Determine the volume change, in ft3, when 1 lb of water, initially saturated liquid, is heated to saturated vapor while pressure
remains constant at 1.0, 14.7, 100, and 500, each in lbf/in.2Comment.a.1.0: 333.58, b.14.7: 26.78, c.100: 4.42, d.500: 0.91
1 answer:
You might be interested in
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
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