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Problem Set on Conduction Gat Granada  Jamie hendrix  Angel Salcedo Kay Villaflor Jan Rannel Alejandro Group 3 4ChEB
Problem No. 3 ,[object Object]
Problem No. 3 ,[object Object],0.23m 315°C 38°C
0.23m 315°C 38°C
Solution: k = 0.006T + 1.4 x 10^-6 km = 0.006(T2 +T1)/2 – 1.4 x 10^-6  (T2^2 + T2T1 + T1^2) T2 = 588.15K T1 = 311.15 K Am = 1m^2 km = 2.4059 Wm/K q/(Am) = kmdT/dx = 2.4059 x  (588.15-311.15) /(.23-0) q/(Am) = 2897.54 w/m^2
Problem No. 11 ,[object Object]
Problem No. 11 ,[object Object],0. 245 0. 245 0. 245 Do = 0.0735 Di = 0.0245 Do = 0.1225 Di = 0.0735
Do = 0.0735 Di = 0.0245 Do = 0.1225 Di = 0.0735 Assume k1 = 1W/mK k2 = 4W/mK q = ∆T/∆R R1 = ∆x / KmAm = 0.0245/(1)(0.1401L) = 0.1748/L Am =  ∏L (Do – Di)  [ln(Do/Di)] R2 = ∆x / KmAm = 0.0245/(4)(0.3014L) = 0.0203/L q = ∆T / [(0.1748/L) + (0.0203/L)] q =  ∆T / [(0.1951/L)]
Geankoplis: 4.3-2, 5.3-3
5.3-3 Cooling of a Slab of Aluminum. A large piece of aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4K. The surface is suddenly exposed to an environment at 338.8K with a surface convection coefficient of 455W/m 2 -K.  Calculate  the time in hours (hr) for the temperature to reach 388.8K at a depth of 25.4mm. The average physical properties are  α =0.340m 2 /hr and k=208W/m-K.
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],T1 = 388.8K T0 = 505.4K 25.4 mm T = 388.8K
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],1588K 299K
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
THANK YOU

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Problem set 2 4b3

  • 1. Problem Set on Conduction Gat Granada Jamie hendrix Angel Salcedo Kay Villaflor Jan Rannel Alejandro Group 3 4ChEB
  • 2.
  • 3.
  • 5. Solution: k = 0.006T + 1.4 x 10^-6 km = 0.006(T2 +T1)/2 – 1.4 x 10^-6 (T2^2 + T2T1 + T1^2) T2 = 588.15K T1 = 311.15 K Am = 1m^2 km = 2.4059 Wm/K q/(Am) = kmdT/dx = 2.4059 x (588.15-311.15) /(.23-0) q/(Am) = 2897.54 w/m^2
  • 6.
  • 7.
  • 8. Do = 0.0735 Di = 0.0245 Do = 0.1225 Di = 0.0735 Assume k1 = 1W/mK k2 = 4W/mK q = ∆T/∆R R1 = ∆x / KmAm = 0.0245/(1)(0.1401L) = 0.1748/L Am = ∏L (Do – Di) [ln(Do/Di)] R2 = ∆x / KmAm = 0.0245/(4)(0.3014L) = 0.0203/L q = ∆T / [(0.1748/L) + (0.0203/L)] q = ∆T / [(0.1951/L)]
  • 10. 5.3-3 Cooling of a Slab of Aluminum. A large piece of aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4K. The surface is suddenly exposed to an environment at 338.8K with a surface convection coefficient of 455W/m 2 -K. Calculate the time in hours (hr) for the temperature to reach 388.8K at a depth of 25.4mm. The average physical properties are α =0.340m 2 /hr and k=208W/m-K.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16.