BCIT 5th class boiler engineer quiz 9

Question 1

In the SI system, the unit of mass is the

a) pound.

b) kilogram.

c) meter.

d) ampere

e) candela

Answer 1

b) kilogram.

B03 CH03 Q004

Quantity Name of Base Unit Symbol
* Length metre m
* Mass kilogram kg
* Time second s
* Electric current ampere A
* Thermodynamic temperature kelvin K
* Amount of substance mole mol
* Luminous intensity candela cd

Question 2

A pump shaft measures 3.75 cm. The equivalent diameter would be

a) 3 3/16 cm.

b) 2 7/8 cm.

c) 3 3/8 cm.

d) 3 3/32 cm.

e) 3 3/4 cm.

Answer 2

e) 3 3/4 cm.

B03 CH03 Q010
Converting, 3.75 cm = 3 3/4cm

Question 3

Subtract the following fractions

Answer 3

B03 CH04 Q012
Answer 1 7/8

Question 4

185 is what percent of 500?

a) 63%

b) 0.63%

c) 37%

d) 0.37%

e) 1.702%

Answer 4

c) 37%

B03 CH04 Q024

Question 5

Multiply the following fractions

Answer 5

B03 CH04 Q014

Question 6

Add the following fractions

Answer 6

B03 CH04 Q010

Question 7

Convert the following fraction to percentage (round off to two decimals, if necessary):

Answer 7

B03 CH04 Q020

Question 8

A tank contains 16 kg of oil. 25% of this quantity is drained off and later 25% of the remainder is drained off. Calculate the quantity of oil remaining in the tank.

a) 4 kg

b) 10 kg

c) 8 kg

d) 12 kg

e) 9 kg

Answer 8

e) 9 kg

B03 CH04 Q026

Question 9

Transpose the following equation and solve for P when X = 12, B = 3, and L = 8:

Answer 9

B03 CH05 Q004

Question 10

A rectangular building is 5.5 meters high by 6.2 meters wide and 5 meters deep. The air in the building should be changed ten times every hour. Determine the capacity of a fan required in m³/min.

a) 170.5 m³/min

b) 1705 m³/min

c) 28.42 m³/min

d) 2.84 m³/min

e) 34.1 m³/min

Answer 10

c) 28.42 m³/min

B03 CH06 Q014

Volume of building = L x W x H
= 5.5 x 6.2 x 5 = 170.5 m³

Therefore the volume air that must be moved per hour
= 170.5 m³x 10 = 1705 m³

and fan requirement
= 1705/60 = 28.4 m³/min

Question 11

Convert 2 000 hp to kW using your “Handbook of Formulae and Physical Constants” as a reference.

a) 2680 kW

b) 268 000 kW

c) 1492 kW

d) 2 000 kW

e) 1 492 000 kW

Answer 11

c) 1492 kW

B03 CH06 Q010

Question 12

Convert 3921.2 mg to kilograms.

a) 3 921 200 kg

b) 392 120 kg

c) 39.212 kg

d) 0.039212 kg

e) 0.0039212 kg

Answer 12

e) 0.0039212 kg

B03 CH06 Q008

3921.2 mg x 1 g/1000 mg x 1 kg/1000 g = 0.003 921 2 kg

Question 13

A cylindrical tank has a diameter of 2 m. If it contains 12 000 L of water when full, calculate its length.

Hint: Transpose the formula for the volume of a cylinder to determine the length.

a) 3820 L

b) 3.819 m

c) 15.27 m

d) 15.27 L

e) 2 m

Answer 13

b) 3.819 m

B03 CH06 Q022

Note that 1 m³ of water = 1 000 L
Therefore 12 000 liters = 12 m³
Volume of a cylinder (V) = 0.7854 d x L

 Transposing for Length


Question 14

A vertical cylindrical water tank is 2 meters in diameter and 3.5 meters high. Determine the amount of water the tank will hold, in liters, when completely filled.

a) 43 982 L

b) 43.98 L

c) 4 398 L

d) 10.99 L

e) 10 995 L

Answer 14

e) 10 995 L

B03 CH06 Q016

From the Handbook of Formulae and Physical Constants –
1 m³= 1000 liters
Therefore the tank volume is 10.99 x 1000 = 10 995 liters.

