We have a parallel-plate capacitor, with each plate having a width W and a length L. The plates are separated by air with a dist
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Answer:
a) The change in Kinetic energy, KE = -1.95 kJ
b) Power output, W = 10221.72 kW
c) Turbine inlet area,
Explanation:
a) Change in Kinetic Energy
For an adiabatic steady state flow of steam:
.........(1)
Where Inlet velocity, V₁ = 80 m/s
Outlet velocity, V₂ = 50 m/s
Substitute these values into equation (1)
KE = -1950 m²/s²
To convert this to kJ/kg, divide by 1000
KE = -1950/1000
KE = -1.95 kJ/kg
b) The power output, w
The equation below is used to represent a steady state flow.
For an adiabatic process, the rate of heat transfer, q = 0
z₂ = z₁
The equation thus reduces to :
w = h₁ - h₂ - KE...........(2)
Where Power output, ..........(3)
Mass flow rate,
To get the specific enthalpy at the inlet, h₁
At P₁ = 10 MPa, T₁ = 450°C,
h₁ = 3242.4 kJ/kg,
Specific volume, v₁ = 0.029782 m³/kg
At P₂ = 10 kPa, , x₂ = 0.92
specific enthalpy at the outlet, h₂ =
h₂ = 3242.4 + 0.92(2392.1)
h₂ = 2392.54 kJ/kg
Substitute these values into equation (2)
w = 3242.4 - 2392.54 - (-1.95)
w = 851.81 kJ/kg
To get the power output, put the value of w into equation (3)
W = 12 * 851.81
W = 10221.72 kW
c) The turbine inlet area
Answer:
The velocity at section is approximately 42.2 m/s
Explanation:
For the water flowing through the pipe, we have;
The pressure at section (1), P₁ = 300 kPa
The pressure at section (2), P₂ = 100 kPa
The diameter at section (1), D₁ = 0.1 m
The height of section (1) above section (2), D₂ = 50 m
The velocity at section (1), v₁ = 20 m/s
Let 'v₂' represent the velocity at section (2)
According to Bernoulli's equation, we have;
Where;
ρ = The density of water = 997 kg/m³
g = The acceleration due to gravity = 9.8 m/s²
z₁ = 50 m
z₂ = The reference = 0 m
By plugging in the values, we have;
50 m + 30.704358 m + 20.4081633 m = 10.234786 m +
50 m + 30.704358 m + 20.4081633 m - 10.234786 m =
90.8777353 m =
v₂² = 2 × 9.8 m/s² × 90.8777353 m
v₂² = 1,781.20361 m²/s²
v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s
The velocity at section (2), v₂ ≈ 42.2 m/s
Answer:
a. 318.2k
b. 45.2kj
Explanation:
Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.
See attachment for detailed analysis
