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S_A_V [24]
3 years ago
12

2–25 Consider a medium in which the heat conduction equation is given in its simplest form as

Engineering
1 answer:
barxatty [35]3 years ago
3 0

Answer:

d) Is the thermal conductivity of the medium constant or variable.

Explanation:

As we know that

Heat equation with heat generation at unsteady state and with constant thermal conductivity given as

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}+\dfrac{\dot{q}_g}{K}=\dfrac{1}{\alpha }\dfrac{dT}{dt}

With out heat generation

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}

In 2 -D with out heat generation with constant thermal conductivity

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}

Given equation

\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{a }\dfrac{dT}{dt}

So we can say that this is the case of  with out heat generation ,unsteady state and with constant thermal conductivity.

So the option d is correct.

d) Is the thermal conductivity of the medium constant or variable.

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An oil reservoir has a average porosity of 20%, an area of 100 acres, and a av- erage thickness of 10 feet. The connate water sa
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Answer:a) recovery factor = 53.3% ,bi) volume of oil per acre-ft = 11090.64bbl/acre-ft

bii) bulk volume of the reservoir in acre-ft = 1000 acre-ft

c) 62214.74 cubic feet

Explanation: a) recovery factor is the percentage amount of oil that can recovered from a reservoir, it is the oil produced divided by oil initially in place

Recovery factor= 1 - (Soi/1-Swi)

= 1 - (0.35/1-0.25)

= 0.533 × 100%

= 53.3%

bi) volume of oil which may be recovered per acre-ft = 7758.porosity.(1- Swi-Soi)

= 7758 x 0.2 x (1-0.25-0.35)

= 620.64 bbl/acre-ft

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= 100 acres x 10 ft

= 1000 acre-ft

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since we have gotten volume as 1000 acre-ft we simply multiply it by the volume of oil gotten in answer bi)

= 1000 acre-ft x 620.64 bbl/ acre-ft

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3 years ago
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