AME 331, Prof. P. Ronney
Midterm Exam #1
March 13, 2008
The first midterm exam
will be open book. You may use any
reference materials you want. There
will be two or three problems similar to the homework problems assigned from
the text, appropriately shortened for exam problems. There may also be a short-answer type of question (e.g.,
define the conduction shape factor, why it is useful and when can it not be
used.)
The exam may cover any
material through the end of chapter 4 (unsteady heat conduction.) This material
includes:
¥ Basic concepts and
tools
¥ Types of heat transfer
- conduction, convection, radiation
¥ Applications
¥ Governing equation for steady or
unsteady conduction
¥ Buoyancy-driven convection
¥ Units
¥ Interpolation
¥ Engineering scrutiny
¥ Steady 1-D conduction
¥ Planar geometry,
constant and variable k
¥ Thermal resistance
¥ Radial system
¥ Critical insulation thickness
¥ Overall heat transfer coefficient
¥ Systems with heat sources
¥ Conduction-convection system (fins)
¥ Thermal contact resistance
¥ Steady conduction in
multiple dimensions
¥ Separation of
variables
¥ Rectangular plate
¥ Conduction shape factors
¥ Numerical methods
¥ Unsteady conduction
¥ Lumped capacity
methods
¥ Semi-infinite solid
¥ Unsteady conduction with convection
boundary conditions - Heisler charts
¥ Multi-dimensional systems
¥ Numerical methods
First midterm exam from the
last time PDR taught the course (2 years ago)
(Note: some of the Greek symbols donÕt show up
properly if youÕre using a web browser other than Internet Explorer; D = Delta,
r = rho, etc.)
Instructions:
55
minutes allowed. Use any non-human
reference materials (book, notes, homework, etc.) you want. Be sure to show all your work; liberal
partial credit will be given if you have the right idea. Note point values (100 points total)
and budget your time accordingly.
Make any reasonable assumptions (e.g. steady, constant thermal
conductivity, lumped capacity, etc.) you want but be sure to state these
assumptions.
Problem
#1 (15 points) (Basic
concepts)
a.
Briefly define
thermal resistance
Thermal resistance is analogous to electrical
resistance, i.e. the ratio of the driving force for heat transfer (the
temperature drop across the thermal element, DT) to the ÒcurrentÓ flow of heat transfer (the
heat rate, q), thus Rth = DT/q.
b.
Explain why it is
useful
It is useful because various types of thermal
resistances (in particular conductive resistance = Dx/kA, convective resistance = 1/hA and contact
resistances) can be combined in series and parallel in the same way as
electrical resistors to produce a total thermal resistance Rth,total,
and thereby produce an equivalent ÒcircuitÓ that can be analyzed in a simple
way, e.g. q = DToverall/Rth,total.
c.
Explain under what
circumstances it is not applicable, or at least is not likely to provide
accurate estimates of temperatures and/or heat transfer rates.
Problem
#2 (30 points) (1D steady
conduction)
A
solid sphere of uranium has radius rs and conductivity k, and (via radioactive
decay) generates (units Watts/m3)
of heat per unit volume. The outer
radius of the sphere is maintained at a constant temperature Ts.
a)
(10 points) Write down the appropriate governing
equation for this system (i.e the equation that ensures conservation of energy
+ FourierÕs Law is satisfied for this system.)
Assuming constant k,
b)
(8 points) How many boundary conditions are
needed, and what are they?
One 2nd order ordinary differential
equation, requires 2 boundary conditions
(1)
Specified temperature at outer radius of sphere: T = Ts at r = rs
(2) Symmetry condition at r = 0:
c)
(12 points) Solve the governing equation with the
boundary conditions to determine the temperature profile T(r) inside the sphere
in terms of the parameters rs, k, and Ts.
Problem
#3 (30 points) (1D unsteady
conduction)
A
plastic bead 10 mm in diameter having k = 0.2 W/mK initially has a uniform
temperature of 350K and is surrounded by air (k = 0.025 W/mûC) at 300K with
convection heat transfer coefficient h = 20 W/m2K. The density (r) and heat capacity (CP) of the plastic
are 200 kg/m3 and 1000 J/kgK, respectively. Assume all properties are independent
of temperature. The plastic does
not radiate.
a)
(15 points) What would the center temperature of
the bead be after 50 seconds? Do
not use a lumped capacitance analysis.
b)
(5 points) Could a
lumped capacity analysis have been used in this situation? Why or why not?
For the purpose of computing Bi to determine
whether the lumped capacity method can be used, the text (p. 245) says to use
the definition Bi* = hLc/k, where Lc = V/A (V
= volume, A = area). For a sphere V
= 4¹r3/3 and A = 4¹r2,
thus V/A = r/3. Then
Bi* = (20 W/m2K)(0.005
m/3)/(0.2 W/mK) = 0.17 > 0.1
So NO, lumped capacity method couldnÕt have been
used anyway.
c)
(10 points) How many Joules of heat have been
transferred from the bead to the air at this time (50 seconds)?
Bi2Fo = (1/2)2 2 = 0.5;
Figure D.9: for Bi2Fo = 0.5, Bi = 0.5, Q/Qo Å 0.9 (itÕs
hard to tell exactly where Bi = 0.5 since the plot is crazy; it has a
logarithmic scale but only 8 lines per decade!)
Qo = mc(Ti – T°)
= rVc(Ti – T°) = (200
kg/m3)((4¹(0.005m)3/3)(1000 J/kgK)(350K –
300K)
= 5.24 J
QÅ 0.9 Qo = 4.7 J
Problem
#4 (25 points) (2D conduction, numerical methods)
Derive
a nodal equation for numerical solution of two-dimensional unsteady heat conduction at a lower right exterior corner
node with a constant heat flux per unit area qlaser (from a laser
heating source) coming in from the bottom side (no convection on the bottom
side), with convection (but no laser heating) on the right side, and having
solid thermal conductivity k, density r and heat capacity c.
Assume a grid spacing Dx = Dy. Note that the control
volume is the diagonal cross-hatched area.
Assuming no heat
generation:
(You donÕt need to do all
the algebra to get full credit!)