Question 15

Calculate the work done to raise a block having a mass of 26 kg by 1.5 meters.

Hint: convert mass of block to force required to lift by multiplying mass by the force of gravity (9.81 m/s2).

a) 3.97 J

b) 382.59 J

c) 39 J

d) 390 J

e) 17.33 J

Answer 15

b) 382.59 J

B03 CH07 Q012

Force required to lift block = mass x gravity
= 26 x 9.81 = 255.06 Newtons

W.D. = f x d
= 255.06 x 1.5 m
= 382.59 N-m

Question 16

The same pulley is used to lift a load set up in two different configurations as shown. The force on the support in configuration

a) A is the same as in configuration B.

b) A is greater than in configuration B.

c) B is greater than in configuration A.

Answer 16

b) A is greater than in configuration B.

B03 CH07 Q026

Since the load is supported by 2 upward forces in configuration B, each force supports 1/2 of the load and therefore the force on the support is less than in A. In configuration A the downward force of the load is the same as the downward force of the effort and therefore the force acting on the support is twice the load.

Question 17

Ignoring friction, state what effort must be applied to lift a load of 400 newtons using the system shown.

a) 400 N

b) 133.3 N

c) 80 N

d) 200 N

e) 100 N

Answer 17

e) 100 N

B03 CH07 Q028

Theoretical mechanical advantage is equal to the velocity ratio which is the number of
ropes supporting the load = 4.
The effort required is 100 N.

Question 18

A 5 m bar of negligible mass is supported on a fulcrum 1 m from the left end. If a force of 25 N is applied to the long end of the bar, what force must be applied to the short end to maintain equilibrium.

a) 6.25 N

b) 0.16 N

c) 10.2 N

d) 0.638 N

e) 100 N

Answer 18

e) 100 N

B03 CH07 Q022

For equilibrium,
Clockwise moments = Counter clockwise moments
25 N x 4 m = F x 1 m
F = 100 N

Question 19

Calculate the power required to lift a 100 kg electric motor using a hoist if the motor must be lifted 6.5 meters in 10 seconds.

Hint: convert mass of motor to force required to lift by multiplying mass by the force of gravity (9.81 m/s²). Calculate work done and then power required.

a) 637.65 W

b) 650 W

c) 63.765 W

d) 6.5 kW

e) 65 W

Answer 19

a) 637.65 W

B03 CH07 Q014

Force required to lift block = mass x gravity
= 100 x 9.81 = 981 Newtons

W.D. = f x d
= 981 x 6.5 m
= 6376.5 Joules

Question 20

To lift a heavy load using a lever, the effort required is reduced if the fulcrum is moved ______________ the load.

a) closer to

b) half way between the effort and

c) with

d) in relation to

e) further from

Answer 20

a) closer to

B03 CH07 Q032

To lift a heavy load using a lever, the effort required is reduced if the fulcrum is moved closer to (closer to/further from) the load.

Question 21

Examples of elements include

a) hydrogen, sodium and salt

b) salt, water, and calcium carbonate

c) carbon, hydrogen and water

d) water, salt and air

e) carbon, hydrogen and oxygen

Answer 21

e) carbon, hydrogen and oxygen

B03 CH08 Q008

Elements - carbon, oxygen, bromine, sodium, chlorine, hydrogen, iron, aluminum

Compounds - common salt, water, carbon dioxide, calcium carbonate, sodium sulfite

Question 22

A mill is supplied with steam from a plant next door. The length of the pipeline supplying steam is 600 meters. Calculate the increase in length, in meters, when the line is put into service after a when the temperature is 0°C if the steam temperature is 450°C.
(take the coefficient of expansion for the steel to be 12 x 10^-6)

a) 0.0324 m

b) 3.24 m

c) 0.324 m

d) 32.4 m

e) 0.00324 m

Answer 22

b) 3.24 m

B03 CH08 Q012

Δl = r L (T2– T1)
where Δl = change in length
r = coefficient of linear expansion
L = original length
T2= final temperature
T1= original temperature
Δl = r L (T2– T1)
Δl = (12 x 10^-6) x 600 x (450 – 0)
Δl = 3.24 meters

Question 23

The density of water is

a) 1000 kg/m³.

b) 9.81 kg/m³.

c) 1 kg/m³

d) 981 kg/m³.

e) 10 kg/m³.

