Answer:
The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Explanation:
Length of curve is given as
is given as
The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as
A is given as
With initial grade, the elevation of PVC is
The station is given as
Low point is given as
The station of low point is given as
The elevation is given as
So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Answer:
Enthalpy at outlet=284.44 KJ
Explanation:
We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation
m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation
By putting the values
So h=284.44 KJ
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
= - ( εˣ / v )
= - ( -0.00088 / 0.34 )
= - ( - 0.002588 )
= 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
