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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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Numerical Analysis Questions
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