Answer to Question #231323 in Molecular Physics | Thermodynamics for Unknown346307

Question #231323

A small electric heating application

uses wire of 2 mm diameter with 0.8 mm thick insulation (k

= 0.12 W/m°C). The heat transfer coefficient (ho) on the in￾sulated surface is 35 W/m2°C. Determine the critical thick￾ness of insulation in this case and the percentage change in

the heat transfer rate if the critical thickness is used, assum￾ing the temperature difference between the surface of the

wire and surrounding air remains unchanged.




1
Expert's answer
2021-08-31T17:13:18-0400

Gives

k=0.12W/m°ck=0.12W/m°c

h0=35W/m2°ch_0=35W/m^2°c

r1=2mmr2=1+0.8=2.8mmr_1=2mm\\r_2=1+0.8=2.8mm

Q1˙=t12πklln(r2r1)+12πr2lh0\dot{Q_1}=\frac{∆t}{\frac{1}{2\pi kl}ln(\frac{r_2}{r_1})+\frac{1}{2\pi r_2l h_0}}

Q1˙=2πlt1kln(r2r1)+1r2h0\dot{Q_1}=\frac{2\pi l∆t}{\frac{1}{k}ln(\frac{r_2}{r_1})+\frac{1}{r_2h_0}}

Put value

Q1˙=2πlt10.12ln(2.82)+10.0028×35\dot{Q_1}=\frac{2\pi l∆t}{\frac{1}{0.12}ln(\frac{2.8}{2})+\frac{1}{0.0028\times35}}

Q1˙=2πlt0.02803+10.2040\dot{Q_1}=\frac{2\pi l∆t}{0.02803+10.2040}

Q1˙=2πlt10.2321\dot{Q_1}=\frac{2\pi l∆t}{10.2321}

Q1˙=2πlt×0.09773\dot{Q_1}=2\pi l∆t \times0.09773

Critical radius


rc=kh0=0.1235=0.0034285m=3.4285mmr_c=\frac{k}{h_0}=\frac{0.12}{35}=0.0034285m=3.4285mm

Q2˙=2πlt1kln(rcr1)+1rch0\dot{Q_2}=\frac{2\pi l∆t}{\frac{1}{k}ln(\frac{r_c}{r_1})+\frac{1}{r_ch_0}}

Put value

Q2˙=2πlt10.12ln(3.42852)+10.0034285×35\dot{Q_2}=\frac{2\pi l∆t}{\frac{1}{0.12}ln(\frac{3.4285}{2})+\frac{1}{0.0034285\times35}}

Q2˙=2πlt4.491+8.333\dot{Q_2}=\frac{2\pi l∆t}{4.491+8.333}

Q2˙=2πlt12.8245\dot{Q_2}=\frac{2\pi l∆t}{12.8245}

Q2˙=2πlt×0.07797\dot{Q_2}={2\pi l∆t}\times0.07797

Percentage increase and heat loss


=Q2˙Q1˙Q1˙×100=0.077970.097730.09773×100=20.21=\frac{\dot{Q_2}-\dot{Q_1}}{\dot{Q_1}}\times100=\frac{0.07797-0.09773}{0.09773}\times100=20.21

Heat loss=20.21%


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!

Leave a comment