Physics 2B Electricity & Magnetism
Reading Quiz 1 Solutions
TA: Samuel May
September 29, 2016
22.01
What is the SI unit of charge?
The SI unit of charge is the Coulomb.
22.03
A rod attracts a positively charged hanging ball. The rod is.
Either negative or
Spring 2016 Physics 2A Problems
The Problem Sets for Physics 2A have been divided into categories of Easy,
Medium and Hard.
This is designed to maximize your learning in the class.
Easy problems should be completed before attempting the other categories.
Do - Di, and the Reynolds number would be,
Re 4m / D0 Di
Re 4 0.5 / 3.45 104 3.142 0.02 0.03
Nu 0.023 Re
where
0.8
Pr
0.33
= 3.69 104, a turbulent flow.
,
Pr c / k 3.45 10 4 907 / 0.07 4.47
0.023 3.69 104
0.8
4.47 0.33 169.8
h o nu k / Do D i 169.
For this purpose, we will introduce the radiation intensity, I, defined as the energy emitted per
unit area, per unit time, per unit solid angle. Before writing an equation for this new property, we
will need to define some of the terms we will be using.
overall heat transfer coefficient IS 80 W/m2K and air is the mixed fluid, calculate the exchanger
effectiveness and the surface area.
Solution: Let the specific heat capacity of air and water be 1.005 and 4.182 kJ/ kgK. By
making an energy balance:
&c cc
Since other parameters remain the same,
25/(100 - 25) = T/(100 - 35)
and T = 21.66; or,
Tc0
= 35 + 21.66 = 56.66C.
3.11. Increasing the Heat Transfer Coefficient
For a heat exchanger, the heat load is equal to Q = UA (LMTD). The effectiveness of
the heat
NTU = AU/Cmin
(3.15)
An another useful parameter in the design of heat exchangers is the ratio the minimum
to the maximum thermal capacity, i.e., R = Cmin/Cmax,
where R may vary between I (when both fluids have the same thermal capacity) and 0
(one of the
Fig 3.17 shows three types of transverse baffles used to increase velocity on the shell
side. The choice of baffle spacing and baffle cut is a variable and the optimum ratio of baffle cuts
and spacing cannot be specified because of many uncertainties and
= (hpkA)
And
fin
1/ 2
0
/hA
= (hpkA)
0
1/ 2
= (kp/hA)1/2
0
(hpL
0
) = (hpkA)1/2 / (hpL)
E
kp / hA hpL pL
Surface area of fin
fin
A Cross-sectional area of the fin
hpkA 1/ 2
1/ 2
i.e., effectiveness increases by increasing the length of the fin but it w
In this example there are three junctions, so we will obtain three equations. This will allow us to
solve for three unknowns.
Radiation problems will generally be presented on one of two ways:
1. The surface net heat flow is given and the surface temperat
or an observer moving with the bulk velocity. For a stationary observer, the absolute
flux of any species A will be equal to the sum of the flux due to the molecular diffusion and that
due to the bulk motion.
Thus, Absolute flux: A VA and, Diffusion flux:
(0.045 - 0.025) = 0.02 m
&/ D o Di 4 0.1/ 3.142 0.07 3.25 10 2 56.0
Re 4m
laminar flow.
Assuming Uniform temperature along the Inner surface of the annulus and a perfectly
insulated outer surface.
Nu = 5.6, by interpolation (chapter 6)
ho = 5.6 0.138/0.02
The efficiency of circumferential fins is also obtained from curves for efficiencies
3
2
1
2
for different values of
b
2h
along Y-axis : r2 r1 r2 r1
2
Kb
b
r2 / r1
2
.
UNIT IV
RADIATION HEAT TRANSFER
4.1RADIATION
Definition:
Radiation is the energy tr
The view factor may be obtained from:
Since dAi is a differential area
Substituting for the cosines and the differential area:
After simplifying:
After integrating,
Substituting the upper & lower limits
This is but one example of how the view factor may b
&A / A 8.7 10 8 1.2 10 2 0.3 10 2 / 0.4 10 3
m
1.957 106
kg/m2s.
