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4ChEB Group 6


Answers to Problem #s 6 and 14 of Problem Set 2 plus problems from Geankoplis..

Answers to Problem #s 6 and 14 of Problem Set 2 plus problems from Geankoplis..

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  • 1. Allan Gabriel Carasco Roent Dune Cayetano Patrick Cristobal Nikko Zandro Reyes Aaron Turingan
  • 2.
    • 4.2-2 Heat Removal of a Cooling Coil
    • A cooling coil of 1.0 ft of 304 stainless-steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40F and I 80F on the outside. The thermal conductivity of 304 stainless steel is a function of temperature:
    • k=7.75+7.78E-3T
    • where k is in btu/h*ft*F and T is in F. Calculate the heat removal in btu/s and watts.
  • 3.
    • Given:
    • k=7.75+7.78E-3T
    • D in = 0.25in =0.0208ft k in =8.0612btu/h*ft*F
    • D out = 0.40in =0.0333ft k out =8.3724btu/h*ft*F
    • Sol’n:
    * * =1308W
  • 4.
    • 5.3-9 Temperature of Oranges on Trees During Freezing Weather.
    • In orange-growing areas, the freezing of the oranges on the trees during cold noghts is of serious economic concern. If the oranges are initially at a temperature of 21.1C, calculate the center temperature of orange is exposed to air at -3.9C for 6 hours. The oranges are 102 mm in diameter and the convective coefficient is estimated as 11.4W/m2-K. The thermal conductivity k is 0.431 W/m-K and α is 4.65E-4m2/h. Neglect any latent heat effects.
  • 5.
    • Given:
    • k = 0.431 W/m-K α = 4.65E-4m2/h
    • T o =21.1C T 1 =-3.9C T(center)=?
    • Solution:
    • assume m=0
    • From the graph we get value of Y=0.05
    • T at center = -2.65C
  • 6.
    • Problem #6
    • A flat slab of rubber, 2.5 cm thick initially at 20°C is to be placed between two heated steel plates maintained at 140°C. The heating is to be discontinued when the temperature at the mid plane of the slab is 132°C. The rubber has a thermal conductivity of 0.16 W/mK and thermal diffusivity of 8.671x10 -6 m 2 /s. Thermal resistance from metal to rubber may be neglected. Calculate:
      • A) The length of the heating period, sec
      • B) The temperature of the rubber 0.65cm from the metal
      • C) The time required for the rubber to reach 132.°C at the plane specified at b).
  • 7.
    • Given:
    • k=0.16W/m-K α =8.671E-6 m2/s
    • T o =20C T 1 =140C T(center)=132C
    • Time @ T(center)=132C, assume m=0
    • using the Gurney graph for plate we get value of X=1.3
  • 8.
    • b) Temp @ 0.65cm from metal
    • Given:
    • x=1.25-0.65 =0.6cm t=23.43 sec
    • n=0.6/1.25 =0.48 X=1.3
    • From the graph we get a value of Y=0.037
  • 9.
    • c) Time when the Temp @ 0.65cm from the metal is 132C.
    • Given:
    • m=0 n=0.48 Y=0.0667
    • We get the value X=1.1 from the graph