A certain working substance receives 100 Btu reversibly as heat at a temperature of 1000℉ from an energy source at 3600°R. Refer
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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:
The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh
However, the average power P, required for a year, t = ? Kw
Expressing their relationship, we will have
U = P x t
Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)
t = 8760 hours
P =
P = 7134.7Kw
Hence, the average power required during the year is 7,135Kw
Now to calculate the energy used by the power plant in a year (in quads)?
Recall, Efficiency, η = Power Output/Power Input (100)
so, we have
η = P₀/P₁, given
0.45 =
P₁ = 15,855Kw
the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules
So to convert this to quads, Note;
1 quads of energy = 10¹⁵ Joules
The total energy used is 0.000000139 quads
Now to find the cubic feet of natural gas required to generate this power?
Note: 0.29Kwh of Power generated = 1 cubic feet of natural gas used
Since, the power plant generated = 62500000000Kwh
The cubic feet of natural gas used =
Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.
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