Orthographic drawings are used to express ideas that are more complicated. Explain the purpose of the different views and the im
2 answers:
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Answer:
a)Wt =25.68 lbf
b)Wt = 150 lbf
F= 899.59 N
Explanation:
Given that
m= 150 lbm
a)
Weight on the spring scale(Wt) = m g
We know that
Wt = 150 x 5.48/32 lbf
Wt =25.68 lbf
b)
On the beam scale
This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.
Wt = 150 lbf
If the plane is moving upward with acceleration 6 g's then the for F
F = m a
We know that
a=6 g's
So
F = 90 x 9.99 N
F= 899.59 N
The radius of the specimen is 60 mm
<u>Explanation:</u>
Given-
Length, L = 60 mm
Elongated length, l = 10.8 mm
Load, F = 50,000 N
radius, r = ?
We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this
elongation using Equation:
ε = Δl / l₀
ε = 10.8 / 60
ε = 0.18
We know,
σ = F / A
Where A = πr²
According to the stress-strain curve of brass alloy,
σ = 440 MPa
Thus,
Therefore, the radius of the specimen is 60 mm
Answer:
Pump factor = Fp = 7.854 gal/cycle
Ev = 82.00 %
P_H = 183.29 hp
Explanation:
Given data:
Dimension of duplex pump
6.5 inch liner
2.5 inch rod
18 inch strokes
Pressure 3000 psig
Pit dimension
7 ft wide
20 ft long
Ls = 18 inch
Velocity = (18)/10
volumetric efficiency is given as E_v = (Actual flow rate)/(Theortical flow rate) * 100
we know that flow rate is given as = Area * velocity
Theoritical flow rate
Ev = 82.00 %
Pump factor
Fp = 1814.22 in^3/cyl
Fp = 7.854 gal/cycle
Flow rate
Power
Answer:
the temperature of the aluminum at this time is 456.25° C
Explanation:
Given that:
width w of the aluminium slab = 0.05 m
the initial temperature = 25° C
h = 100 W/m²
The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;
density ρ = 2702 kg/m³
thermal conductivity k = 231 W/m.K
Specific heat c = 1033 J/Kg.K
Let's first find the Biot Number Bi which can be expressed by the equation:
Bi = 0.0108
The time constant value is :
Considering Lumped capacitance analysis since value for Bi is less than 1
Then;
where;
which correlates with the change in the internal energy of the solid.
So;
The maximum value for the change in the internal energy of the solid is :
By equating the two previous equation together ; we have:
Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75
Thus;
So;
t = 1.386294361 × 697.79
t = 967.34 s
Finally; the temperature of Aluminium is determined as follows;
T - 600 = -575 × 0.25
T - 600 = -143.75
T = -143.75 + 600
T = 456.25° C
Hence; the temperature of the aluminum at this time is 456.25° C