 V.

10. In a dual slop integrating type digital voltmeter, the first integrating is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V, the total conversion time for an input of 1 V is _________
A. 0.01 s
B. 0.05 s
C. 0.1 s
D. 1 s

Answer: C
Clarification: In a dual slope integrating digital voltmeter,
((frac{t_1}{t_2})) V_in = V_ref
Where, t₁ = first integration time = 10 × (frac{1}{50}) = 0.25
But V_in = 1 V and V_ref = 2 V
∴ t₂ = (frac{V_{in} t_1}{V_{ref}}) = 0.1 s.

11. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. The internal resistance and a full-scale deflection produced by a DC current are 100 Ω and 1 mA respectively. A voltage of 100 V (rms) is applied to the input terminals. The value of R required is _________
A. 63.56 Ω
B. 89.83 Ω
C. 89.83 kΩ
D. 141.3 kΩ

Answer: C
Clarification: V_OAverage = 0.636 × (sqrt{2})V_rms = 0.8993 V_rms
The deflection with AC is 0.8993 times that with DC for the same value of voltage V
S_AC = 0.8993 S_DC
S_DC of a rectifier type instrument is (frac{1}{I_{fs}}) where I_fs is the current required to produce full scale deflection, I_fs = 1 mA; R_m = 100 Ω; S_DC = 10³ Ω/V
S_AC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R_S = S_AC V – R_m – 2R_d, where R_d is the resistance of diode, for ideal diode R_d = 0
∴ R_S = 899.3 × 100 – 100 = 89.83 kΩ.

12. A 1 μF capacitor is connected across a 50 V battery. The battery is kept closed for a long time. The circuit current and voltage across capacitor is __________
A. 0.5 A and 0 V
B. 0 A and 50 V
C. 20 A and 5 V
D. 0.05 A and 5 V

Answer: B
Clarification: We know that, when the capacitor is fully charged, it acts as an open circuit.
That is when the capacitor is fully charged.
So, the circuit current and voltage across capacitor are 0 A and 50 V respectively.

13. In the circuit given below, the switch is closed at t = 0. At t = 0⁺ the current through C is ___________

A. 2 A
B. 3 A
C. 4 A
D. 5 A

Answer: D
Clarification: We know that,
When the capacitor is fully charged, it acts as a short circuit.
The equivalent resistance R_eq = (10 + 10) Ω
= 20 Ω
Given voltage = 100 V
So, current through C = (frac{100}{20}) A = 5 A.

14. A digital Multimeter reads 10 V when fed with a triangular wave, which is symmetric about the time-axis. If the input is same, the rms reading meter will read?
A. (frac{20}{sqrt{3}})
B. –(frac{10}{sqrt{3}})
C. –(frac{20}{sqrt{3}})
D. (frac{10}{sqrt{3}})

Answer: D
Clarification: For triangular wave Average value = (frac{V_m}{3}), rms value = (frac{V_m}{sqrt{3}})
∴ (frac{V_m}{3}) = 10 V or, V_m = 30 V.

15. A resistance R is measured using the connection shown in the below figure.

The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω. The value of R is _________
A. 12.75 Ω
B. 12.0 Ω
C. 12.25 Ω
D. 12.5 Ω

Answer: C
Clarification: Percentage error in ammeter = ± (frac{1}{10×100}) × 100 = ± 0.1%
Percentage error in voltmeter= ±(frac{1}{10 ×150}) × 100 = ± 0.067%
So, δI = ± 0.3 ± 0.1 = ± 0.4%
δV = ± 0.4 ± 0.067 = ± 0.467%
R = (frac{V}{I})
So, error = ± δV ± δI = ± 0.867
Measured value of resistance = R_m = (frac{125}{10}) = 12.5
∴ True value = R_m(1-(frac{R_a}{R_m})) = 12.25 Ω.

Steam and Gas Turbines Multiple Choice Questions on “Fundamentals of Thermodynamics”.

1. Steam turbines work on __________
a) Otto cycle
b) Dual cycle
c) Rankine cycle
d) None of the mentioned
Answer: c
Clarification: Otto cycle petrol is used as fuel, Rankine cycle is the process followed in steam turbines.

2. Gas turbines work on ___________
a) Rankine cycle
b) Diesel cycle
c) Brayton cycle
d) None of the mentioned
Answer: c
Clarification: Rankine cycle is used in steam turbines. Brayton cycle is followed in gas turbines.

3. Generally, water is used as working fluid in rankine cycle.
a) True
b) False
Answer: a
Clarification: Rankine cycle is followed in steam turbines. In steam turbines generally, water is used as working fluids.

4. Ideal Brayton cycle consists of ________
a) Isoentropic process and Isobaric process
b) Adiabatic process and Isobaric process
c) Adiabatic process
d) None of the mentioned
Answer: a
Clarification: Ideal brayton cycle works on isoentropic process and isobaric process.

5. Actual Brayton cycle works on _________
a) Isoentropic process and Isobaric process
b) Adiabatic process and Isobaric process
c) Isoentropic process
d) None of the mentioned
Answer: b
Clarification: Actual brayton cycle works on Adiabatic process and Isobaric process slight deviations from actual brayton cycle.

6. Brayton cycle is also known as __________
a) Joule cycle
b) Regenerative cycle
c) Carnot cycle
d) None of the mentioned.
Answer: a
Clarification: Regenerative cycle is a closed loop usage of working fluid in rankine cycle. Brayton cycle is also called a joule cycle.

7. Brayton cycle is also called an Ericsson cycle.
a) True
b) False
Answer: a
Clarification: Brayton cycle is also called an ericsson cycle.

8. Brayton cycles are further divided into ______ types.
a) 2
b) 3
c) 4
d) None of the mentioned
Answer: a
Clarification: Brayton cycle is further divided into two types. They are open cycle and closed cycle.

9. Internal combustion chamber is used in _______ Brayton cycle.
a) open
b) closed
c) depends
d) none
Answer: a
Clarification: Internal combustion chambers are present in open cycles to heat the working fluid.

10. Heat exchanger is used in a closed Brayton cycle.
a) True
b) False
Answer: a
Clarification: Heat exchangers are used in closed cycles to raise the temperature of working fluid.