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marissa [1.9K]
3 years ago
14

2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.

a. Estimate drawdown in this well if the radius of influence is 100 ft. b. Estimate the hydraulic gradient in the aquifer at distances of 20 and 75 ft, both directly upstream and downstream of the well.

Engineering
1 answer:
hodyreva [135]3 years ago
8 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.

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Read 2 more answers
Estimate the chronic daily intake of toluene from exposure to a city water supply that contains a toluene concentration equal to
vredina [299]

Answer: the total chronic daily intake is  0.03296 mg/ kg.d  

Explanation:

First of  we have to determine the CDI due to ingestion of drinking water

so

CDI = (C x IR x EF x ED) / (BW x AT)     let leave this as equation 1

where C is the concentration of the life time risk of drinking water(1.0mg/L)

ED is the risk dying per year((70),

EF is the number of days per year(365 day/year)

BW is the weight of body(65.4)kg

AT is the average tenure time of life(75years)

IR = 2.3 L/days

so we substitute

CDI₁ = (1(mg/L) x 2.3(L/day) x 365 x 70) / (65.4 x 365 x 75)

= 3.29 x 10⁻² mg/kg.d  

next we determine the CDI due to dermal contact with water.  

find the ingestion rate which is equal to the exposure time in hour/day.

IR = ET

(20 min/day) /  (60 min/hour)

= 0.333 h/day

now lets substitute for CD1₂

1.0 mg/L for C , 0.333 h/day for m , 365 day/year for EF, 70 year for ED, 65.4 kg for Bw , and 75 year for AT

0.8 for submergence and 1.69m² for area of skin of adult female in our equation 1

CDI₂ = [(1 mg/1) × (1.69 m²) × (9.0 x 10⁻⁶(m/h)) × 0.333(h/day) x 365 x 70)) / (65.4 x 365 x 75)]  × ( 0.8 x 10³ L/m³)

CDI₂ = 5.41 x 10⁻⁵ mg/kg.d  

now we find the CDI due to inhalation during bath

we substitute 1.0 mg/L for C, 11.3m³/day for IR , 365 day/year for EF , 70 year for ED. 65.4 kg for BW , and 75 year for AT in our equation 1

CDI₃ = [( 1ug/m³ × 1mg/10³ug) x (11.3m³/day x 1day/24 hr) x 365 x 70)] / (65.4 kg x 365 x 75 )

= 6.71 x 10⁻⁶ mg/kg.d

finally we Calculate the total chronic daily intake value

CDI_total = CDI₁ + CDI₂ + CDI₃  

we substitute

CDI_total = 3.29 x 10⁻² + 5.41 x 10⁻⁵ + 6.71 x 10⁻⁶

= 0.03296 mg/kg.d  

so the total chronic daily intake is  0.03296 mg/ kg.d  

4 0
3 years ago
The rate of heat transfer between a certain electric motor and its surroundings varies with time as , Q with dot on top equals n
tekilochka [14]

Answer:

the required expression is E_2 - E_1 = 4[ 1 - e^{(-0.05t)} ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - e^(-0.05t) ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w'  ------ let this be equation 1

w' is the net power on the system

w' = w_{elect - w_{shaft

w_{shaft = T × ω

we substitute

w_{shaft= 18 × 100

w_{shaft = 1800 W

w_{shaft = 1.8 kW  

so

w' = w_{elect - w_{shaft

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - e^{(-0.05t) ] + 0.2

dE/dt = -0.2 + 0.2e^{(-0.05t) ] + 0.2

dE/dt = 0.2e^{(-0.05t)

Now, the change in total energy, increment E, as a function of time;

ΔE = \int\limits^t_0}\frac{dE}{dt}  . dt

ΔE = \int\limits^t_0} 0.2e^{(-0.05t)} dt

ΔE = \int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0

E_2 - E_1 = 4[ 1 - e^{(-0.05t)} ]

Therefore, the required expression is E_2 - E_1 = 4[ 1 - e^{(-0.05t)} ]

5 0
2 years ago
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