BTEC Level 3 Diploma In Mechanical Engineering

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