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Download GTU BE/B.Tech 2018 Winter 4th Sem New 2140505 Chemical Engineering Maths Question Paper

Download GTU (Gujarat Technological University) BE/BTech (Bachelor of Engineering / Bachelor of Technology) 2018 Winter 4th Sem New 2140505 Chemical Engineering Maths Previous Question Paper

This post was last modified on 20 February 2020

GTU BE/B.Tech 2018 Winter Question Papers || Gujarat Technological University


Enrolment No.

Subject Code: 2140505

GUJARAT TECHNOLOGICAL UNIVERSITY

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BE - SEMESTER-IV (NEW) EXAMINATION - WINTER 2018

Subject Name: Chemical Engineering Maths

Date: 22/11/2018

Time: 02:30 PM TO 05:30 PM

Total Marks: 70

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Instructions:

  1. Attempt all questions.
  2. Make suitable assumptions wherever necessary.
  3. Figures to the right indicate full marks.

Q1 (a) Explain false position method. 03

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(b) Differentiate between bracketing and open methods to solve non-linear algebraic equations. 04

(c) Find root of the equation x3 —2x—5 = 0 using bisection method. 07

Q2 (a) Define: (1) Coefficient of determination, (2) Correlation coefficient, and (3) standard error of estimate 04

(b) Explain Gauss elimination method with its pitfalls. 07

(c) Use Gauss-Jordan technique to solve the following three equations. 07

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3x1-0.1x2 - 0.2x3 = 7.85

0.1x1 + 7x2 - 0.3x3 = -19.3

0.3x1 — 0.2x2 + 10x3 = 71.4

OR

(c) Solve following equations using Newton-Raphson technique, starting with X = [0.5 0.5]. Carry out two iterations.

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f1(x1,x2)=4-8x1+4x2-x12 = 0

f2(x1,x2)=1-4x1+3x2+x12 = 0

Q3 (a) Given a value of x = 2.5 with an error of ?x = 0.01, estimate the resulting error in the function, f(x) = x3 04

(b) Explain the following terms with suitable example: (1) Significant figures, (2) Relative error 07

(c) Use Jacobi’s method to solve the following three equations with initial values X1 = x2 = x3 = x4 = 0. Carry out three iterations. 03

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10x1 — 2x2 — x3— x4 = 3

-2x1 + 10x2 — x3 — x4 = 15

-x1 — x2 + 10x3 — 2x4 = 27

—x1 — x2 — 2x3 + 10x4 = -9

OR

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(a) Explain Gauss-Seidel method. 04

(b) Suggest method to plot the variables y and x, given in the following equation, so that data fitting the equation will fall on straight line. 07

y = ax / (1+x(x-1))

(c) For temperatures (T) and length (L) of heated road. If find the linear relationship between T and L. 07

T, °C 20 30 40 50 60 70
L, mm 800.3 800.4 800.6 800.7 800.9 801

Q.4 (a) Explain Simpson’s 3/8th rule. 03

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(b) Explain Newton’s divided difference interpolation method. 04

(c) Using Newton’s forward difference formula and data given in the table below, estimate vapor pressure of ammonia vapor at 23°C. The latent heat of ammonia is 1265 kJ/kg. 07

Temperature, °C 20 25 30 35
Pressure, KN/m2 810 985 1170 1365

OR

Q.4 (a) From the following table of values of x and y, obtain dy/dx for x=1.2 07

X 1 1.2 1.4 1.6 1.8 2 2.2
Y 2.7183 3.3201 4.0552 4.9530 6.0496 7.3891 9.0250

(b) Water is flowing through a pipe line 6 cm in diameter. The local velocities (u) at various radial positions (r) are given below: 07

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u, cm/s 2 1.94 1.78 1.5 1.11 0.61 0
r, cm 0 0.5 1 1.5 2 2.5 3

Estimate the average velocity U, using Simpson’s 1/3rd rule.

The average velocity is given by: U= (2/R2) ?0R U r dr, where R is radius of pipe.

Q.5 (a) Explain Milne’s predictor corrector method. 03

(b) Explain procedure to solve following heat conduction equation using finite difference technique. 04

?2T / ?x2 = ?T / ?t

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(c) Solve the following 3rd order ordinary differential equation using Euler method. At time, t= 0, initial guess values are X0=2, X'0 =16, X''0 =4. Use time interval from 0 to 1 second, with step size h = 0.5 sec. 07

d3x/dt3 + 9 d2x/dt2 + 6 dx/dt + 8x = 21

OR

Q.5 (a) Consider general linear 2nd order partial differential equation given below. 03

a ?2u/?r2 + b ?2u/?r?z + c ?2u/?z2 + d ?u/?r + e ?u/?z + fu + g = h

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where, a, b, d, e, f, g and h are functions of r, z and their derivatives. How to check, whether given partial differential equation is parabolic, hyperbolic or elliptic?

(b) Explain modified Euler’s method. 04

(c) Solve the following set differential equations using fourth order Runge-Kutta method assuming that at x=0, y1=4 and y2=6. Integrate to x=1 with a step size of 0.5. 07

dy1/dx = -0.5y1 + y2

dy2/dx = -0.3y1 - 0.1y2

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