TUTORIAL SHEET I ERRORS IN NUMERICAL CALCULATIONS 1. 2. 3. 4. Find the quotient q=x/y, where x=4.536 and y=1.32, both x and y being correct to the digits given. Find also the relative error in the result. Find the sum of the numbers 105.5, 27.25, 6.56, 0.1568, 0.000256, 208.6, 0.0235, 0.538, 0.0571, where each number is correct to the given decimal places. Estimate the absolute error in the sum. Find the product of the numbers 56.54 and 12.4 which are both correct to significant digits given. The Maclaurin’s expression of sin(π₯) is given by sin(π₯) = π₯ − π₯3 3! + π₯5 5! − π₯7 7! + …, Where x is in radians. Use the series to compute the value of sin(25ο°) with an accuracy of 0.001. 5. Find the number of terms of the exponential series such that their sum gives the value of π π₯ correct to five decimal places for all values of x in the range 0 ≤ π₯ ≤ 1. 6. Given f(x) = sin(π₯), construct the Taylor series approximation of order 5 at π₯ = π/2 to compute sin(π/2). Hint: Let π₯π+1 =π/2 and consider β = π/12. 7. If π’ = 3π£ 7 − 6π£, find the percentage error in u at π£ = 1, if the error in ‘π£’ is 0.05. 8. The strain in an axial member of a square cross-section is given by πΉ π = β2 πΈ Where F=axial force in the member, N H=length of width of the cross-section, m E=Young’s modulus, Pa πΉ = 72 ± 0.9 π β = 4 ± 0.1 ππ πΈ = 70 ± 1.5 πΊππ Find the maximum possible error in the measured strain. A defense system consists of an electronic detection device called range gate. It calculates the area in the air space where it should look for a moving target object. To find out where it should aim next, it calculates the velocity of the moving object and the last time the radar detected the object. To store the time, the internal clock of the system is to be measured at regular intervals. This is measured at every one-tenth of a second. Therefore, binary the number representing one-tenth should be stored in a register whose length is 24 (say). Compute the error in the measurement of time after 100 hours, due to Inaccuracy in the representation of ‘one-tenth’ using 24 bits, caused due to truncation of digits. 9. TUTORIAL SHEET II SOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS 1. Obtain a root correct to 3 decimal places for the following polynomial. π₯ 3 + π₯ 2 + π₯ + 7 = 0. 2. Find the root of π(π₯) = (π₯ − 4)2 (π₯ + 2) = 0, in the range (-2.5, -1) using false position method. Assume the tolerance to be 0.001 Find the real root of the equation π₯π π₯ = 1, using the secant method. Assume the tolerance to be 0.001. The search range is (0,1). Show that in bisection method the number of steps required reduce the final search interval to ε from the initial search range (b, a), is given by |π − π| ππππ ( π ) π≥ ππππ 2 3. 4. 5. 1 π Establish the formula π₯π+1= 2 (π₯π + π₯ ) and hence compute the value of √2 correct to six π 6. decimal places. Thermistors are temperature measuring devices based on the principle that the thermistor material exhibits a change in electrical resistance with a change in temperature. By measuring the resistance of the material one can then determine the temperature. Find the value of the resistance R for which the measured temperature 19ο°C. The relationship between the resistance R and the temperature of the thermistor is given by 1 = 1.129241 × 10−3 + 2.341077 × 10−4 ln(π ) + 8.775468 × 10−8 {ln(π )}3 π Where T is in Kelvin and R is in ohms. Consider the relative error in R as 0.01. 6.. Solve the following systems of nonlinear equations using Newton-Raphson method. Assume relative error to be 0.001. Take initial guess as (1,-1). π₯ 2 − π¦ = 11 π₯ + π¦2 = 7 Solve the set of nonlinear equations using Newton-Raphson method. π₯ 3 − 2.5π¦ 2 + 4π₯ − 3.7 = 0 0.7π₯ 2 + 3π¦ − 1.8π₯ + 1 = 0 Initial guess π₯ = 0, π¦ = 0. Perform two iterations. 7. The output voltage and current waveforms of a half wave rectifier circuit is drawn below. The transcendental equation which relates the extinction angle ο’ and the time πΏ constant ο΄=(π ) of the circuit is ο’ π ππ (ο’ − ο¦) + π ππο¦ π −ο·ο΄ = 0. R=10 ο and L=10 mH. Find the extinction angle ο’ using Newton-Raphson method.. Assume relative error to be 0.001. 