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3 years ago

Which of the following describes what occurs when energy is lost in efficient transformation?

Engineering

1 answer:

8090 [49]3 years ago

8 0

Answer: It takes the form of heat or thermal power.

Explanation: I got it right on the test.

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You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is

kipiarov [429]

14256000. Kanjiuijhgg

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3 years ago

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Most pedestrian fatalities occur

Makovka662 [10]

Answer: Outside an intersections

Explanation:

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3 years ago

An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove

Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

$n=Q out / W in$

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

$12000 BTU/hr = 4.72 hp$

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]

6 0

3 years ago

Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.

lisov135 [29]

Answer:

specific energy = 2.65 ft

y2 = 1.48 ft

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + $\frac{v^2}{2g}$ ...............1

put here value and we get

specific energy = $2 + \frac{6.5^2}{2\times 9.8\times 3.281}$

specific energy = 2.65 ft

and

alternate depth is

y2 = $\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})$

and

here Fr² = $\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}$

Fr² = 0.8025

put here value and we get

y2 = $\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})$

y2 = 1.48 ft

7 0

3 years ago

Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma

Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

$F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}$

The bouyancy force is:

$F_{B} =V*p_{w} *g$

The weight is:

$W=V*p_{c} *g$

Laminar flow:

$v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s$

Reynold number:

$Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1$

Not flow region

For Newton flow region:

$v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s$

$Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4$

$Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31$

$Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99$

$Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K$

$\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s$

$e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m$

8 0

3 years ago