Example 5.15 In order to maintain a pressure close to I bar, a pipeline carrying ammonia gas is
vented to atmosphere. Venting is achieved by tapping the pipe and inserting a 4
mm diameter
Let us consider a flat plate where the concentration at the surface of the species A is
different than its concentration in the free stream. The species A will diffuse into the fluid and a
concentration boundary layer will develop as shown in Fig. 5.8. Th
In the analysis of such fins, it is assumed that:
Fig 3.18
(For increasing the heat transfer rate by fins, we should have (i) higher value of thermal
conductivity, (ii) a lower value of h, fins are therefore generally placed on the gas side, (iii)
perimet
The saturation temperature corresponding to 20 kPa is 60C and as such the temperature of the
cooling water at the pinch pint is 50C. The condensing unit may be considered as a combination
of three sections:
(i) desuperheater - the superheated steam is con
4. A diffusely emitting surface is exposed to a radiant source causing the irradiation on the
surface to be 1000W/m2.The intensity for emission is 143W/m2.sr and the reflectivity of the
surface is 0.8.Determine the emissive power ,E(W/m2),and radiosity ,J
or
DAB = DBA = D
This fact is known as the equivalence of diffusion coefficients or diffusivities in binary
mixtures, and is a property of the binary mixture.
By integrating Eq. (12.10), we can obtain the mass flux of the species A as;
(5.15)
&A
m
DM A
p
Fig 3.11 Heat exchanger effectiveness for parallel flow
Example 3.13 A single pass shell and tube counter flow heat exchanger uses exhaust gases on
the shell side to heat a liquid flowing through the tubes (inside diameter 10 mm,
outside diameter 12.5 mm,
Heat capacity rate of water = 1 4.2; 4.2 kW/K
Cmin 3.0 kW / K
and
R Cmin / C max 3 / 4.2 0.714
For a counter flow arrangement,
Effectiveness,
NTU 1/ R 1 ln 1 / R 1
65 42 / 65 28 0.6216
and NTU = 1.346 =
A U / Cmin
; A = 1.346 3000 / 700 = 5.77 m2
By mak
occupied by the mixture at the temperature of the mixture.
5.4.
Molar Density, Mass Density, Mass Fraction and Mole Fraction
There are a number of ways by which the concentration for a species in a
multicomponent mixture can be defined:
(i) Molar Density
Module 9: Worked out problems
1. A spherical aluminum shell of inside diameter D=2m is evacuated and is used as a radiation
test chamber. If the inner surface is coated with carbon black and maintained at 600K, what is the
irradiation on a small test surf
Since the relative humidity is 0.4, the partial pressure of water at 25C = 0.4 0.03166
bar, and the concentration is then, 0.4 x 0.023 = 0.0092 kg/m3
Rate of mass transfer,
&/ A h D Cs C
m
0.01405 0.023 0.0092 3600 kg / m .hr
2
= 0.698 kg/m2.hr
Example
(ii) When the humidity is 0.5, partial pressure of water vapour would be
0.5 0.03169 = 0.0158 bar.
and
&w
m
0.256 104 1105 18 0.75
8314 298 10
2
ln 1 0.0158 / 1 0.03169
= 5.33 108 kg/s
Example 5.10 A 50 rom deep pan contains water to a level of 20 nun
O2
5.9.
N2
0.18 104
273
Main Points of Fick's Law of Diffusion
These can be summarised as:
(i) It cannot be derived from first principles because it is based on experimental
evidence.
(ii) It is valid for all phases of matter. Since mass transfer is stron
(i) the system is isothermal,
(ii) the total pressure remains constant,
(iii) the system is in steady state. Since there has to be a little movement of air over the
top of the tank to remove the water vapour that diffuses to that point, the air movement d
From our conservation rule we have:
Since surface 1 is not convex F1,1 = 0. Then:
4.17 Reciprocity
We may write the view factor from surface i to surface j as:
Similarly, between surfaces j and i:
Comparing the integrals we see that they are identical so
In the development of a new design, the following factors are important:
(a) Fluid Temperature - the temperature of the two fluid streams are either specified for
a given inlet temperature, or the designer has to fix the outlet temperature based on flow r