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Amiraneli [1.4K]

3 years ago

9

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio

n 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.

1 answer:

saveliy_v [14]3 years ago

3 0

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

$K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}$

Put the value into the formula

$K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}$

$K_{H}=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}$

Put the value into the formula

$K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}$

$K_{V}=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

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A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of

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Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end + area of the curve surface

$=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}$

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Write using about 10-15 lines for each of the six materials (metals, ceramics, glasses, polymers, composites, and semiconductors

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Answer:

See Answer below- Explanation is the entire answer

Explanation:

Metals:

Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

Ceramics:

Properties: Brittle, poor heat conductors, poor electrical conductors, high wear resistance, corrosion resistance;

Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

Examples: concrete, tungsten carbide, diamond

Applications: bricks for constructions, clay pots to keep heat, cutting tools for metals;

Glasses:

Properties: amorphous, transparent, high weight

Drawbacks: poor conductors of heat and electricity; brittle; low shock resistance;

Examples: Silica, lead glass, glaze;

Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

Composites:

Properties: high strength to weight ratio, can get combination of properties from the used materials, rarely conductive, good shock resistance;

Drawbacks: high cost, hard to recycle, expensive;

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Applications: buildings, bullet proof vests, body of the Formula 1 cars, rockets, roof panels.

Semiconductors:

Properties: brittle, change conductive behavior under certain scenario, poor heat conductors;

Drawbacks: hard to manufacture, expensive;

Examples: Silicon-based semiconductors, Germanium-based semiconductors, Ga-based semiconductors;

Applications: chips, LED, diodes, transistors, op-amps, microprocessors.

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3 years ago

134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.

vichka [17]

Answer:

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(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = $- 10^{\circ}C$ = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = $60^{\circ}C$ = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T = $60^{\circ}C$ :

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at P = 140 kPa, T = $- 10^{\circ}C$ :

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = $s'_{s}$ and $h'_{s} = 278.06 kJ/kg.K$ at 700kPa

(a) Isentropic efficiency of the compressor, $\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}$

$\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%$

(b) The temperature of the environment, $T_{e} = 27^{\circ}C$ = 273 + 27 = 300 K

Availability at state 1, $\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg$

Similarly for state 2, $\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg$

Now, the efficiency of the compressor as per the second law;

$\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757$ = 67.57%

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They are as follows-

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Diligence
Automation
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