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lilavasa [31]
3 years ago
14

During a long run a very well-trained dog can use up to 1000 ‘cal’/hour (Note: Food calories differ by a factor of one thousand

from the scientific calorie). If it is assumed that the dog has a mass of twenty five kilograms and the physical properties of water, determine the uniform heat generation per unit volume under the assumption that the heat generation is uniform in the dog’s body(Note: You must determine the dog’s volume as part of the calculation.).
Engineering
1 answer:
cricket20 [7]3 years ago
5 0

Answer:

46488.8 W / m³

Explanation:

Given:

calories used by the dog = 1000 cal/ hour

1 Food calories = 1000 scientific calories

also,

1 scientific calorie = 4.184 J

thus,

total Power used by the dog per second = ( 1000 × 1000 ×  4.184 ) / 3600

or

total power used by the dog per second =  1162.22 W

Mass of the dog = 25 kg

since, the physical properties of the dog is similar to water

thus,

density of the dog = 1000 kg/m³

thus,

the volume of the dog = Mass / Density = 25 / 1000 = 0.025 m³

Now,

we have the relation

Heat supplied (Q) = volume × Heat generated per unit volume (q)

or

1162.22 = 0.025 × q

or

q = 46488.8 W / m³

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3 years ago
In 2002 Acme Chemical purchased a large pump for
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Answer:

$151094

Explanation:

Solution

Recall that:

Acme Chemical in 2002 purchased a large pump worth of = 112,000

The estimation for the pump to the industrial pump index is =100

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The current index is = 286

k = is the  reference year for which cost or price is known.

n =  the year for which cost or price is to be estimated (n>k).

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We find the estimated cost of the new pump which is stated as follows:

Cn = (112,000 * 286) /212

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3 years ago
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

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= 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)

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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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A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
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Answer:

Force per unit plate area is 0.1344 N/m^{2}

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