Title: Transient Conduction
1Chapter 5
2Transient Conduction
- Many heat transfer problems are time dependent
- Changes in operating conditions in a system cause
temperature variation with time, as well as
location within a solid, until a new steady state
(thermal equilibrium) is obtained. - In this chapter we will develop procedures for
determining the time dependence of the
temperature distribution - Real problems may include finite and
semi-infinite solids, or complex geometries, as
well as two and three dimensional conduction - Solution techniques involve the lumped
capacitance method, exact and approximate
solutions, and finite difference methods. - We will focus on the Lumped Capacitance Method,
which can be used for solids within which
temperature gradients are negligible (Sections
5.1-5.3)
3Lumped Capacitance Method
- Consider a solid that is initially at a uniform
temperature, Ti , and at t0 is quenched by
immersion in a cool liquid, of lower temperature - The temperature of the solid will decrease for
time tgt0, due to convection heat transfer at the
solid-liquid interface, until it reaches
T
t0
4Lumped Capacitance Method
- If the thermal conductivity of the solid is very
high, resistance to conduction within the solid
will be small compared to resistance to heat
transfer between solid and surroundings. - Temperature gradients within the solid will be
negligible, i.e. the temperature of the solid is
spatially uniform at any instant.
T
5Lumped Capacitance Method
- Starting from an overall energy balance on the
solid
where
(5.1)
- Lets define a thermal time constant
Rt is the resistance to convection heat
transfer Ct is the lumped thermal capacitance of
the solid
(5.2)
6Transient Temperature Response
- Based on eq. (5.1) the temperature difference
between solid and fluid decays exponentially.
7Transient Temperature Response
From eq. (5.1) the time required for the solid
to reach a temperature T is
(5.3)
The total energy transfer, Q, occurring up to
some time t is
(5.4)
8Validity of Lumped Capacitance Method
Ts,1
qcond
qconv
Ts,2
(5.5)
T?
9Validity of Lumped Capacitance Method
(Rearranging 5.5)
- What is the relative magnitude of DT solid versus
DT solid/liquid for the lumped capacitance method
to be valid?
10Biot and Fourier Numbers
- The lumped capacitance method is valid when
where the characteristic length LcV/AsVolume
of solid/surface area
We can also define a dimensionless time, the
Fourier number
where
Eq. (5.1) becomes
(5.6)
11True or False?
- A hot solid will cool down faster when it is
cooled by forced convection in water rather than
in air. -
- For the same solid, the lumped capacitance method
is likely more applicable when it is being cooled
by forced convection in air than in water. - The lumped capacitance method is likely more
applicable for cooling of a hot solid made of
aluminum (k237 W/m.K) than copper (k400 W/m.K) - The transient response is accelerated by a
decrease in the specific heat of the solid. - The physical meaning of the Biot number is that
it represents the relative magnitude of
resistance due to conduction and resistance due
to convection.
12Example (Problem 5.7 Textbook)
- The heat transfer coefficient for air flowing
over a sphere is to be determined by observing
the temperature-time history of a sphere
fabricated from pure copper. The sphere, which is
12.7 mm in diameter, is at 66C before it is
inserted into an air stream having a temperature
of 27C. A thermocouple on the outer surface of
the sphere indicates 55C, 69 s after the sphere
is inserted in the air stream. - Calculate the heat transfer coefficient, assuming
that the sphere behaves as a spacewise isothermal
object. Is your assumption reasonable?
13What if?
- What happens to the rate of cooling if h
increases? - What happens to the rate of cooling if the
diameter of the sphere increases? - What happens if we have a huge sphere?
14General Lumped Capacitance Analysis
- In the general case we may have convection,
radiation, internal energy generation and an
applied heat flux. The energy balance becomes
Tsur
qrad
qs
T?, h
qconv
- Numerical solutions are generally required
- Simplified solutions exist for no imposed heat
flux or generation (see equations (5.19, 5.25)
textbook).
As(c,r)
As,h
15Example 5.2
- Calculation of the steady state temperature of
the thermocouple junction. - How much time is needed for the temperature to
increase from 25C to within 1C from its steady
state value?
16Example 5.2
17Example (5.33)
- Microwave ovens operate by rapidly aligning and
reversing water molecules within the food,
resulting in volumetric energy generation. - Consider a frozen 1-kg spherical piece of ground
beef at an initial temperature of Ti-20C.
Determine how long it will take the beef to reach
a uniform temperature of T0C, with all the
water in the form of ice. Assume that 3 of the
oven power (P1kW total) is absorbed by the food. - After all the ice is converted to liquid,
determine how long it will take to heat the beef
to Tf80C, if 95 of the oven power is absorbed.
18Other transient problems
- When the lumped capacitance analysis is not
valid, we must solve the partial differential
equations analytically or numerically - Exact and approximate solutions may be used
- Tabulated values of coefficients used in the
solutions of these equations are available - Transient temperature distributions for commonly
encountered problems involving semi-infinite
solids can be found in the literature
19Summary
- The lumped capacitance analysis can be used when
the temperature of the solid is spatially uniform
at any instant during a transient process - Temperature gradients within the solid are
negligible - Resistance to conduction within the solid is
small compared to the resistance to heat transfer
between the solid and the surroundings - The Biot number must be less 0.1 for the lumped
capacitance analysis to be valid. - Transient conduction problems are characterized
by the Biot and the Fourier numbers.