Answer 23

a) 1000 kg/m³

03 CH08 Q010

On cubic meter of water has a mass of 1000 kg. Therefore the density of water is 1000 kg/m³.

Question 24

When two substances are combined or separated in a chemical reaction, the substances produced may have totally different chemical and physical properties.

True
False

Answer 24

True

B03 CH08 Q006
True

Question 25

**A cast iron base plate for a motor and pump has a volume of 0.76 m²  when at a temperature of 20°C. Calculate its volume when heated to 600°C.** *Take the coefficient of expansion for cast iron to be 31.2 x 10^-6 m³/m³ °C.* a) 0.0138 m³ b) 0.142 m³ c) 0.7738 m³ d) 0.0142 m³ e) 0.138 m³

Answer 25

**c) 0.7738 m³** B03 CH08 Q014 Δv = r V (T2 – T1) where Δv = change in volume r = coefficient of volumetric expansion V = original volume T2 = final temperature T1 = original temperature   Δv = r V (T2 – T1) Δv = (31.2 x 10^-6) x 0.76 x (600 – 20) Δv = (31.2 x 10^-6) x 0.76 x (580) Δv = 0.0138 m³   Final Volume = 0.76 + 0.0138 = 0.7738 m³

Question 26

**In the SI system the prefix used to indicate a multiplication factor of 1/1000 is** a) kilo b) deci c) deca d) milli e) mega

Answer 26

**d) milli** B03 CH03 Q008

Question 27

**In the SI system the prefix used to indicate a multiplication factor of 1/10 is** a) kilo b) deci c) deca d) milli e) mega

Answer 27

**b) deci** B03 CH03 Q006

Question 28

**Reduce the following fraction to its simplest terms:**

Answer 28

**27/31** B03 CH04 Q022

Question 29

**Evaluate using your calculator: 25x12x64x42 / 18x80x7** a) 1.25 b) 25 088 000 c) 6.667 d) 80 e) 1.904

Answer 29

**d) 80** B03 CH04 Q018

Question 30

**A manufacturer employs 500 females and 1500 males. What is the ratio of females to the total number of employees?** a) 1 to 4 b) 4 to 1 c) 3 to 1 d) 3 to 4 e) 1 to 3

Answer 30

**a) 1 to 4** B03 CH04 Q028

Question 31

**Divide the following fractions:**

Answer 31

B03 CH04 Q016

Question 32

**Calculate: 185 is what percent of 500?** a) 0.63% b) 0.37% c) 1.702% d) 63% e) 37%

Answer 32

**e) 37%** B03 CH04 Q024

Question 33

**Convert the following fraction to percentage (round off to two decimals, if necessary): 2/3** a) 6 b) 60 c) 0.67 d) 6.67 e) 66.67

Answer 33

**e) 66.67** B03 CH04 Q020

Question 34

**The formula *v = u + at* gives a final velocity, V, if the initial velocity (u), acceleration (a), and time (t) are known. Transpose this equation to solve for “a”.**

Answer 34

B03 CH05 Q002 Subtract u from both sides: v - u = u + at - u v - u = at Divide both sides by t

Question 35

**Convert 118.3 dam to centimetres.** a) 118 300 cm b) 11 830 cm c) 0.1183 cm d) 11.83 cm e) 11830 cm

Answer 35

**a) 118 300 cm** B03 CH06 Q006 118.3 dam x 10 m/dam x 100 cm/1 m = 118 300 cm

Question 36

**The steam drum on a water tube boiler is 1.6 meters in diameter with hemispherical heads (1/2 sphere at each end) and the drum is 5 meters long as shown. Find volume of the drum with no internals installed.** a) 8.98 m³ b) 7.91 m³ c) 10.05 m³ d) 10.18 m³ e) 19.23 m³

