# quench by wanghonghx

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```									Ordinary Differential Equations

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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