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lutik1710 [3]
3 years ago
12

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.

Engineering
1 answer:
Trava [24]3 years ago
3 0

Answer:

a. Heat removal rate will increase

b. Heat removal rate will decrease

Explanation:

Given that

One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.

We know that heat transfer in fin given as follows

Q_{fin}=\sqrt{hPKA}\ \Delta T

We know that area

A=\dfrac{\pi}{4}d^2

Now when diameter will triples then :

A_f=\dfrac{\pi}{4}{\left (3d \right )}^2

A_f=9A

Q'_{fin}=\sqrt{9hPKA}\ \Delta T

Q'_{fin}=3\sqrt{hPKA}\ \Delta T

Q'_{fin}=3Q

So the new heat transfer will increase by 3 times.

Now when copper rod will replace by aluminium rod :

As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.

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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 °
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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

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