Solution :
Given data :
p = 315612 Pa
At exit of B,
p =
At exit of A,
We need to determine X component of force () to hold in its place.
From figure,
Substitute all the values,
Therefore, the force required to hold the nozzle in its place along horizontal direction.
Answer:
Detailed solution is given in the attached diagram
Answer:
209.55 ft
Explanation:
Given Data:
Benchmark:
Reduced Level or Elevation = 210.50
Height of Instrument = Reduced Level + Back sight Reading
Height of Instrument = 210.50 + 3.57 = 214.07 ft
Turning Point:
Back sight Reading = 2.91 ft
Fore Sight Reading = 4.52
Reduced Level or Elevation of Turning Point = Height of Instrument – fore sight Reading
Reduced Level or Elevation of Turning Point = 214.07 – 4.52 = 209.55 ft
Height of Instrument at Turning Point = Reduced Level + Back sight Reading
Height of Instrument at Turning Point = 209.55 + 2.91 = 212.46 ft
Answer:
<em>a) 42 mm</em>
<em>b) 144.4 MPa</em>
<em></em>
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * *
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x x
= 7.638 x 10^-5
d = = 0.042 m = <em>42 mm</em>
b) Normal stress = P/A
where A is the area
A = = = 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>
Answer:
net power = 108.89kw
Explanation:
check the attachment for explicit explanation.