Download f21/2509/2009 githua scolastica njoki heat and mass transfer

F21/2509/2009
GITHUA SCOLASTICA NJOKI
HEAT AND MASS TRANSFER
QUESTION ONE.
Define heat transfer giving the three modes and the laws that govern each.
Heat transfer is energy in transition due to temperature differences. This transfer occurs by
interaction of a system and its surroundings.
The three modes of heat transfer include and the laws that govern them include:

Conduction: where heat is transferred due to motion of free electrons in metals or atoms
in non-metals. Conduction is governed by Fourier’s Law which states that the heat flux
‘q’ is proportional to the temperature gradient in the direction of the outward normal.

Convection: this is energy transfer between a surface and a fluid moving over the
surface. Convection is governed by Newton’s Law.

Radiation: this is energy transfer from one object to another without the use of an
intervening medium i.e vacuum. Radiation is governed by Stefan - Boltzmann Law.
State the equation (s) that describe each mode and explain the terms used.
Conduction.
𝑑𝑇
Q = −𝐾 𝐴 𝑑𝑥
Where; Q is the heat flow by conduction (W)
K is the thermal conductivity in watts
A is the area through which the heat flows (m2)
𝑑𝑇
𝑑𝑥
is the temperature gradient in the x direction
Convection
𝑄 = ℎ 𝐴 (𝑇1 − 𝑇2)
Where Q is the convective heat tranfered (W)
h is heat transfer coefficient (W/ M2 K)
A is the area (M2)
T1 is the temperature of the surface receiving or giving heat (K/℃)
T2 is the average temperature of the stream of fluid.(K/℃)
Radiation.
𝑄 = 𝜎 A1 ( T14 – T24)
Where Q is the heat transferred
σ is the Stefan – Boltzmann constant (W/ M2 K4)
A1 is the surface area of the body (M2)
T1 is the absolute temperature to the surrounding (K/℃)
T2 is the uniform absolute temperature. (K/℃)
QUESTION TWO
Show that for cylindrical systems the heat flow is given by
𝒒=𝒌
𝟐𝝅𝑳
𝐥𝐧
𝒓𝟐
𝒓𝟏
(T1 – T2)
For 1 dimensional when temperature is a function of radii only.
The gorverning equation for steady state one dimensional conduction in a radial system is:
1 𝑑𝑇
d2T/dr2 + 𝑟
𝑑𝑅
=0
Integrating twice:
T = C1 ln (x) + C2
If T = T1 at r = r1 and T = T2 at r = r2 then
𝑟
ln (𝑟1)
T – T1
=[
]
𝑟2
𝑇2 − 𝑇1
ln (𝑟1)
Similarly since the heat flow Q = k A ( dt/dr ) then for a length, L (in the axial or ‘z’ direction)
the heat flow can be found by differentiating
𝑟
ln (𝑟1)
T – T1
=[
]
𝑟2
𝑇2 − 𝑇1
ln (𝑟1)
Therefore:
𝑞=𝑘
2𝜋𝐿
ln
𝑟2
𝑟1
(T1 – T2)
Derive an expression for the critical radius of insulation for a radial system
Adding insulation to a cylindrical piece or a spherical shell increases the conduction resistance of
the insulation layer but decreases the convection resistance of the surface because of the increase
in the outer surface area for the convection. The heat transfer from the pipe may increase or
decrease depending on which effect dominates.
Consider a cylindrical pipe of outer radius r1 whose outer surface temperature T1 is maintained
constant. The pipe is now insulated with a material whose thermal conductivity is k and outer
radius r2. Heat is lost from the pipe to the surrounding medium at temperature T∞ with a
convection heat transfer coefficient h. the rate of heat transfer from the insulated pipe to the
surrounding air is expressed:
The two significant resistances for a cylinder of unit length are:
Rtconduction = ln (ro/r1)/2𝜋kl
Rtconvection = 1/ 2 π ro h
1
𝑟𝑜
dq/ dro = T1- T∞/ 2𝜋𝑟𝑜ℎ + ln (𝑟1) /2𝜋𝑘)2
Bi = hrcrit/k
QUESTION THREE
The hot combustion gases of a furnace r separated from the ambient air and its
surroundings which are at 250C by a brick wall 0.15m thick. The brick has a thermal
conductivity of 1.2 W/m.K and a surface emissivity of 0.8. Under steady state conditions an
outer surface temperature of 1000C is measured. Free convection heat transfer in the air
adjoining this surface is characterized by a convection of h=20 W/m2.K .What is the brick
inner surface temperature.
Using the sketch below
T1 =?
T2 = 100oC
0.15m
T3 = 25oC
The brick conducts heat to the surrounding. The surface of the brick emits heat to the
surrounding.This heat must balance the heat transferred through convection in the surrounding
air.Thus the following equations will hold true:
𝑄𝑐𝑜𝑛𝑣 = ℎ̅𝐴(𝑇2 − 𝑇𝑎𝑖𝑟 )……..1
𝑑𝑇
𝑄𝑐𝑜𝑛𝑑 = −𝐾𝐴 𝑑𝑥 =
−𝐾𝐴
𝑑𝑥
(𝑇2 − 𝑇1 )……………2
𝑄𝑟𝑎𝑑 = 𝐴𝐸𝜎(𝑇24 − 𝑇14 )……………..3
From conservation of energy:
𝑄𝑖𝑛 = 𝑄𝑜𝑢𝑡 …………4
𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 = 0………5
Consolidating the 3 equations and using 4 and 5
𝐾𝐴
𝑑𝑋
(𝑇1 − 𝑇2 ) − ℎ𝐴(𝑇2 − 𝑇3 ) − 𝐴𝐸𝜎(𝑇24 − 𝑇14 )……….6
𝑘(𝑇1 −𝑇2 )
𝑑𝑥
− ℎ(𝑇2 − 𝑇3 ) − 𝐸𝜎(𝑇24 − 𝑇34 ) = 0………………..7
𝑘(𝑇1 − 𝑇2 )
= ℎ(𝑇2 − 𝑇3 ) + 𝐸𝜎(𝑇24 − 𝑇34 )
𝑑𝑥
(𝑇1 − 𝑇2 ) = [20(110 − 25) + 0.8 ∗ 5.67 ∗ 10−8 (3734 − 2984 ] ∗
(𝑇1 − 𝑇2 ) = 2020.31 *
0.15
1.2
𝑇1 = 645.54𝐾 𝑂𝑅 372.54℃
0.15
1.2