Find the mechanical average of a wheel axle System of the wheel has a radius of 1.5 feet in the accident has a radius of 6 inche
1 answer:
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Answer:
Explanation:
Obtain the following properties at 6MPa and 600°C from the table "Superheated water".
Obtain the following properties at 10kPa from the table "saturated water"
Calculate the enthalpy at exit of the turbine using the energy balance equation.
Since, the process is isentropic process
Use the isentropic relations:
Calculate the enthalpy at isentropic state 2s.
a.)
Calculate the isentropic turbine efficiency.
b.)
Find the quality of the water at state 2
since at 10KPa <
<
at 10KPa
Therefore, state 2 is in two-phase region.
Calculate the entropy at state 2.
Calculate the rate of entropy production.
since, Q = 0
Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
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