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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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5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025
Radda [10]

Answer:

LMC wall thickness= 5.05

Explanation:

Given:

Minimum inner diameter = 25 - 0.05 = 24.95

Maximum outer diameter = 35 + 0.05 = 35.05

Find:

LMC wall thickness

Computation:

LMC wall thickness = (maximum outer diameter - minimum inner diameter) / 2

LMC wall thickness = (35.05 - 24.95) / 2

LMC wall thickness= 5.05

3 0
3 years ago
Question 2 (Multiple Choice Worth 3 points)
ololo11 [35]

Answer:

the car to the right

Explanation:

its in the name the RIGHT of way hope it helps good luck

6 0
3 years ago
From the article "Time Travel Is A Fun Science Fiction Story But Could It Be Real?", Albert Einstein's Theory of Relativity is d
zhannawk [14.2K]

Answer:

Option A

Explanation:

As per the article "Time Travel Is A Fun Science Fiction Story But Could It Be Real?" the gravitational field is a representation of curving space and time. As the gravity becomes strong, the space-time get more curved and hence the time gets slower.

Hence, option A is correct

8 0
3 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

4 0
3 years ago
The thermal conductivity of a solid depends upon the solid’s temperature as k = a T+b where a and b are constants. The temperatu
Alexandra [31]

Answer:

Answer is Option (e) - None of these

Explanation:

The step by step derivation from the fourier's law of heat conduction is as shown in the attachment below.

Where K = thermal conductivity

7 0
4 years ago
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