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2 years ago

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Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s

treet repair area of 2000 ft by 48 ft by 6 in. The aggregate in the stockpile contains 3.1 percent moisture. If the required compaction is 95% of the target, how many tons of aggregate will be needed?

1 answer:

djyliett [7]2 years ago

3 0

Answer:

total weight of aggregate = 5627528 lbs = 2814 tons

Explanation:

we get here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate = 5627528 lbs = 2814 tons

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Answer:

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Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W$

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$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275$

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