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:
Mechanical average of a wheel = 3
Explanation:
Given:
Radius of wheel = 1.5 ft = 1.5 x 12 = 18 inches
Radius of axle = 6 inches
Find:
Mechanical average of a wheel
Computation:
Mechanical average of a wheel = Radius of wheel / Radius of axle
Mechanical average of a wheel = 18 / 6
Mechanical average of a wheel = 3
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
![w_{b, out} =m([tex]u_{1} -u_{2](https://tex.z-dn.net/?f=w_%7Bb%2C%20out%7D%20%3Dm%28%5Btex%5Du_%7B1%7D%20-u_%7B2)
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
I believe the answer is D: brazing
Hope this helps you have a good night
