Question

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 °C and 50 °C. (a) In the units of Btu and J, how much heat flows along the rod each second? (b) What is the temperature of the welding rod at its midpoint?

Answer 1

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Step-by-step explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)=
`(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) * (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=(k*A*(T_1-T_2))/(L) \\Q=(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=(k*A*(T_1-T_2))/(L) \\Q=(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=(k*A*(T_1-T_2))/(L)`

On rearranging:

`T_2=T_1-(Q*L)/(KA)`

`T_2=500-(1.467*0.1)/(51.88*0.00001256) \\T_2=274.866\ C`