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An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

Answers

Answer 1

Answer:

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

$K_(H)=(H_(A)K_(A)+H_(A)K_(A)+H_(A)K_(A))/(H_(A)+H_(B)+H_(C))$

Put the value into the formula

$K_(H)=(8.0*25+2,0*142+34*40)/(8.0+2.0+34)$

$K_(H)=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_(V)=(H_(A)+H_(B)+H_(C))/((H_(A))/(K_(A))+(H_(B))/(K_(B))+(H_(C))/(K_(C)))$

Put the value into the formula

$K_(V)=(8.0+2.0+34)/((8.0)/(25)+(2.0)/(142)+(34)/(40))$

$K_(V)=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

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Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T

Answers

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)

Q = 95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C

The compressibility factor provides a quick way to assess when the ideal gas law is valid. Use a solver to find the minimum temperature where the fluid has a vapor phase compressibility factor greater than 0.95 at 3 MPa. Report the value in oC, without units.

Answers

Answer:

The answer is

Explanation:

The compressibility factor

Under the right conditions, it is possible, due to surface tension,to have metal objects float on water. Consider placing a shortlength of a small diameter steel ( γ = 490 lb/ft3)rod on a surface of water. What is the maximum diameter that therod can have before it will sink? Assume that the surface tensionforces act vertically upward. Note: A standard paper cliphas a diameter of 0.036 in. Partially unfold a paper clip and seeif you can get it to float on water. Do the results of thisexperiment support your analysis?

Answers

Answer:

A) 0.0614 inches

b) The standard steel paper clip should float on water

Explanation:

The maximum diameter that the rod can have before it will sink

we can calculate this using this formula :

D = $((8\alpha )/(\pi y ) )^{(1)/(2) }$ ----- 1

∝ = value of surface tension of water at 60⁰f = 5.03×10^−3 lb/ft

y = 490 Ib/ft^3

input the given values into equation 1 above

D = $((8*(5.3*10^(-3)) )/(\pi *490 ) )^{(1)/(2) }$

= 5.11 * 10^-3 ft convert to inches

= 5.11 *10^-3 ( 12 in/ 1 ft ) = 0.0614 inches

B) The diameter of a standard paper Cliphas = 0.036 inches

and the diameter of the rod = 0.0614. Hence the standard steel paper clip should float on water

can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and angle. thanks

Answers

Answer:

branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
branch 2: i = 21.466∠63.435°; pf = 0.447 leading
total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

$X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915*10^(-3)\approx j4.99984\,\Omega$

The capacitive reactance is ...

$X_C=(1)/(j\omega C)=(-j)/(100\pi\cdot 318.31*10^(-6)F)\approx -j10.00000\,\Omega$

Branch 1

The impedance of branch 1 is ...

Z1 = 8 +j4.99984 Ω

so the current is ...

I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

Z2 = 5 -j10 Ω

so the current is ...

I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mechanical resonance frequency is [Equation] = 100 Hz. If the mass is to be made of cube of `silicon, what would its physical dimensions be?

Answers

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

Explanation:

Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg

What is the average distance in microns an electron can travel with a diffusion coefficient of 25 cm^2/s if the electron lifetime is 7.7 microseconds. Three significant digits and fixed point notation.

Answers

Answer: The average distance the electron can travel in microns is 1.387um/s

Explanation: The average distance the electron can travel is the distance an exited electron can travel before it joins together. It is also called the diffusion length of that electron.

It is gotten, using the formula below

Ld = √DLt

Ld = diffusion length

D = Diffusion coefficient

Lt = life time

Where

D = 25cm2/s

Lt = 7.7

CONVERT cm2/s to um2/s

1cm2/s = 100000000um2/s

Therefore D is

25cm2/s = 2500000000um2/s = 2.5e9um2/s