BCIT 5th class boiler engineer quiz 9 Flashcards
In the SI system, the unit of mass is the
a) pound.
b) kilogram.
c) meter.
d) ampere
e) candela
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
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.
e) 3 3/4 cm.
B03 CH03 Q010
Converting, 3.75 cm = 3 3/4cm
Subtract the following fractions
B03 CH04 Q012
Answer 1 7/8
185 is what percent of 500?
a) 63%
b) 0.63%
c) 37%
d) 0.37%
e) 1.702%
c) 37%
B03 CH04 Q024
Multiply the following fractions
B03 CH04 Q014
Add the following fractions
B03 CH04 Q010
Convert the following fraction to percentage (round off to two decimals, if necessary):
B03 CH04 Q020
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
e) 9 kg
B03 CH04 Q026
Transpose the following equation and solve for P when X = 12, B = 3, and L = 8:
B03 CH05 Q004
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
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
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
c) 1492 kW
B03 CH06 Q010
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
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
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
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
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
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.
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
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
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.
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.
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
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.
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
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
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
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
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
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.
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
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
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
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
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.
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.
When two substances are combined or separated in a chemical reaction, the substances produced may have totally different chemical and physical properties.
True
False
True
B03 CH08 Q006
True
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 10-6m3/m3°C.
a) 0.0138 m3
b) 0.142 m3
c) 0.7738 m3
d) 0.0142 m3
e) 0.138 m3
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
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
d) milli
B03 CH03 Q008
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
b) deci
B03 CH03 Q006
Reduce the following fraction to its simplest terms:
27/31
B03 CH04 Q022
Evaluate using your calculator: 25x12x64x42 / 18x80x7
a) 1.25
b) 25 088 000
c) 6.667
d) 80
e) 1.904
d) 80
B03 CH04 Q018
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
a) 1 to 4
B03 CH04 Q028
Divide the following fractions:
B03 CH04 Q016
Calculate: 185 is what percent of 500?
a) 0.63%
b) 0.37%
c) 1.702%
d) 63%
e) 37%
e) 37%
B03 CH04 Q024
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
e) 66.67
B03 CH04 Q020
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”.
B03 CH05 Q002
Subtract u from both sides:
v - u = u + at - u
v - u = at
Divide both sides by t
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
a) 118 300 cm
B03 CH06 Q006
118.3 dam x 10 m/dam x 100 cm/1 m = 118 300 cm
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
a) 8.98 m3
B03 CH06 Q018
From the Handbook of Formulae and Physical Constants the formula for
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
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
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
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
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
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
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
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
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
b) 245.25 N
B03 CH07 Q030
Weight of load = 100 x 9.81 = 981 N
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.
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).
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.
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).
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
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.
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.
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.
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.
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
The unit of heat in the SI system is
a) watt.
b) btu.
c) secand.
d) farad.
e) joule.
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).
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
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
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
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
When two substances are combined or separated in a chemical reaction, the substances produced may have totally different chemical and physical properties.
True False
True
B03 CH08 Q006>
True
