The initial temperature distribution of a 5 cm long stick is given by thefollowing function. The circumference of the rod in question is completelyinsulated, but both ends are kept at a temperature of 0 °C. Obtain the heatconduction along the rod as a function of time and position ? (x =1.752 cm²/s for the bar in question)100A) T(x1) = 1 Sin ().e(-1,752 (³¹)+(sin().e (-1,752 (²) ₁ +13πTC3.....)100t + ··· .......13) T(x,t) = 200 Sin ().e(-1,752 (²t) + (sin (3). e (-1,752 (7) ²) tB)3/3t + …............)C) T(x.t) = 200 Sin ().e(-1,752 (²t) (sin().e(-1,752 (7) ²) t–D) T(x,t) = 200 Sin ().e(-1,752 (²)-(sin().e (-1,752 (²7) ²) tE) T(x.t)=(Sin().e(-1,752 (²t)-(sin().e(-1,752 (²) t+t + ··· .........)t +....t + ··· .........)…..)

The initial temperature distribution function, TS0(x):TS0(x) = -1752 sin(pi*x/5)Next, we can use the…

A) T(xt)=10Sin ().e(-1,752 (²+)+(sin().e(-1,752 (3²)t + B) T(x.t) = 200 (Sin ().e(-1,752 (²t) + (sin().e(-1,752 (37)²) t. t + ··· ... ····.·.) ....) TC3 C) T(x.t) = 20 Sin ().e(-1,752 (t)-(sin().e(-1,752 (²²) t t +...........) D) T(x.t) = 200 Sin ().e(-1,752 (²t)-(sin().e (-1,752 (²) ²) t+-..) 73 E) T(x.t) = 100 Sin ().e(-1,752 (²t)-(sin().e(-1,752 (²) ²) t+ -

A) T(xt)=10Sin ().e(-1,752 (²+)+(sin().e(-1,752 (3²)t + B) T(x.t) = 200 (Sin ().e(-1,752 (²t) + (sin().e(-1,752 (37)²) t. t + ··· ... ····.·.) ....) TC3 C) T(x.t) = 20 Sin ().e(-1,752 (t)-(sin().e(-1,752 (²²) t t +...........) D) T(x.t) = 200 Sin ().e(-1,752 (²t)-(sin().e (-1,752 (²) ²) t+-..) 73 E) T(x.t) = 100 Sin ().e(-1,752 (²t)-(sin().e(-1,752 (²) ²) t+ -

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.10P: 5.10 Experiments have been performed on the temperature distribution in a homogeneous long cylinder...

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