quench by wanghonghx

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									Ordinary Differential Equations

   Everything is ordinary about them
         How long should I wait
           before I quench?

                                              2000 K


2500 K
                             Quenching Bath
             Time = ??????
Ordinary Differential Equations
 Problem: You are working for a ball bearing company -
  Ralph’s Bearings. For long lasting life of some
  spherical bearings made in this company, they need to
  be quenched in water for 20 minutes. The water is
  maintained at a room temperature of 300K while the ball
  bearings need to be heated to a high temperature of
  2000K. However, it takes time to take the ball from the
  furnace to the quenching bath and its temperature falls.
  If the temperature of the furnace is 2500K and it takes
  10 seconds to take the ball to the quenching bath. Is less
  or more time needed to get to a temperature of 2000K?
Assumptions
 The spherical ball bearing is a lumped mass
  system.
 What does a lumped system mean? It
  implies that the internal conduction in the
  sphere is large enough that the temperature
  throughout the ball is uniform.
 This allows us to make the assumption that
  the temperature is only a function of time
  and not of the location in the spherical ball.
Energy Conservation




Heat In – Heat Lost = Heat Stored
Heat Lost
  When the ball is taken out of the furnace at the initial
  temperature of To and is cooled by radiation to its
  surroundings at the temperature of Ta, the rate at which
  heat is lost to radiation is
      Rate of heat lost due to radiation = A  (T4-Ta4)
  where
      A = surface area of ball, m2
       = Emittance
       = Stefan-Boltzmann constant,
                   5.67 x 10-8 J/(s-m2-K4)
      T= temperature of the ball at a given time, K
Heat Stored

  The energy stored in the mass is given by

     Energy stored by mass = mCT

  where
     m = mass of ball, kg
     C = specific heat of the ball, J/(kg-K)
Energy Conservation


  Rate at which heat is gained – Rate at which heat is lost

               =Rate at which heat is stored

Gives

                - A (T4-Ta4) = m C dT/dt
Putting in The Numbers
Given the

   Radius of ball, r = 1.0 cm =0.01 m
   Density of ball,  = 3000 kg/m3
   Specific heat, C = 1000 J/(kg-K)
   Emittance,  = 0.5
   Stefan-Boltzmann Constant,  = 5.67 x 10-8 J/(s-m2-K4)
   Initial temperature of the ball, T(0) = 2500 K,
   Ambient temperature, Ta = 300 K,
The Differential Equation
  Surface area of the ball
         A = 4 r2
           = 4 (0.01)2
           = 1.25663x10-3 m2


  Mass of the ball
        M=V
            =  (4/3  r3)
            = (3000)*[4/3  (0.01)3]
            = 0.0125663 kg
 The Differential Equation
      Hence

                                                        dT
          A   (T                  Ta         )  mC
                                4            4

                                                        dt
                                                                                 dT
 (1.25663  10 3 ) (0.5) (5.67  10 8 ) (T 4  300 4 )  (0.0125663) (1000)
                                                                                 dt


dT              12
    2.8349 10 (T  8110 ),T (0)  2500
                    4      8

dt

								
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