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Alekssandra [29.7K]

3 years ago

8

A confined aquifer with a transmissivity of 300 m2/day and a storativity of 0.0005 and a well radius of 0.3 m. Find the drawdown

in the well at 100 days if the following pumping schedule is followed after a long period of time of no pumping.
Period
1 2 3 4
Time (days) 0-20 20-50 50-90 90-100
Q (m3/day) 500 300 800 0

1 answer:

Olin [163]3 years ago

8 0

Answer:

8.4627 m

Explanation:

Transmissivity( T ) = 300 m^2/day

Storativity( S ) = 0.0005

well radius ( r ) = 0.3m

<u>Determine the drawdown in well at 100 days </u>

Drawdown at 100 days = ∑ Drawdown at various period

We will use the equation : S = Q / U*π*T [ -0.5772 - In U ] ----- ( 1 )

where : Q = discharge , T = transmissivity

S = drawdown ,

U = r^2*s / 4*T*t --- ( 2 )

r = well radius , S = Storativity, t = time period

i) During 0-20

U1 = r^2*s / u*π*t = 1.875 * 10^-9

Input values into equation 1

S1 = 2.5885

ii) During 20-50

U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9

input values into equation 1

S2 = 1.5854 m

iii) During 50 -90

U3 = r^2*s / 4*π*t = 9.375 * 10^-10

input values into equation 1

S3 = 4.2888 m

iv) During 90-100

U4 = 0

s4 = 0

<em>Drawdown at 100 days = ∑ Drawdowns at various period </em>

<em> = s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0</em>

<em> = 8.4627 m</em>

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Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

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PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

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PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

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3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

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ANSWERS:

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Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

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Isothermal process $T=C$

a)

$-V_{2} =2V_{1}$ ( double)

b)

$-V_{2} =.5V_{2}$ (reduced by half)

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state(1)

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then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

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assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

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$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

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then the state exist in the wet zone

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3 years ago

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Answer:

using System;
public class Program
{
public static void Main()
{
Console.WriteLine("Enter number of students: ");
int num = Convert.ToInt32(Console.ReadLine());
string [] firstName = new string[num];
string [] lastName = new string[num];
for(int i=0 ; i < num; i++){
Console.WriteLine("Enter first name: ");
firstName[i] = Console.ReadLine();
Console.WriteLine("Enter last name: ");
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}
}
}

Explanation:

Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).

Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).

Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).

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AnnyKZ [126]

Answer:

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Explanation:

Please see attachment

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4 years ago

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

irina [24]

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.

So; we can say :

$D_{pursuit} =D_{police}$

By using the second equation of motion to find the distance S;

$S= ut + \dfrac{1}{2}at^2$

$D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)$

$D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)$

$D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)$

$D_{police} = ut _P + \dfrac{1}{2}at_p^2$

where ;

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$D_{police} = \dfrac{1}{2}at_p^2$

$D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2$

$D_{police} = 0.98*(t+12)^2$

$D_{police} = 0.98*(t^2 + 144 + 24t)$

$D_{police} = 0.98t^2 + 141.12 + 23.52t$

Recall that:

$D_{pursuit} =D_{police}$

$(187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t$

$(187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0$

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR t = 3.02

Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs

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Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s

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