## Transcribed Text

2.2 secant method
You need find the temperature at which the volumetric heat capacity of mercury 8520 2.47T -
=1
106.
and
you
know
that solution close 55°C density mercury giver par(T) 13595
the
0,00038T° units kg/m³, and the heat capacity mercury given by Cp.ilg finding
139.68
0.031T
units o
J/kg (all units here are consistent so just ignore them). Put the equation the standard form for root
(as : function f(x)):
16 /(x).
24
Using your initial guess and second guess n=20+ 0.001, find the solution using three steps in
method (complete table below with values the form 123456, that each
secant decimal digits) Hint: the exact solution very nearly exact beyond the first decimal digit (like
28 500028 3.099922).
o
question 2.2:
1
Ar appropriate f(x) would be f(x): 3.87ln(x) 13.3e^-0.73x=0
x0= 000000 f(x0) -0.406263
2
x1 2.001000 f(x1)==0.402074
2.096995
f(x2)=-0.011846
2.099909
3
f(x3)= -0.000357
2.100000
solution
4
useful (partial) relation
₹1+1
3.2 polynomial interpolation
From the table (preceding page). estimate the heal capacity of methane al MPa and 271 K again, this time by each
of quadratic and cubic Lagrange interpolation Complete the table below for the coefficients for each interpolation,
appropriate. Don't lorget check the suin coefficients (very easy check for calculation errors). Give
the
as
interpolated values to the same precision as the closest corresponding values in the table (or you will not receive
credit)
30
Complete the table as appropriate for quadratic, cubic and quartic interpolation, with numbers only
1/
Li
L1
quadratic
cubic
1
2
3
4
5
10 heat capacity (quadratic)
10
heat capacity (cubic):
There no way to determine the errors in your above solutions from the information provided. However, you
could improve interpolation finding data more closely spaced around target temperature Determine
factors which the error would reduced (ideally) the data were given in 5 K increments instead 01 25 K
increments for each of quadratic and cubic interpolation
10 error reduction factor for quadratic interpolation
10
error reduction factor for cubic interpolation:
1.1 RLC circuit with amplifier
current.
In
the
circuit
AC illustrated circnits below have overall there are resistances three components: (Z) which depend capacitor on the (C), frequency arange inductor f frequencies (w) of the (L), alternating and (40 Hz custom 105 Hz). resistor the resistance (R). The
resistor
unit
has
an
integrated
filter
and
amplifier,
that
over
higher:
105
Hz
R(w) 35 2 35 ? Klw)
where K(w) defined by
K(w) 10800H21-9
The overall resistance given by
Z(w)
Given
that C= 1.2 10 F and L 0.50H, find either the frequencies side al 100 which Hz, but Z(w) only 35.9, one an using actual bisection. solution. There All
are too solutions* which can be found by bisection, may safely ignore them. false solution, in the range
units consistenti this problem, range soyou 10 Hz to 100Hz within +2 Hz), and Lhe terms of a plot Z(w)
Find the 500 actual Hz (to solution, within = the Hz). Explain what is wrong with and second at least solution, one discontinuity)
100 the range 10 Hz to 500 Hz (which should shuw exactly one root
over
1.2 properties of air
temperature
and
pressure.
Normally,
such
properties
are
As with most materials, temperature the properties and pressure of air by depend looking them tables data (such there handbooks), application and perhaps where
on
found interpolating specific for property given those required, values interpolation and the material, which the next temperature both section kinematic and/or this viscosity, course). pressure However must the be heat chosen caparity match per unit the volume desired
values. oCe must Consider have specific heat transfer values system 2.11 X 10-5 m²/s, and pCp 1490 J/m³ K
The kinematic viscosity air well-approximated by the expression
v
where
the units are as follows v, m²/s; T. K: and P, alm. The heat capacity per unit volume of air well-
approxinated by the expression
pC, (6.229 105T 1.13 847T 0.3 42.35) P
where the units are as follows: PC, J/m³-K;T. K; and P. atm The units here are all consistent and may safely
be ignored There ar two unknowns in the above system K) T and P, beginning but the with equations an initial can guess be combined of 310 to K. eliminate Substitute your
the solution resulting for equation into one for of the the temperature original equations to determine via secant method.
1.3 flow of air through pipe
diameter
of
D
=9.7
em.
Air flows velocity U 2.6 m/s through round pipe with length 10-5 L m²/s. 4.1 m The and pipe walls have roughness
The €= kinematic 0.11 mm. The viscosity method solve the for operating the friction kg/m³. conditions factor, Determine f, starting the with pressure an initial drop guess over the length of the pipe
by
density of the air is 1.478 of -0.031. Maloc sure to
first
using
secant
work
consistent
units.
Pressure drop through round pipe given by
AP 2D
where the dinuensionless friction factor, f the solution to the following equation
and the Reynolds number (Re) is
UD

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