On a day in which the local atmospheric pressure is 99.5 kPa, answer each of the following: (a) Calculate the column height of m
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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:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S =
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M