A hydrological system consists of five horizontal formations with an equal thickness of 10 m each. The hydraulic conductivities of the formations are 20, 10, 15, 50, and 1 m/day, respectively. Calculate equivalent horizontal and vertical hydraulic s conductivities. If the flow in the uppermost layer is at an angle of 30 away from the normal direction relative to the boundary, calculate flow directions in all the formations.
Answers
Answer: Hello The required formula is attached below
answer:
A) KH = 19.2 m/day , Kv = 4.045 m/day
B) flow directions : ∝1 = 30°, ∝2 = 16.102°, ∝3 = 23.62, ∝4 = 55.55°,
∝5 = 1.66°
Explanation:
Given data :
Thickness of five horizontal formations ( K ) = 10 m each
Hydraulic conductivities of formations ( b ) =
b1= 20, b2= 10, b3= 15, b4= 50 and b5 = 1 (m/day)
a)Determine the equivalent horizontal and vertical conductivities
kH ( equivalent horizontal conductivity )
= ∑ Kb / b = ( K1b1 + K2b2 + k3b3 + k4b4 + k5 b5 ) / b1 + b2 +b3+b4+b5
input given values into the above equation
kH = 960 / 50 = 19.2 m/day
Kv ( equivalent vertical conductivity )
= ∑( b / (b/k) ) = ( b1 + b2 + b3 + b4 + b5 ) / ( b1/k1 + b2/k2 + b3/k3 + b4/k4 + b5/k5 )
input given values into equation above
Kv = 50 / 12.36 = 4.045 m/day
b) Determine the flow directions in all the formations
given that ∝1 = 30° and Ki / kj = bi / bj
k1 /k2 = tan ∝1 / tan ∝2
20 / 10 = tan 30 / tan ∝2
∴ ∝2 = 16.102°
K2/k3 = tan 16.102 / tan ∝3
= 10 / 15 = tan 16.102 / tan ∝3
∴ ∝3 = 23.62
K3/k4 = tan ∝3 / tan ∝4
= 15 / 50 = tan 23.62 / tan ∝4
∴∝4 = 55.55°
k4 /k5 = tan ∝4 / tan∝5
= 50 / 1 = tan 55.55 / tan ∝5
∴ ∝5 = 1.66°
Related Questions
A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 19,815 lb, a wingspan of 53.3 ft, and an average chord length of 6 ft. The Oswald efficiency factor is 0.81 and the zero-lift drag coefficient is equal to 0.02. The jet has two turbofan engines, each producing a maximum thrust of 3,650 lb at sea level.
Required:
a. Create a plot of the drag polar for this aircraft for CL from 0 to 5. Plot CL on the vertical axis, CD on the horizontal axis, and do not include negative CL values.
b. What is the total drag coefficient at the flight condition described above?
c. What is the required thrust for level flight at this altitude in lb?
d. If the pilot runs the engines at maximum thrust, what is the instantaneous rate of climb at this altitude and velocity?
Answers
Answer:
a) attached below
b) 0.0337
c) 2730.206 Ib
d) 2320.338 ft/min
Explanation:
a) Plot of the drag polar for this aircraft
first we will calculate :
Wing area (s) = Wing span (b) * Average chord length(c)
= 53.3 * 6 = 319.8 ft^2
Aspect ratio = b^2 / s = 8.883
K = 1 / [tex]\pi[/tex]eAR = 1 /
Drag polar ( Cd ) = 0.02 + 0.044 C^2L
attached below is a plot of the drag polar
Attached below is the detailed solution of the remaining part of the question
7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 veh/h. The approach is controlled by a pretimed signal with a cycle length of 60 seconds and D/D/1 queuing holds. Local standards dictate that signals should be set such that all approach queues dissipate 10 seconds before the end of the effective green portion of the cycle. Assuming that approach capacity exceeds arrivals, determine the maximum length
Answers
Answer:
23.34 seconds
Explanation:
Flow rate = 1500
Arrival = 800 vehicle per hour
Cycle c = 60 seconds
Dissipation time = 10 seconds
Arrival time = 800/3600 = 0.2222
Rate of departure = 1500/3600 = 0.4167
Traffic density p = 0.2222/0.4167 = 0.5332
Real time = r
r + to + 10 = c
to = c-r-10 ----1
t0 = p*r/1-p ----2
Equate both 1 and 2
C-r-10 = p*r/1-p
60-r-10 = 0.5332r/1-0.5332
50-r = 0.5332r/0.4668
50-r = 1.1422r
50 = 1.1422r + r
50 = 2.1422r
r = 50/2.1422
r = 23.34 seconds