Question #210464

A brass bar, 15 cm in length, is welded to an aluminum bar, 10 cm in length. Both have a cross section of 3 cm on a side. If the free end of the brass bar is at 50 degrees Celcius and the the free end of the aluminum bar is at 25 degrees Celcius, what is the steady-state temperature at the junction of the 2 bars? *




1
Expert's answer
2021-06-24T16:14:45-0400

Solution.

lb=0.15m;l_b=0.15m;

la=0.1m;l_a=0.1m;

d=0.03m;d=0.03m;

tb=50oC;t_b=50^oC;

ta=25oC;t_a=25^oC;

ρb=8500kg/m3;\rho_b=8500kg/m^3;

ρa=2700kg/m3;\rho_a=2700kg/m^3;

cb=380J/kgoC;c_b=380J/kg^oC;

ca=920J/kgoC;c_a=920J/kg^oC;

mbcbΔTb=macaΔTa;m_bc_b\Delta T_b=m_ac_a\Delta T_a;

mb=ρbSblb;ma=ρaSala;m_b=\rho_bS_bl_b; m_a=\rho_aS_al_a;

ρblbcbΔTb=ρalacaΔTa;\rho_bl_bc_b\Delta T_b=\rho_al_ac_a\Delta T_a;

ΔTbΔTa=ρalacaρblbcb;\dfrac{\Delta T_b}{\Delta T_a}=\dfrac{\rho_al_ac_a}{\rho_bl_bc_b};

ΔTbΔTa=27000.192085000.15380=0.5;\dfrac{\Delta T_b}{\Delta T_a}=\dfrac{2700\sdot0.1\sdot920}{8500\sdot0.15\sdot380}=0.5;

tbtftfta=0.5    tf=tb+0.5ta1.5;\dfrac{t_b-t_f}{t_f-t_a}=0.5\implies t_f=\dfrac{t_b+0.5t_a}{1.5};

tf=50+12.51.5=41.7oC;t_f=\dfrac{50+12.5}{1.5}=41.7^oC;

Answer: tf=41.7oC.t_f=41.7^oC.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS