Question

4. Determine the amount of pig iron, steel scrap, and ferrosilicon to produce a gray iron having the composition (wt. %) 3.2 C, 3.0 Si, using a coreless induction melting furnace (30 points). 0,032C0.03 S Typical losses for C and Si elements in an induction furnace are as follows: Element Si Loss, % The following raw materials are available (composition in wt. %) pig iron: 3.8 C, 0.8 Si steel scrap: 0.8 C, 0.5 Si -20% returns: 3.2 C, 3.0 Si ferrosilicon (75FeSi): 75% Si, balance Fe. Note: You must use 20% returns in your charge calculation. Pi Iron 3.c? 3. 800%

Answer 1

Before arriving at the solution, we will go through some basic concepts related to the problem so that it will be easier to understand the problem

The Gray iron is made of 3 elements, namely carbon C (3.2 %), Silicon Si (3.0 %) and the rest is iron Fe (93.8%). I.e. for every 100 gm block of Gray iron, we would have 3.2 gms of C, 3 gms of Si and remaining 93.8 gms (100-(3.2+3)) of Fe. [Note: gms means grams]

Same goes for all other elements such as Pig iron, steel scraps, and Ferrosilicon as mentioned in the problem.

Now what required is we need to produce let's say 100 gms of Gray iron by melting and mixing 'X' gms of pig iron, 'Y' gms of steel scrap, 'Z' gms of Ferrosilicon and 20 gms of returns (Here 20% returns. 20% wt implies 20% of the weight of Gray irons produced (20/100)*100 = 20 gms ). Hence our job is to find X, Y, and Z. Here there are 3 unknowns and we need to set up 3 equations to find those unknowns.

The 3 equations will be, 1. Carbon balance equation, 2. Silicon balance equation and 3. Fe balance equation.

So now, amount of carbon present in 'X' gms of pig iron = ((3.8/100) * X) gms.

Similarly, we can find the amount of C, Si, and Fe in X, Y, and Z gms.

Equation 1: Carbon balance

(3.8/100) * X + (0.8/100) * Y + (0/100) * Z + (3.2/100) * 20 = (3.2/100) * 100.

Before, writing the other two equations we need to note that we incur some loss of C and Si in induction melting furnace. While melting in the induction furnace we suffer a loss of 7% C and 5% Si. This implies that 7% of the carbon present in the melt will be lost. Hence to be prepared for that loss we need to have extra C and Si in the mixture such that after the losses we are left with 3.2 gms C and 3 gms Si in the final block.

This part is tricky. How much additional C and Si we need?. First, consider only carbon for simplicity. It says that 7% of the total carbon in the mixture will be lost. Now if I want 100 gms of carbon in my final product what gms of Carbon should I start with so that I am left with 100 gms carbon in the end?

it will NOT be 107 gms, because

If I start with 107 gms, its 7% will be (7/100) * 107 = 7.49 gms, Hence I am left with (107 - 7.49) = 99.51 gms which is less than 100 gms.

Let us denote the start amount as A gms. Hence the equation will be,

A - (7/100) * A = 100

(93/100) * A = 100

A = (100/.93) = 107.526 gms.

To cross check, 7% of 107.526 = (7/100) * 107.526 = 7.526.

Final mass 107.526 - 7.526 = 100 gms

Solution to the problem begins:

The total % of carbon required during melting and mixing phase to compensate for the 7% loss occurring in the furnace = 3.2 / 0.93 = 3.4408 %.

The total % of silicon required during melting and mixing phase to compensate for the 5% loss occurring in the furnace = 3.0 / 0.95 = 3.15789 %.

Equation 1 i.e carbon balance equation becomes,

(3.8/100) * X + (0.8/100) * Y + (0/100) * Z + (3.2/100) * 20 = (3.4408/100) * 100.

Multiplying throughout by 100, we get

$\dpi{100} \therefore$ 3.8 X + 0.8 Y + 0 Z + 64 = 344.08

$\dpi{100} \therefore$ 3.8 X + 0.8 Y + 0 Z = 280.08 ............................................................................................(1)

Equation 2: Silicon balance equation

(0.8/100) * X + (0.5/100) * Y + (75/100) * Z + (3/100) * 20 = (3.15789/100) * 100

Multiplying throughout by 100, we get

$\dpi{100} \therefore$ 0.8 X + 0.5 Y + 75 Z + 60 = 315.789

$\dpi{100} \therefore$ 0.8 X + 0.5 Y + 75 Z = 255.789 .........................................................................................(2)

Equation 3: Fe balance equation

% Fe present in Pig iron = (100 - (3.8 + 0.8)) = 95.4%

% Fe present in steel scrap = (100 - (0.8 + 0.5)) = 98.7%

% Fe present in Ferrosilicon = (100 - 75) = 25 %

% Fe present in returns = (100 - (3.2 + 3.0)) = 93.8%

% Fe present in Melt = (100 - (3.4408 + 3.15789)) = 93.40131 %

The Equation will then be,

(95.4/100) * X + (98.7/100) * Y + (25/100) * Z + (93.8/100) *20 = (93.40131/100) * 100

Multiplying throughout by 100, we get

$\dpi{100} \therefore$ 95.4 X + 98.7 Y + 25 Z +1876 = 9340.131

$\dpi{100} \therefore$ 95.4 X + 98.7 Y + 25 Z = 7464.131...............................................................................(3)

Now, have three equations and 3 unknowns, writing it down,

3.8 X + 0.8 Y + 0 Z = 280.08

0.8 X + 0.5 Y + 75 Z = 255.789

95.4 X + 98.7 Y + 25 Z = 7464.131

Solving using Calculator, we obtain,

X = 72.7209 gms, Y = 4.6753 gms, Z = 2.6036 gms. and 20 gms returns for 100 gms og Gray iron

Hence we need to use, 72.7209 % Pig iron, 4.6753 % Steel scrap, 20% returns and 2.6036% Ferrosilicon to obtain 100% Gray iron after considering all the C and Si losses in the induction melting furnace.

(check: 72.7209 + 4.6753 + 20 + 2.6036 = 100 ).

Note: The 3 equations can be solved manually by writing it in matrix form AX = B, where A (3 x 3 )is the coefficient matrix, X (3 x 1) is the matrix of unknowns and B (3 x 1) is the matrix consisting of right hand side values.

Reducing matrix A into its echelon form by using Gaussian elimination, will lead you to the desired result.