STUPIDSID
1(a)
Explain how to use 'for' loop in SciLab with appropriate example.
5 M
1(b)
Use Crank-Nicholson Scheme to solve,
uxx = ut 0≤x≤1 t>0 \( h=\dfrac{1}{4} \) \( k=\dfrac{1}{4} \)
Given u(x, 0) = 0, u(0, t) = 0, u(1, t) =50t Compute u for one step in t-direction.
uxx = ut 0≤x≤1 t>0 \( h=\dfrac{1}{4} \) \( k=\dfrac{1}{4} \)
Given u(x, 0) = 0, u(0, t) = 0, u(1, t) =50t Compute u for one step in t-direction.
5 M
1(c)
Show progress of bisection method using graphical representation.
5 M
1(d)
Solve following system of equation,
x+y+z=7
z+2y+3z=16
z+3y+4z=22
x+y+z=7
z+2y+3z=16
z+3y+4z=22
5 M
2(a)
Liquid Molar volume of n-butane at 350 K and 9.4573 bar may be calculated using Redlich-Kwong equation as given below
\[V^l =\dfrac{zRT}{p}\]
where,
\[z=\beta +z(z+\beta)\left ( \dfrac{1+\beta-z}{q\beta} \right )\]
\[\beta=0.08664\dfrac{P_r}{T_r}\ \ \ Pr=\dfrac{P}{P_c}\ \ \ T_r=\dfrac{T}{T_c}\ \ \ q=6.6.48\]
for n-butane, Tc = 425.1Ki, pc = 37.96bar
Calculate liquid molar volume for n-butane at given condition using Newton-Raphson method starting with z=β.
\[V^l =\dfrac{zRT}{p}\]
where,
\[z=\beta +z(z+\beta)\left ( \dfrac{1+\beta-z}{q\beta} \right )\]
\[\beta=0.08664\dfrac{P_r}{T_r}\ \ \ Pr=\dfrac{P}{P_c}\ \ \ T_r=\dfrac{T}{T_c}\ \ \ q=6.6.48\]
for n-butane, Tc = 425.1Ki, pc = 37.96bar
Calculate liquid molar volume for n-butane at given condition using Newton-Raphson method starting with z=β.
20 M
3(a)
Solve following set of equations using Gauss-Seidel and Gauss-Jordan Method
2x1-3x2+x3=-11
3x1+4x2-3x3=-34
x1+5x2-2x3=-17
2x1-3x2+x3=-11
3x1+4x2-3x3=-34
x1+5x2-2x3=-17
14 M
3(b)
Write Laplace equation and express it in difference form using Taylor's series expansion.
6 M
4(a)
A chemical reactor that has a single second order reaction and a outlet flowrate that is a linear function of height has the following model
\[\dfrac{dVC}{dt}=F_{in}C_{in}-FC-kVC^2\]
\[\dfrac{dV}{dt}=F_{in}-F\]
where, F = βV.
The parameters and variables are as given below.
Fin = inlet flowrate, (2 LPM)
Cin = inlet concentration, (1 gmol/lit)
k = reaction rate constant, (2 lit/(gmol-min))
β = 1 min-1
V = reaction mixture volume, (at t = 0, 1 lit)
C = concentration in reactor, (at t = 0, 0.5 gmol/lit)
Find the concentration and volume after one minute using Runge-kutta second order method.
\[\dfrac{dVC}{dt}=F_{in}C_{in}-FC-kVC^2\]
\[\dfrac{dV}{dt}=F_{in}-F\]
where, F = βV.
The parameters and variables are as given below.
Fin = inlet flowrate, (2 LPM)
Cin = inlet concentration, (1 gmol/lit)
k = reaction rate constant, (2 lit/(gmol-min))
β = 1 min-1
V = reaction mixture volume, (at t = 0, 1 lit)
C = concentration in reactor, (at t = 0, 0.5 gmol/lit)
Find the concentration and volume after one minute using Runge-kutta second order method.
20 M
5(a)
Friction factor is commercial pipe for turbulcnt flow can be calculated using Colebrook equation. If roughness factor(k) for carbon steel pipe is 0.00015 m for a pipe with ID (D) 0.315 m, using suitable numerical method calculate friction Colebrook equation,
\[\dfrac{1}{\sqrt{f}}=-2.0\log\left ( \dfrac{k/D}{3.7}+\dfrac{2.51}{Re\sqrt{f}} \right )\] .
\[\dfrac{1}{\sqrt{f}}=-2.0\log\left ( \dfrac{k/D}{3.7}+\dfrac{2.51}{Re\sqrt{f}} \right )\] .
12 M
5(b)
Find the root of the function \( f(x)=\dfrac{3x^2}{16}-\dfrac{27}{4} \) using Regula-Falci method. Consider the span [0, 10].
8 M
6(a)
Solve the following system by Gaussian Elimination with and without partial pivoting and comment on the results:
\[\begin{bmatrix} 2 & 1 & 1 & -2\\ 4 & 0 & 2 & 1\\ 3 & 2 & 2 & 0\\ 1 & 3 & 2 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 8\\ 7\\ 3 \end{bmatrix}\]
\[\begin{bmatrix} 2 & 1 & 1 & -2\\ 4 & 0 & 2 & 1\\ 3 & 2 & 2 & 0\\ 1 & 3 & 2 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 8\\ 7\\ 3 \end{bmatrix}\]
10 M
6(b)
Solve the following system by LU decomposition:
\[A=\begin{bmatrix} 4 & 0 & -1 & 3\\ 2 & 1 & -2 & 0\\ 0 & 3 & 2 & -2\\ 1 & 1 & 0 & 5 \end{bmatrix}\ \ \ \ b_1=\begin{bmatrix} 0\\ 1\\ 4\\ -2 \end{bmatrix}\]
\[A=\begin{bmatrix} 4 & 0 & -1 & 3\\ 2 & 1 & -2 & 0\\ 0 & 3 & 2 & -2\\ 1 & 1 & 0 & 5 \end{bmatrix}\ \ \ \ b_1=\begin{bmatrix} 0\\ 1\\ 4\\ -2 \end{bmatrix}\]
10 M
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