Engineering Science N4

Engineering Science N4 Question Papers


August 2012

Answers of selected questions follow below ...

Question 1

1.1

1.2

1.3     vertical motion
1.3.1 Considering upward motion, we are given that: v = o m/s; u = 50 m/s; and of course for upward motion we have deceleration, so that g = - 9,8 m/s^2
So, if v = u + gt
t = 5,102 s
therefore the time taken for the stone to reach the ground = 2 x 5,102 s = 10,204 s

1.3.2 Consider upward motion again. We are give that: u = 50 m/s; t = 5,102 s; g = -9,8 m/s^2; 
Therefore, if h = ut + 1/2at^2
then substituting known values, we calculate h = 127,551 metres

1.4 : Projectile motion

Question 2

2.1

2.2 Angular motion
We are given that: d = 40 cm; w1 = 4 rad/s; w2 = 10 rad/s; t = 20 s
2.2.1 Angular acceleration = w2 - w1/t
                                         = 0,3 rad/s^2

2.2.2  Angular displacement  = 140 radians

2.2.3 Angular displacement in revolutions = 22,282 revs

2.3 Power = 3769,911 W

Question 3

3.1 Defifition of Newton's Third law: For every action, there is an equal and opposite recaction.

3.2 Force of the engine to pull train uphill = 263 262,821 N

3.3

Question 4  - Beam

4.1
4.1.1 Load Diagram of Beam
4.1.2 Rleft = 53,5 kn and Rright = 81,5 kN
4.1.3 Bending Moment values
BM at point  = - 32,5 kNm
BM at point D = - 40 kNm
BM at point C = 1,5 kNm
4.1.4 Shear Force and Bending Moment Diagrams (to be uploaded)
4.1.5 To be calculated in conjunction with 4.1.4

4.2
4.2.1 height from base up  = h/3, position horizontally from vertical side = 2/3 x h from vertical side
4.2.2 position of centre of gravity, vertically, measured from the base up = h/3 on the vertical centre line of the cone. (the line that runs perpendicular to the base and through the apex.)

Question 5 Density, pressure, volume, height, head

5.1 The density of a body or substance is the ratio of mass compared to volume, and is expressed in kg / m^3.
5.2
  1.  the pressure exerted at a particular point is the same intensity in all directions
  2.  the pressure exerted at any point in a volume of liquid is directly proportional to the density of the liquid, as well as the depth of the point in the liquid.
5.3    Single-acting hydraulic press
       we are given: cross sectional area plunger = 30% of cross sectional area ram; stroke length of plunger = 0,2 m; force on plunger = 600 N; cross sectional area of ram = 0,2 m^2

5.3.1 volume displaced by plunger in 12 strokes = 12 x cross sectional area of plunger  x stroke length
                                                                         = 12 x (0,3 x 0,2) x 0,2
                                                                         =  0,144       m^3 (metres cubed)

5.3.2     volume received by ram = volume displaced by plunger in 1 stroke
             area of ram x height raised by ram =      0,144  / 12  [ / means divided by ]
            0,2 x height raised = 0,012
therefore height raised =  0,012  / 0,2
                                  =  0,060               m =       60   mm

5.3.3  Force exerted by ram   W / A = F / a
  therefore       W = F x A / a
                           = 600 x 0,2 / ( 0,3 x 0,2 )
                           =     2000 N            N

5.3.4   mechanical advantage of press = Load / Effort = W / F
                                                          =     2000    /   600         
                                                          =  3,333

5.3.5 pressure in liquid = F / a = 600 / (0,3 x 0,2)
                                                  = 10 kPa

5.4 single-acting pump
We are given that d = 8 cm; stroke length = 30 cm; delivery stroke pressure = 1 MPa
5.4.1 Force per stroke per cylinder  = pressure x cross sectional area
                                              = 1 x 10^6 x pi/4 (0,08)^2
                                             =   5026,548 N

Work done per stroke/ cylinder = Force x stroke length
                                                = 5026,548 x 0,3
                                                 = 1507,964 J/stroke per cylinder
                                                = 1507,964 x 2 J/ stroke for the pump
                                                =  3015,929 J/stroke

 Work done per second (Power) =  Joules / second
                                                   = Joules / stroke x no of strokes/second
                                                   = 3015,929 x 200 / 60
                                                   =    10,053  kW
with an efficiency of 85 %           = driving power required = 10,053 /0,85
                                                      =   10,053  kW
                                                       = 11,827 kW

5.4.2 volume of water delivered per minute = volume per stroke x no of strokes/ minute
                                                                  = 2 x pi/4 (0,08)^2 x 0,3 x 200
                                                                  = 0,603 m^3 per minute [means metres cubed per minute]


Question 6

6.1 Tensile stress is the stress that is induced by a tensile (pulling) force, while compressive stress is the stress induced by a compressive (pushing) force.

6.2 stress = load / area
therefore the load = stress x cross sectional area
                           = 5 x 10^6 x pi/4 x 60^2 x 10^-4
                           =   1413716,694 N

6.3 we are given load = 100 kN, 25 mm x 25 mm profile original length = 330 mm, extension = 0,3 mm
6.3.1 stress = load / area
                   = 100 000 / 0,025 x 0,025
                   =  160      MPa

6.3.2    strain = change in length / original length
                     = 0,3 / 330
                     = 909,091 x 10^-6

6.3.3   Young's modulus of elasticty = stress / strain
                                                      =    160 x 10^6 / 909,091 x 10^-6  GPa
                                                       = 176 GPa

Question 7 Gas Laws, Boyle's Law, Charles' law, Universal gas Law

7.1 Definition of Boyle's Law

7.2 we are given : original length of wire = 10 cm; change in temp = 60 deg C; final length = 10,01722 cm;
       change in length = linear coefficient of expansion x original length x change in temperature
therefore linear coeff. of expansion = change in length / (original length x change in temp)
                                                       = 0,01722 / ( 10 x 60)
                                                      =     28,7 x 10^-6   / deg C

7.3       volume of nitrogen = 10 litre at 15 deg C T1 = 291 K; pressure = 1 600 kPa; new temp T2  = 5 deg C = 281 K
7.3.1

7.3.2                    mRT1 = P1V1
therefore               m = P1 x V1/(R x T1)
                                =  1600 000 x 0,010/ ( 260 x 291)
                               =    211,472 x 10^-3   kg                                         


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