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Answer:
the difference in pressure between the inside and outside of the droplets is 538 Pa
Explanation:
given data
temperature = 68 °F
average diameter = 200 µm
to find out
what is the difference in pressure between the inside and outside of the droplets
solution
we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is
σ = 2.69 × N/m
so average radius = = 100 µm = 100 ×
m
now here we know relation between pressure difference and surface tension
so we can derive difference pressure as
2π×σ×r = Δp×π×r² .....................1
here r is radius and Δp pressure difference and σ surface tension
Δp =
put here value
Δp =
Δp = 538
so the difference in pressure between the inside and outside of the droplets is 538 Pa
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:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = .................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q =
Q =
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W