8. The circuit for realizing the pulse-width modulator used in the control of power electronic converters is shown below. Find the instant ‘t’ at which the comparator switches. The search range is (0.0025,0.075). 9. In a single phase full-wave controlled rectifier supplying a purely resistive load, the rms value of the output voltage is given by π − ο‘ sin 2ο‘ ) π0 = ππ √( + 2π 4π where Vm is the peak value of the supply voltage and ο‘ is the firing angle for the thyristors. (a) Establish the formula. (b) Find the value of ο‘ for which the output rms voltage is 175 V. The available supply voltage is 220 V (rms). The search range is (0, π/2). Use false position method. TUTORIAL SHEET III SOLUTION OF SYSTEM OF LINEAR ALGEBRAIC EQUATIONS 1. Solve the network using mesh current analysis method. 2. Solve the following system of equations using gauss elimination method correct to three decimal places. 9.617π₯ + 3.502π¦ − 2.111π§ = 12.345 2.972π₯ + 4.405π¦ − 3.954π§ = 8.671 3.248π₯ + 7.199π¦ + 1.924π§ = 6.227 3. Find the inverse of the following matrix using Gauss-Jordan elimination method. 6 −4 −0 π΄ = (−4 16 −8) 0 −8 10 4. Find the inverse of the matrix using gauss-Jordan method. 1 1 −1 π΄ = [ 1 2 −2] −2 1 1 Determine whether the following matrices are Ill/well conditioned. 0.4033 0.9166 9.617 3.502 −2.111 (a) (b) π΄=( ) π΅ = (2.972 4.402 −3.954) 0.9166 −0.4033 3.248 7.199 1.924 (c) 6 −4 −0 πΆ = (−4 16 −8) 0 −8 10 Solve the following system of equations using Jacobi and Gauss-Seidel method correct to three decimal places. 6π₯1 − 4π₯2 − 10 = 0 −4π₯1 + 16π₯2 − 8π₯3 = 0 −8π₯2 + 10π₯3 − 15 = 0 5. 6. 7. Solve the following system of equations using (i) Gauss-Seidel (ii) Jacobi’s method. In each case carry your computations to two decimal places and proceed up to 10 iterations. Does the solution converge? 17 65 −13 50 π₯1 84 π₯2 25 12 16 17 18 [ ] [π₯ ] = [36] 3 56 23 11 −19 18 3 −5 47 10 π₯4 Check the convergence using the following condition. π ∑ππ=1,π≠π | ππ | ≤ 1, πππ π = 1,2, … , π where the ‘<’sign should be valid in case of ‘at least πππ 8. 9. one equation. Fin the ‘L’and ‘U’ components of the following matrix. 9 3 −2 π΄ = (2 4 −3) 3 7 1 An electrical network is represented by the following system of equations. 2π₯ + π¦ = 2 2π₯ + 1.01π¦ = 2.01 Solve for the unknowns π₯ and π¦. One of the system parameters is changed by a small value and the resulting system is represented by the following set of equations. 2π₯ + π¦ = 2 2. 01π₯ + π¦ = 2.05 Solve for the unknowns π₯ and π¦. What changes you observe in the responses (x and y). Write your comments. TUTORIAL SHEET IV REGRESSION 1. For Ahmedabad the monthly averaged daily global solar radiation (Hg) is a function of number of sun-shine hours per day (s) and is related as Hg=a+bs. At Ahmnedabad Hg and s data for 12 months are taken as follows. Find the constants ‘a’ abd ‘b’ using least squares method. Month s Hg JAN 0.8837 0.7044 FEB 0.8978 0.7111 MAR 0.7757 0.6527 APRIL 0.7895 0.6588 MAY 0.8162 0.6723 JUNE 0.6273 0.5809 JULY 0.3391 0.4426 AUG 0.3052 0.4261 SEP 0.5606 0.5492 OCT 0.8384 0.6821 NOV 0.9007 0.7117 DEC 0.8954 0.7110 2. A separately excited DC motor is driving a load whose load-torque characteristics is given by TL =k1N2+ k2N where k1 is the windage coefficient and k2 is friction coefficient. k1=2.1*10-5 k2=0.02. The motor is supplied by a 220 V DC source. The maximum mechanical power that the motor can deliver is 3 HP. If the motor is operating under full load, find the speed of operation. Use secant method. 3. The speed-torque characteristic of a separately excited DC motor is given in the following table. Obtain the speed-torque relation using linear regression. Torque (N-m) 0 25 50 75 100 ω (rad/s) 157.0796 156.2419 154.7758 154.2522 153.2050 The motor is supplying a mechanical load whose speed-torque characteristic is given in the following table. Obtain the speed-torque relation using Lagrangian interpolation formula. Torque (N-m) 60 72 84 ω (rad/s) 76.5 115.6 164.79 Obtain the operating point of the system using the Newton-Raphson method with initial guess T=80 N-m (Hint: The operating point is nothing but the point of intersection of speedtorque characteristics of motor and the load). 