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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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At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing
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Answer:

vB = - 0.176 m/s   (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒  (OA) = Sin 35°*(AB)

⇒   (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒  (OB) = Cos 35°*(AB)

⇒   (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒  (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒  vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒  vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒  vB = - 0.176 m/s   (↓-)

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3 years ago
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A bathtub faucet has a maximum flow rate of 3 gal/min. The tub is rectangular (3 ft long x 2 ft wide x 1.5 ft deep). Although th
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5. Name two health problems that fume can cause?<br> a)....<br> b)......
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One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal
7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

 

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m  

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 )  = ( (4h/kD)^1/2 )  

m =  ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04    

now lets find the value of h/mk    

h/mk = 10⁵ / ( 942809.04 × 30) =  0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

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the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ 1 ]  / [cosh mL+  (h/mk) sinh mL]

so we substitute

0.893 =  [ 1 ]  / [cosh (942809.04 × L) +  (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

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