Answer:
Yes equation is valid.
Explanation:
Given:
h = (0.04 to 0.09)(D/d)^4*V^2/2*g
Using SI units to assign dimensions to every quantity as follows:
Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]
Hose diameter D = [m]
Nozzle tip diameter d = [m]
Fluid velocity in the hose V = [ m s^-1 ]
Acceleration of gravity g = [ m s^-2 ]
Using the Given Equation and plug the SI units of respective quantities:
h = (0.04 to 0.09)(D/d)^4*V^2/2*g
[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]
Simplify the equation above:
[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]
[m] = [m]
Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.
Answer:
The original length of the specimen 
Explanation:
Original diameter
= 30 mm
Final diameter
= 30.04 mm
Change in diameter Δd = 0.04 mm
Final length
= 105.20 mm
Elastic modulus E = 65.5 G pa = 65.5 ×
M pa
Shear modulus G = 25.4 G pa = 25.4 ×
M pa
We know that the relation between the shear modulus & elastic modulus is given by



This is the value of possion's ratio.
We know that the possion's ratio is given by


0.00476

Final length
= 105.2 m
Original length


This is the original length of the specimen.
Answer:
The surface temperature is 921.95°C .
Explanation:
Given:
a=25 cm ,P=350 hp⇒P=260750 W
Power transmitted
W and remaining will lost in the form of heat.This heat transmitted to air by the convection.
h=230
Actually heat will be transmit by the convection.
In convection Q=hA
So 
T=921.95°C
So the surface temperature is 921.95°C .
Answer:
b. 10A
Explanation:
Using the formula, E= k × r×I
200= 0.5 ×2000×0.02×I
200=20×I
Dividing with 20
I = 200/20= 10A
Answer:

Explanation:
From the information given:
Life requirement = 40 kh = 40 
Speed (N) = 520 rev/min
Reliability goal
= 0.9
Radial load
= 2600 lbf
To find C10 value by using the formula:

where;


The Weibull parameters include:



∴
Using the above formula:


![C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}](https://tex.z-dn.net/?f=C_%7B10%7D%20%3D%203640%20%5Ctimes%20%5Cbigg%5B%5Cdfrac%7B1248%7D%7B0.9933481582%7D%5Cbigg%5D%5E%7B%5Cdfrac%7B3%7D%7B10%7D%7D)

Recall that:
1 kN = 225 lbf
∴

