BTEC Level 3 Diploma In Mechanical Engineering

This document provides solved assignments for BTEC Level 3 Diploma In Mechanical Engineering. It includes solutions for topics like concrete reservoir dam, up thrust on an immersed body, continuity of volume and mass flow for an incompressible fluid, and more. The document is relevant to the course Uni.t 5: Mechanical Principles and Applications.

BTEC Level 3 Diploma In Mechanical Engineering
Uni.t 5: Mechanical Principles and Applications

Assignment 1
PS Task 5 A concrete reservoir dam has a vertical face on the water side and dimensions as
shown below. The top water level reaches 1.5m below the crest of the dam, Determine the
resultant thrust at the centre of Pressure and the overturning moment about the base.
Assume that the density of the water is 1010 kg m-3 .The dam has a width of 20m.
Solution
Height of dam H= 13 meters
a = 1.5 meters
Depth of water h = 13-1.5 = 11.5 meters
Total water pressure per meter length .
p =1/2 * w* h^2
p = 1 / 2 * 1010* 9.81 * 11.5^2
p =655.17 kN / meter
Weight per meter length of the dam(W),
Weight of concrete = 24 kN/m^3 (considered)
Base = 20 m
Top = 10 m
W = 24 * ((1/2) *(10+20)*13)
W= 4680 kN/ meter
a
h/3
h

Resultant thrust on the base per meter length of the dam,
R =( p^2 + W^2)^(1/2)
R =( 655.17^2 + 4680^2)^(1/2)
R =4725.6 KN/ meter
Overturning moment per meter of length
M =p * h/3
M =655.17*11.5/3
M =2511.485 KNm per meter of length
P6 Task 6 Determine the up thrust on an immersed body
A steel ball of diameter 350mm and density 8150 kg/m3 .is immersed in to water with a
density of 1020 kg/ m^3 .What is the upthrust and the stress cable of 5 mm supporting the
ball .
Solution
Upthrust on the cable = Weight of steel ball – weight of ball balanced by water .
Upthrust on the cable = ρ1∗g∗v−ρ2∗g∗v
Upthrust on the cable =gv ( ρ¿¿ 1− ρ2)¿
Upthrust on the cable = 9.81∗4 π
3 ( 350
2∗1000 )3
∗(8150−1020)
Upthrust on the cable =1569.42 Newtons

Stress in cable
Stress = Upthrust / area of cable
Stress =1569.42 / ( π∗ ( d /2 )2
Stress =1569.42 / ( π∗( 5
2∗1000 )
2
Stress = 79.97 MPa
P7 Task 7 Use the continuity of volume and mass flow for an incompressible fluid to
determine the design characteristics of a gradually tapering pipe.(Draw a sketch of the pipe
system)
1.A square duct has sides of 1.5m and carries liquid at a velocity of 2m s·1. The duct reduces
to a cross sectional area of 1.5m2. Determine the volumetric flow rate and the velocity in
the smaller section.
Solution
As pipes are in series, therefore volumetric flow rate is same in both pipes.
Q = A x v
Q = 1.5 * 1.5 * 2
Q = 4.5 m^3 / sec
Velocity in smaller section
Q =A X v
4.5 = 1.5 x v
V = 3 m / sec
2A liquid flows in a pipe at a rate of 300kg min ·1. The pipe has a bore of 75mm diameter ,
The liquid has density of 800kg/ m3• Determine the volumetric flow rate and velocity of the
liquid?
Solution
a b

Mass flow rate m = ρ∗a∗v
300/60 = 800∗a∗v
0.00625 m^3/ sec = Q
Q =6.25 L /sec
Velocity
0.00625 m^3/ sec = Q
0.00625 = ( π∗( 75
2∗1000 )
2
x v
V = 1.415 m/sec
3.If the pipe diameter decreases to 50 mm, what will be the new velocity?
Solution
As pipes are in series, therefore volumetric flow rate is same in both pipes.
0.00625 m^3/ sec = Q
0.00625 = ( π∗( 50
2∗1000 )
2
x v
V = 3.184 m/sec
4. How will mass flow rate change because of the reduction in the pipe diameter?
Solution
Mass flow rate m = ρ∗a∗v
m∝ a
m∝ d2
As mass flow rate is square of directly proportional to diameter of pipe, therefore reduction
in diameter will reduce the mass flow rate.

Assignment 2
1. A quantity of gas in an engine cylinder occupies a volume of 2m^3 at a pressure of
300kPa. A piston slides in the cylinder and compresses the gas without changing the
temperature, until the volume is 0.5m^3 . If the area of the piston is 0.02m^2 ,
calculate the force on the piston when the gas is compressed.
Solution
P1 =300 kPa
V1 = 2 m^3
V2 = 0.5 m^3
Area a = 0.02 m^2
At constant temperature
P1∗V 1 =P2∗V 2
300∗2=P2∗0.5
P2=1200 kPa
Force on piston
F = Pressure (P2 * Area )
F = 1200 * 0.02
F = 24 Kn
2. A cylinder 200mm in diameter and 1.5m long contains 1.25 kg of oxygen a
temperature of 20°C. Determine the pressure of oxygen in the cylinder. Assume the
characteristic gas constant for oxygen is 260J kg^-1 K-1
Solution
D = 200 mm , L = 1.5 m
Mass of oxygen = 1.25 Kg
Temperature T = 20 deg C
r = 260J kg^-1 K-1
Ideal gas equation

PV = mrT
Volume V= π∗( 200
2∗1000 )
2
∗1.5
Volume V=0.0471 m^3
PV = mrT
P X 0.0471 = 1.25 X 260 X (20+273)
P = 20.21 Bar
3. (a) A compressed air cylinder has a volume of 0.6m3 and contains air at a pressure of
1.2MPa and temperature of 37°C. After use the pressure is 800kPa and the temperature is
17°C.
Calculate (a) the mass of air removed from the cylinder
Solution
P1 = 1.2 MPa
V1 = 0.6 m^3
T1 = 37 dec C
P2 = 800 kPa
T2 = 17 dec C
Ideal gas equation
P1∗V 1
T 1
= P2∗V 2
T 2
1.2∗1000∗0.6
( 37+273) = 800∗V 2
(17+ 273)
V2= 1.403 m^3
Mass of air removed
m1−m2= P1∗V 1
RT 1
− P2∗V 2
R T 2
m1−m2= 1.2∗1000∗0.6
0.287(37 +273)− 800∗1.403
0.287 (17+273)
Mass of air removed m¿=m1−m2=5.39 kgs

4. (b) The volume the mass of air removed from the compressed air cylinder above would
occupy at STP conditions. Take the characteristic gas constant of air (R) as 287 J kg-1 K-1 and
atmospheric pressure as l00kPa and standard temperature is 15°C.
Solution
Ideal gas equation
PV = mrT
100 x v = 5.39 x 287 x (15+273)
V = 0.224 m^3
volume the mass of air removed = 0.244 m^3

Assignment 3
P1 Calculate the magnitude, direction and position of the line of action of the resultant and
equilibrant of a non-concurrent coplanar Force system containing a minimum of four forces
acting in different directions
The mechanism shown is subject to the forces as shown below
A=4kN
B= 3 kN
C=5 kN
D =8 kN
Note when calculating
moments, the pivot

point is x
Find the resultant force acting on the structure about point X and therefore the force
required for equilibrium. Also determine the distance perpendicular to point X that this
force must act.
Solution
Forces (kN) H = F cosα V = F sinα H comp Moment V comp Moment
A 2.828 2.828 0 3.535
B -2.598 1.5 -2.598 0.375
C 0 5 0 0
D 8 0 2 0
SUM 8.23 9.328 -0.598 3.91
As per given formula , all the horizontal , vertical forces and horizontal , vertical moments at
points A , B ,C ,D are calculated and added in table .From the values obtain , sum of
horizontal and vertical values are also added in last row .
Resultant forces
R = √ ¿ ¿
R = √ ¿ ¿
R = 12.44 Kn
α = tan−1 ∑ FV
∑ FH
α = tan−1 9.328
8.23
α = 48.57 deg
The equilibrium force = resultant moment acting on structure =∑ FH moment +∑ FV moment
Resultant moment = -0.598 +3.91
Resultant moment =3.312 kNm

Distance perpendicular to X
Distance = Moment / Resultant force
Distance =3.312/12.44
Distance =0.266 meters from X