{"id":432550,"date":"2018-04-02T08:15:12","date_gmt":"2018-04-02T08:15:12","guid":{"rendered":"https:\/\/essaypaper.org\/thermodynamics-solution-690952\/"},"modified":"2018-10-24T08:59:44","modified_gmt":"2018-10-24T08:59:44","slug":"thermodynamics-solution-690952","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/thermodynamics-solution-690952\/","title":{"rendered":"Thermodynamics Solution : 690952"},"content":{"rendered":"
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Question:<\/p>\n

Discuss about the Thermodynamics and give brief solution.<\/p>\n

Answer:<\/p>\n

Given data<\/p>\n

Fin length L =110 mm<\/p>\n

Fin thickness t = 1 mm<\/p>\n

Thermal conductivity\u00a0 K =325 W\/mk<\/p>\n

Heat transfer coefficient h = 90 W\/m^2k<\/p>\n

1) Find position and length of minimum temperature .<\/p>\n

Consider the equation<\/p>\n

\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261)<\/p>\n

ta\u00a0 = atmospheric temperature<\/p>\n

C1 , C2 =constant<\/p>\n

m =<\/p>\n

P =Perimeter =2*2 =4mm^2<\/p>\n

A =Area =1 mm^2<\/p>\n

Put values<\/p>\n

m =<\/p>\n

m = 33.28 m^-1<\/p>\n

Use boundary condition<\/p>\n

At x = 0 t =100 deg C , At x = 0.11 m , t = 70 deg C<\/p>\n

Put above condition in equation 1)<\/p>\n

We get<\/p>\n

90 = C1 + C2 \u2026\u2026..2)<\/p>\n

30 =38.89 C1 +0.00257 C2 \u2026\u2026\u2026..3)<\/p>\n

\u00a0<\/p>\n

Solve the equation 2) , 3) We get ,<\/p>\n

C1 =0.702 , C2 = 89.29<\/p>\n

Put C1 , C2\u00a0 values in equation 1)<\/p>\n

\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20264)<\/p>\n

For minimum value of temperature =0<\/p>\n

Now differentiate equation 4 w.r.t x<\/p>\n

Apply log on both sides<\/p>\n

x\u00a0 = 0.0728 m<\/p>\n

x = 72.8 mm from the left side <\/strong><\/p>\n

Apply value of x at equation 4)<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Heat loss<\/p>\n

\u00a0<\/strong><\/p>\n

Fin efficiency<\/p>\n

2) Calculate the minimum length<\/p>\n

\u00a0<\/p>\n

Consider the equation<\/p>\n

\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20265)<\/p>\n

ta\u00a0 = atmospheric temperature<\/p>\n

C1 , C2 =constant<\/p>\n

m =<\/p>\n

P =Perimeter =2*2 =4mm^2<\/p>\n

A =Area =1 mm^2<\/p>\n

Put values<\/p>\n

m =<\/p>\n

m = 33.28 m^-1<\/p>\n

Use boundary condition<\/p>\n

At x = 0 t =100 deg C , At x = 0.11 m , t = 70 deg C<\/p>\n

Put above condition in equation 5)<\/p>\n

We get<\/p>\n

90 = C1 + C2 \u2026\u2026..2)<\/p>\n

30 =38.89 C1 +0.00257 C2 \u2026\u2026\u2026..6)<\/p>\n

\u00a0<\/p>\n

Solve the equation 5) , 6) We get ,<\/p>\n

C1 =0.702 , C2 = 89.29<\/p>\n

Put C1 , C2\u00a0 values in equation 5)<\/p>\n

\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20267)<\/p>\n

For minimum value of temperature =0<\/p>\n

Now differentiate equation 7 w.r.t x<\/p>\n

Apply log on both sides<\/p>\n

x\u00a0 = 0.0728 m<\/p>\n

x = 72.8 mm from the left side <\/strong><\/p>\n

Apply value of x at equation 7)<\/p>\n

The temperature is at lowest value of 25.84 \u00a0,So it is considered that the length of fin should be of length = 72.8 , after that the temperature would rise due to effect of other plate .After this length , further increase in length is of no use, as it do not cause any reduce in temperature .<\/p>\n

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