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Arisa [49]
3 years ago
14

An uncovered swimming pool loses 1.0 inch of water off its 1,000 ft^2 surface each week due to evaporation. The heat of vaporiza

tion for water at the pool temperature is 1050 btu/lb. The cost of energy to heat the pool is $10.00 per million btu. A salesman claims that a S500 pool cover that reduces evaporation losses by two-thirds will pay for itself in one 15-week swimming season. Can it be true?
Engineering
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

The affirmation is true, the cover will be worth buying

Explanation:

The equation necessary to use is

E = m*cv,

Where

cv: the heat of vaporization.  

Finding the rate at which the water evaporates (m^3/week).  

The swimming pool loses water at 1 inch/week off its 1,000 ft^2

Than,

1000 ft² * 1 in/wk * 1 ft/12 in = 83.33 ft³/week

To obtains the rate of mass loss it is necessary to multiply it for the density of water

83.33 ft³/week * 62.4 lb/ft³ = 5200 lb/week

Knowing the vaporization heat it is possible to find the rate of heat which is leaving the swimming pool  

5200 lb/week * 1050 BTU/lb = 5460000 btu/week

Over a 15-week period, the pool loses 81.9 million BTU.  

Knowing the cost of energy to heat the pool is $10.00 per million btu

The price = $819

This way, the affirmation is true, the cover will be worth buying

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Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED
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Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

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2 years ago
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