Answer 36

**a) 8.98 m³** B03 CH06 Q018 From the Handbook of Formulae and Physical Constants the formula for

Question 37

**An air duct is 65 cm by 45 cm and has to be blocked off using a rectangular sheet that just fits into the end of the duct. Calculate the length of angle bracket required to form a tight frame around the sheet (perimeter of the sheet).** a) 0.2925 m b) 1.1 m c) 0.55 m d) 2.2 m e) 2.925 m

Answer 37

**d) 2.2 m** B03 CH06 Q012 1 meter = 100 cm therefore: 65 cm = 0.65 m 45 cm = 0.45 meters Perimeter of a rectangle = 2(Length + Width) = 2 (0.65 + 0.45) = 2.2 m

Question 38

**A cast iron column has an outside diameter of 25 cm and an inside diameter of 17 cm. Find the cross sectional area of metal.** *Hint: The crossectional area is the area of the outer circle minus the area of the inner circle.* a) 640 cm² b) 6.28 cm² c) 50.26 cm² d) 36.91 cm² e) 263.90 cm²

Answer 38

**e) 263.90 cm²** B03 CH06 Q020 From the Handbook of Formulae and Physical Constants the formula for area of circle. Area = 0.7854 d² Area of Cross Section = 0.7854 D² - 0.7854 d² = 0.7854 (25)² - 0.7854 (17)² = 490.88 - 226.98 = 263.90

Question 39

**Convert 3921.2 mg to kilograms.** a) 3 921 200 kg b) 392 120 kg c) 39.212 kg d) 0.039212 kg e) 0.0039212 kg

Answer 39

**e) 0.0039212 kg** B03 CH06 Q008 3921.2 mg x 1 g/1000 mg x 1 kg/1000 g = 0.003 921 2 kg

Question 40

**In the lever diagram, a load is positioned 200 cm from the fulcrum and an effort is applied 1 meter from the fulcrum. If the load is 50 kg, what effort is required to keep the beam in equilibrium.** a) 10 000 N b) 0.4 N c) 2.45 N d) 10 N e) 981 N

Answer 40

**e) 981 N** B03 CH07 Q020 Force due to the load: *F = m x g = 50 x 9.81 = 490.5 N* Distance from load to fulcrum = 200 cm = 2 m Clockwise Moments = Counter Clockwise Moments 490.5 N x 2 m = Effort x 1 m Effort = 981 N

Question 41

**In the figure, if the side of the triangle labelled A is 2 meters and the side labelled C is 8 meters, what effort must be applied to move a load with a mass of 100 kg.** *(ignore friction)* a) 25 N b) 245.25 N c)12.5 N d) 3 924 N e) 400 N

Answer 41

**b) 245.25 N** B03 CH07 Q030 Weight of load = 100 x 9.81 = 981 N

Question 42

**In the figure, the item identified by the letter "D" is** a) a force creating counter clockwise moment. b) the lever. c) a force creating clockwise moment. d) the reaction force. e) the fulcrum.

Answer 42

**b) the lever.** B03 CH07 Q018 a) A turning moment is an action due to applied force that causes or tends to cause rotation around some point or fulcrum. b) A clockwise moment tends to cause rotation in a clockwise direction. c) A counterclockwise moment tends to cause rotation in a counterclockwise direction. To calculate the value of a moment, multiply the force applied to the perpendicular distance (in meters) between that force (in Newtons) and the center of rotation. The units are therefore Newton-meters (N-m).

Question 43

**In the figure, the item identified by the letter "B" is** a) the lever. b) a force creating clockwise moment. c) the fulcrum. d) the reaction force. e) a force creating counter clockwise moment.