4. 5. 6. The stress Vs strain data for a composite material is given in the following table. Strain (m/m) stress 0.0000 0.0000 1.8300 × 10−3 3.0600 × 108 3.600 × 10−3 6.1200 × 108 −3 5.3240 × 10 9.1700 × 108 7.0200 × 10−3 1.2230 × 109 8.6700 × 10−3 1.5290 × 109 −2 1.0244 × 10 1.8350 × 109 1.1744 × 10−2 2.1400 × 109 1.3290 × 10−2 2.7520 × 109 1.4790 × 10−2 2.7670 × 109 −2 1.5000 × 10 2.8960 × 109 use π¦ = π΄π₯ model. Find the equation to solve ‘A’ using least squares criterion. Find the longitudinal modulus E of the material using linear regression Many patients get concerned when a test involves injection of a radioactive material. For example for scanning a gall-bladder, a few drops of Technitium-99m isotope is used. Half of the technetium-99m would be gone in about 6 hours. It, however, takes about 24 hours for the radiation levels to reach what we are exposed to in day-to-day activities. The relative intensity of radiation (ο§) as a function of time is given in the table below. Table1: Relative intensity of radiation as a function of time t (hours) 0 1 3 5 7 9 1.000 0.891 0.708 0.562 0.447 0.355 ο§ If the level of the exponential is related to time via an exponential formula πΎ = π΄π ο¬π‘ , find (a) the value of the regression constants A and ο¬. (b) the half-life of Technitium99m, and (c) the radiation intensity after 24 hours (c) the radiation intensity after 24 hours. Assume the initial interval of uncertainty as (-0.120, -0.110). Find the parameters thermal expansion model of a steel cylinder. Coefficient of thermal Temperature, T (ο°F) expansion, ο‘ (in/in/ο°F) 80 6.47ο΄10-6 40 6.24ο΄10-6 -40 5.72ο΄10-6 -120 5.09ο΄10-6 -200 4.30ο΄10-6 -280 3.33ο΄10-6 -340 2.45ο΄10-6 2 Fit the above data to πΌ = π0 + π1 π + π2 π . Also find the straight line model for the above data. Using the coefficient of determination ‘r2’show that the quadratic model gives a better fit. 7. The height of a child is measured at different ages as follows. t (yrs) 0 5 8 12 16 18 H (in) 20 36.2 52 60 69.2 70 Estimate the height of the child as an adult of 30 years of age using the growth model π π»= 1 + ππ −ππ‘ 8. A long hollow steel shaft called the trunnion is ‘shrink fit’ into a steel hub. The resulting steel trunnion-hub assembly is then shrink fit into the girder of a bridge. This is done by first immersing the trunnion in a cold medium such as dry/ice alcohol mixture or liquid nitrogen. After the trunnion reaches the steady state temperature of the cold medium, the trunnion outer diameter contracts. The trunnion is taken out of the medium and slid through the hole of the hub. When the trunnion heats up to reach the room temperature of 80ο° F (Troom), it expands and creates an interference fit with the hub. A hollow trunnion of outside diameter (D) 12.363” is to be fitted in a hub of inner diameter 12.358”. The trunnion was put in the cold medium at a temperature of -108ο° F (Tfluid) to contract the trunnion so that it can be slid through the hole of hub. To slide the trunnion without sticking, a diametrical clearance of at least 0.01” is required between the trunnion and the hub. The required contraction is given by οD = trunnion outside diameter – hub inner diameter + diametric clearance Formula for calculating the estimated contraction is given by βπ· = π·πΌ(βπ) Where βπ = ππππ’ππ − πππππ Α = 6.47 × 10-6 in/in/ο°F at room temperature. Using this model estimate the contraction. The process failed when this approximate model is used. What could be the resons? Try to find οD using the formula ππππ’ππ βπ· = ∫ πππππ For this α as s function of T is required. ∝ ππ We can use the following experimental data. Temperature ο°F 80 60 40 20 0 -20 -40 -60 -80 -100 -120 -140 -160 -180 -200 -220 -240 -260 -280 -300 -320 -340 Instantaneous Thermal Expansion µin/in/ ο°F 6.47 6.36 6.24 6.12 6.00 5.86 5.72 5.58 5.43 5.28 5.09 4.91 4.72 4.52 4.30 4.08 3.83 3.58 3.33 3.07 2.76 2.45