Michelle is the general manager of a power plant. This morning, she will meet with city officials to discuss environmental issue
1 answer:
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Answer:
a. 4.279 MPa
b. 3.198 MPa to 4.279 MPa
c. 0.939 MPa
d. Below 3.198 MPa
Explanation:
From the given parameters
M = 1.75 MPa
M at 1.6 MPa gives A/A* = 1.2502
M at 1.8 MPa gives A/A* = 1.4390
Therefore, by interpolation, we have M = 1.75 MPa gives A
However, we shall use M = 1.75 MPa and A
Similarly,
P/P₀ = 0.1878
a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have
A/A* = 1.387. by interpolation M
Therefore P = P₀ × P
Which shows that the nozzle is choked for back pressures lower than 4.279 MPa
b) Where there is a normal shock at the exit of the nozzle, we have;
M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa
Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406
Where the normal shock occurs at the nozzle exit, we have
Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as = 4.279 MPa
Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa
c) At M = 1.75 MPa and P
d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane
Answer:
0.867
Explanation:
The driver population factor ()can be estimated using the equation below:
The value of the heavy vehicle factor () is determined below:
The values of the = 2 and
= 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:
= 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833
Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:
= 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867
Answer:
Magnitude of force P = 25715.1517 N
Explanation:
Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.
To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Proof -
Given that,
Diameter = 12 mm = 0.012 m
Length = 0.6 m
= 0.015°
Youngs modulus of elasticity of 34 stainless steel is 193 GPa
Now,
By applying the conditions of equilibrium, we have
∑fₓ = 0, ∑ = 0, ∑M = 0
If ∑ = 0
⇒×0.9 - P × 0.6 = 0
⇒×3 - P × 2 = 0
⇒ =
If ∑ = 0
⇒×0.9 = P × 0.3
⇒ ×3 = P
⇒ =
Now,
Area, A = = 1.3097 × 10⁻⁴ m²
We know that,
Change in Length , =
Now,
= 9.1626 × 10⁻⁹ P
= 1.83253 × 10⁻⁸ P
Given that,
= 0.015°
⇒ = 2.618 × 10⁻⁴ rad
So,
⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9
⇒P = 25715.1517 N
∴ we get
Magnitude of force P = 25715.1517 N