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Part 1: Fluids Question 1
(a) A body is 2m wide, 2.5m long and 3m high and weighs 980N.
i. Calculate the specific gravity of the body. [ 3 Marks]
ii. Calculate the density of the body. [ 3 Marks]
iii. If placed inside water, will the body sink? Please justify your answer. (The density of
water is 1000 kgm-3). [4 Marks]
(b) Atankis5mwide,10mlongand4mhigh.Thetankisgoingtobefilledwithafluidwith specific gravity of 0.75. The hose available to fill the tank has exit diameter 100 mm and exit velocity 5 m s-1.
i. Calculate the maximal volume of fluid that the tank can contain. [2 Marks]
ii. How long does the tank take to fill from empty? [3 Marks]
iii. When fully filled, calculate the weight of the fluid in the tank and the pressure at any position on the bottom. (The density of water is 1000 kg m-3). [5 Marks] Question 2
(a) The pressure in a car tyre depends on the temperature of the air in the tyre. When the air temperature is 270 oC, the gage pressure is 175 KPa.
(b)
(c)
ii.
WriteBernoulli'sequationintermsofenergyperunitmassandstatetheformsofenergy involved. [3 Marks]
Water flows in a horizontal pipe tapering from the inlet diameter 150 mm to outlet diameter 50 mm. The flow rate is 50 litres per second and the inlet pressure is 5 bar.
Apply Bernoulli’s equation to solve the following questions:
i. Assuming that there are no energy losses, calculate the pressure and Reynolds
i. If the volume of the tyre is constant 0.025 m3, determine the pressure rise in the tyre when the air temperature rises to 570 oC. [3 Marks]
ii. With the tyre temperature at 570 oC, determine the mass of air, which must be bled off in order to decrease the pressure to its original value. [4 Marks] Take the molar mass of air as 28.8 g mol-1.
number at the outlet. Indicate whether outlet flow is laminar or turbulent.
iii. Calculatethemassflowrate(kgs-1).
(Density of water is 1000 kg m-3 and dynamic viscosity is 1.002 x 10-3 Pa.s)
Hints:
[6 Marks] [1 Marks] [3Marks]
• Use Bernoulli's equation and eliminate relevant terms.
• For continuity Qin = Qout. We can omit the density term from the Continuity equation
since it will remain the same on both sides of the equation.
Part 3: Mechanics
Question 5
(a) A steel wire is 0.75m long and has a cross-section area of 3 mm2. The wire is hanging vertically from a support at its upper end and at its lower end there is a weight giving a force of 1kN acting
on the wire, as seen in Figure Q5a. Calculate:
i. The tension (tensile force) in the wire
ii. The stress in the wire
iii. Theextensioncausedbythetensileforce
(Take Young's modulus for steel to be 210 GNm-2)
[10 Marks]
(b) State which of the following statements are True or False:
[3 Marks]
Figure Q5a
i. It is possible for an object to be in motion even if there is no resultant force acting on the object
ii. According to Newton’s third law of motion, forces of action and reaction always act on the same object
iii. Workisaproductofforce(vector)anddisplacement(vector)andisascalarquality
(c) Friction force between AB (F1) is 8 N and between BC (F3) is 6 N as shown in Figure Q5c. If the work done by friction force between AB and BC are equal, find the distance between
BC.
[7 Marks]
2
Figure Q5c
Question 6
(a) A body of 2000kg is pulled up a track inclined at 30° above the horizontal line, by a force of 20kN as shown in Figure Q6a. If the friction of the surface is 6kN, find the work done by all applied on the body over a distance of 40 m along the track. (assume g= 10 m𝑠−2).
[12 Marks]
Figure 6a
(b) A constant force of 2 kN pulls an object along a level floor a distance of 10 m in 50 s as shown in Figure Q6b. What is the power used? [8
Marks]
Figure Q6b
3
FORMULA SHEET - FLUIDS
Hydrostatic pressure Archimedes Principle
Reynolds Number
The Continuity Equation
Bernoulli's Equation
p = pgh
Upthrust = pgV where V is
displaced volume
R e =
ρ v D μ
ρ1v1A1 = ρ2v2A2
Q= Av where A is area of
cross - section
gZ +v2 /2 + P/ρ = Constant
4
CONSTANTS
1 bar= 105 Pa
1 atmosphere = 101325 Pa = 760 mm of mercury Water density = 1000 kg m-3
Air density = 1.2 kg m-3
Gravitational acceleration g = 9.81 m s-2

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