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djyliett [7]
3 years ago
8

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

12,000 Btu/h of energy by heat transfer from the freezer compartment at 0°F and discharge energy by heat transfer to a kitchen at 70°F. Evaluate this claim.
Engineering
1 answer:
Lynna [10]3 years ago
6 0

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]

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2 years ago
A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside sur
SOVA2 [1]

Answer:

the rate of heat loss is 2.037152 W

Explanation:

Given data

stainless steel K = 16 W m^{-1}K^{-1}

diameter (d1) = 10 cm

so radius (r1)  = 10 /2 = 5 cm = 5 × 10^{-2}

radius (r2)  = 0.2 + 5 = 5.2 cm = 5.2 × 10^{-2}

temperature = 25°C

surface heat transfer coefficient = 6 6 W m^{-2}K^{-1}

outside air temperature = 15°C

To find out

the rate of heat loss

Solution

we know current is pass in series from temperature = 25°C to  15°C

first pass through through resistance R1  i.e.

R1  = ( r2 -  r1 ) / 4\pi  × r1 × r2 × K

R1  = ( 5.2 - 5 ) 10^{-2}  / 4\pi  × 5 × 5.2 × 16 × 10^{-4}

R1  = 3.825 ×  10^{-3}

same like we calculate for resistance R2 we know   i.e.

R2 = 1 / ( h × area )

here area = 4 \pi r2²

area = 4 \pi (5.2 × 10^{-2})²  =  0.033979

so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979  )

R2 = 4.90499

now we calculate the heat flex rate by the initial and final temp and R1 and R2

i.e.

heat loss = T1 -T2 / R1 + R2

heat loss = 25 -15 / 3.825 ×  10^{-3} + 4.90499

heat loss =  2.037152 W

8 0
3 years ago
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6 0
2 years ago
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A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

                                                 = 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

8 0
2 years ago
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Answer:

I don't really know but i have some info for you...

Explanation:

The cold forging manufacturing process increases the strength of a metal through strain hardening at a room temperature. On the contrary the hot forging manufacturing process keeps materials from strain hardening at high temperature, which results in optimum yield strength, low hardness and high ductility.

7 0
3 years ago
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