Answer:
222.5 Gpa
Explanation:
From definition of engineering stress,
where F is applied force and A is original area
Also, engineering strain, where l is original area and
is elongation
We also know that Hooke's law states that
Since A=20 mm* 20 mm= 0.02 m*0.02 m
F= 89000 N
l= 100 mm= 0.1 m
By substitution we obtain
Answer:
a). 8.67 x m
b).0.3011 m
c).0.0719 m
d).0.2137 N
e).1.792 N
Explanation:
Given :
Temperature of air, T = 293 K
Air Velocity, U = 5 m/s
Length of the plate is L = 6 m
Width of the plate is b = 5 m
Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X Pa-s
We know density of air is ρ = 1.21 kg /
Now we can find the Reyonld no at x = 1 m from the leading edge
Re =
Re =
Re = 332052.6
Therefore the flow is laminar.
Hence boundary layer thickness is
δ =
=
= 8.67 x m
a). Boundary layer thickness at x = 1 is δ = 8.67 X m
b). Given Re = 100000
Therefore the critical distance from the leading edge can be found by,
Re =
100000 =
x = 0.3011 m
c). Given x = 3 m from the leading edge
The Reyonld no at x = 3 m from the leading edge
Re =
Re =
Re = 996158.06
Therefore the flow is turbulent.
Therefore for a turbulent flow, boundary layer thickness is
δ =
=
= 0.0719 m
d). Distance from the leading edge upto which the flow will be laminar,
Re =
5 X =
x = 1.505 m
We know that the force acting on the plate is
=
and at x= 1.505 for a laminar flow is =
=
= 1.878 x
Therefore, =
=
= 0.2137 N
e). The flow is turbulent at the end of the plate.
Re =
=
= 1992316
Therefore =
=
= 3.95 x
Therefore =
=
= 1.792 N