Answer 43

**b) a force creating clockwise moment.** B03 CH07 Q018 a) A turning moment is an action due to applied force that causes or tends to cause rotation around some point or fulcrum. b) A clockwise moment tends to cause rotation in a clockwise direction. c) A counterclockwise moment tends to cause rotation in a counterclockwise direction. To calculate the value of a moment, multiply the force applied to the perpendicular distance (in meters)between that force (in Newtons) and the center of rotation. The units are therefore Newton-meters (N-m).

Question 44

**For a simple machine, the ratio of the effort applied to the load raised is the ______________ _______________ of the machine.** a) work output b) velocity ratio c) simple mechanics d) bottom line e) mechanical advantage

Answer 44

**e) mechanical advantage** B03 CH07 Q024 When using simple machines to perform a task such as lifting, the effort that is applied in usually less that the load being lifted. That is why simple machines can be used to allow a modest effort to move a very heavy load. *The ratio of the effort applied to the load raised is the mechanical advantage of the machine.* When using a lever, if the distance from the load to the fulcrum is less than the distance from the fulcrum to applied effort, the effort required will be less than the load. The greater this difference the more mechanical advantage the lever has.

Question 45

**The same pulley is used to lift a load set up in two different configurations as shown. The force on the support in configuration** a) B is greater than in configuration A. b) A is the same as in configuration B. c) A is greater than in configuration B.

Answer 45

**c) A is greater than in configuration B.** B03 CH07 Q026 Since the load is supported by 2 upward forces in configuration B, each force supports 1/2 of the load and therefore the force on the support is less than in A. In configuration A the downward force of the load is the same as the downward force of the effort and therefore the force acting on the support is twice the load.

Question 46

**The specific heat of water is about** a) 2.1 kJ/kg/°C. b) 4.2 kJ/kg/°C. c) 1.9 kJ/kg/°C. d) 11.2 kJ/kg/°C. e) 8.4 kJ/kg/°C.

Answer 46

**b) 4.2 kJ/kg/°C.** B03 CH08 Q004 The approximate specific heat for: * liquid water is 4.2 kJ/kg/°C * ice is 2.1 kJ/kg/°C * steam is 2.1 kJ/kg/°C

Question 47

**The unit of heat in the SI system is** a) watt. b) btu. c) secand. d) farad. e) joule.

Answer 47

**e) joule.** B03 CH08 Q002 The unit of heat is the Joule (J) although this is a small amount of heat and kilojoules (kJ) are more commonly used. The unit of temperature is degrees celcius °C) (or degrees Kelvin, K for absolute temperatures).

Question 48

**A cast iron base plate for a motor and pump has a volume of 0.76 m² when at a temperature of 20°C. Calculate its volume when heated to 600°C. Take the coefficient of expansion for cast iron to be 31.2 x 10-6 m³/m³ °C.** a) 0.138 m³ b) 0.142 m³ c) 0.0138 m³ d) 0.0142 m³ e) 0.7738 m³

Answer 48

**e) 0.7738 m³** B03 CH08 Q014 Δv = r V (T₂ – T₁) where Δv = change in volume r = coefficient of volumetric expansion V = original volume T₂ = final temperature T₁ = original temperature Δv = r V (T₂ – T₁) Δv = (31.2 x 10^-6) x 0.76 x (600 – 20) Δv = (31.2 x 10^-6) x 0.76 x (580) Δv = 0.0138 m³

Question 49

**Examples of elements include** a) carbon, hydrogen and oxygen b) salt, water, and calcium carbonate c) hydrogen, sodium and salt d) carbon, hydrogen and water e) water, salt and air

Answer 49

**a) carbon, hydrogen and oxygen** B03 CH08 Q008 **Elements** - carbon, oxygen, bromine, sodium, chlorine, hydrogen, iron, aluminum **Compounds** - common salt, water, carbon dioxide, calcium carbonate, sodium sulfite

Question 50

**When two substances are combined or separated in a chemical reaction, the substances produced may have totally different chemical and physical properties.** True False

Answer 50

**True** B03 CH08 Q006> True