Conduction

Question 1

define heat FLOW

Answer 1

the RATE of energy transfer in Watts
Q

Question 2

define heat FLUX

Answer 2

flow of energy normalised by the surface area the energy crosses. W/m^2
q

Question 3

define Newton’s law of cooling in words and the equation

Answer 3

rate of change of a temperature of an object is proportional to the difference between the temp of object and ambient temp

q=h(Tb-Ta)

Question 4

Fourier’s Law (eqn). explain in terms of heat travelling through rod

Answer 4

q = -k dT/dx
heat travels through rod, Temp decreases with distance x along rod linearly –> gradient -k

where k is thermal conductivity, W/mk

Question 5

typical thermal conductivity of metals

Answer 5

10-500 W/mK

Question 6

typical thermal conductivity of non-metal solids

Answer 6

1
glass, concrete, soil

Question 7

typical thermal conductivities of liquids and insulators

Answer 7

between 0 and 1

Question 8

Define the Nusselt number

Answer 8

Nu = ha/k
a is some characteristic length.

ratio of convective to conductive heat transfer across a boundary.

Question 9

htc of wide slab (finite)

Answer 9

h(slab) = k/L
where L is slab thickness

Question 10

Nu of wide slab

Answer 10

Nu = hL/k = 1
as h=k/L

Question 11

define thermal resistance and conductance

Answer 11

1/Ah is thermal resistance
hence Ah is conductance

from analogy to Ohm’s law
I = 1/R (V2-V1)
Q = Ah (T2-T1)

Question 12

htc for cylindrical tube

Answer 12

h = k / (r ln(r2/r1))
depends on position r.

Question 13

Nu for cylindrical tube

Answer 13

Nu = 1/ ln(r2/r1)

Question 14

htc for cylindric tube - thin shell case

Answer 14

𝜏=r2-r1 –> 0
using LOG MEAN AVERAGE for r:
r = (r2-r1) / ln(r2/r1)

h = k/𝜏
approximates a slab!

Question 15

htc for spherical “shell” as infinite medium

Answer 15

h(shell, inf) = k / r1

Question 16

Nu of infinite spherical shell

Answer 16

Nu = 2

Question 17

Internal heat generation steady state balance

Answer 17

V * Q(v) = A * q = A k (-dT/dx)
all the heat generated in the volume V gets conducted away via area A

Q(v) is the heat generated PER UNIT VOL

Question 18

Overall htc is found by

Answer 18

1/U = 1/h1 + 1/h2 + …

Question 19

for a lagged pipe, why might adding insulation actually INCREASE heat loss up to a point? what is that point?

Answer 19

multiple mechanisms of HT here. adding lagging up to a point increases the SA available for heat loss through CONVECTION, if the radius r1 < r1*
where r1* is the critical radius,
r1* = k / hA

Question 20

Define the Biot number

Answer 20

Bi = hL/k
ratio of thermal resistance for conduction inside a body vs resistance for convection at surface of body.

Question 21

difference between Biot and Nusselt number

Answer 21

Bi compares INTERNAL conduction resistance to EXTERNAL convection resistance within a solid.
Used to determine T distribution within a solid subject to convection.

Nu compares the convection to conduction heat transfer across a fluid-solid interface.
higher Nu = more effective convective HT compared to conduction.

*** k in Bi for SOLID, k in Nu for FLUID!

Question 22

What is the characteristic length L in the Biot number usually?

Answer 22

L = volume of solid / surface area of solid

Question 23

Bi «_space;1

Answer 23

large k
external control. assume solid is a uniform / lumped body (no T grad in solid)

Question 24

Bi = 1 approx

Answer 24

mixed control
neither conduction nor convection dominates, must consider both

Question 25

Bi << 1

Answer 25

internal resistance dominates. ignore convection, consider conduction only. large T gradient in solid

Question 26

Governing equation for transient heat conduction

Answer 26

𝛼∇^2 T = ∂T/∂t

Question 27

define 𝛼 in the transient heat conduction governing eqn.

Answer 27

𝛼 = k / 𝜌Cp thermal diffusivity

Question 28

Define the Fourier number

Answer 28

Fo = 𝛼t / L^2 dimensionless time ratio heat conduction to heat storage

Question 29

semi-infinite slab solution

Answer 29

θ = erfc ( z / sqrt(4𝛼t) ) θ is dimensionless temp

Question 30

error function small argument simplification

Answer 30

for small argument (usually below 0.6), it is approximately LINEAR erf(x) = x erfc(x) = 1-x

Question 31

midway temp for semi-infinite slab solution

Answer 31

θ = 1/2 using erf linear approx, arg = 1/2 z = √𝛼t

Question 32

penetration depth solution for semi-infinite slab

Answer 32

penetration depth = arbitrary 1% change θ = 0.01 arg = 2 hence z = 4 √𝛼t

Question 33

for a solid in a fluid, how is the surface temperature found?

Answer 33

instantaneous heat balance, as Ts will not be equal to Tf q(conv) = h(conv) (Tf - Ts) = -k ∂T/∂z = q(cond) ie. newton's for conv = Fourier's for cond

Question 34

For a lump body in a fluid, what is the balance?

Answer 34

enthalpy balance (no phase change) dH/dt = A h (Tf - Ts) = d/dt (V 𝜌 Cp Ts) where Temp of body = Ts as it is uniform.