STP stands for standard temperature pressure and NTP stands for normal temperature pressure
Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.
Answer:
The flux (volume of water per unit time) through the hoop will also double.
Explanation:
The flux = volume of water per unit time = flow rate of water through the hoop.
The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.
This means that
Flow rate = AV
where A is the area of the hoop
V is the velocity of the water through the hoop
This flow rate = volume of water per unit time = Δv/Δt =Q
From all the above statements, we can say
Q = AV
From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number (
) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:
where 
is the density of the fluid, 
is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:
For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:
