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)= [tex]\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)[/tex]
Thermal Conductivity is SI units:
[tex]k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K[/tex]
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)
[tex]A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2[/tex]
Heat Q is:
[tex]Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J[/tex]
In Btu:
[tex]A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2[/tex]
Heat Q is:
[tex]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[/tex]
PArt B:
At midpoint Length=L/2=0.1 m
[tex]Q=\frac{k*A*(T_1-T_2)}{L}[/tex]
On rearranging:
[tex]T_2=T_1-\frac{Q*L}{KA}[/tex]
[tex]T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C[/tex]
