Material Selection For A Product/Substitution Of Raw Material
The cost of a product can be reduced greatly by substitution of the raw materials.
Among various elements of cost, raw material cost is most significant and it forms a major portion of the total cost of any product.
So, any attempt to find a suitable raw material will bring a reduction in the total cost in any one or combinations of the following ways:
ü Reduced machining/process time
ü Enhanced durability of the product
ü Cheaper raw material price
Therefore, the process of raw material selection/substitution will result in finding an alternate raw material which will provide the necessary functions that are provided by the raw material that is presently used.
In this process, if the new raw material provides any additional benefit, then it should be treated as its welcoming feature. This concept is demonstrated with numerical problem given below
In the design of a jet engine part, the designer has a choice of specifying either an aluminium alloy casting or a steel casting. Either material will provide equal service, but the aluminium casting will weigh 1.2 kg as compared with 1.35 kg for the steel casting.
The aluminium can be cast for Rs. 80.00 per kg. and the steel one for Rs. 35.00 per kg. The cost of machining per unit is Rs. 150.00 for aluminium and Rs. 170.00 for steel. Every kilogram of excess weight is associated with a penalty of Rs. 1,300 due to increased fuel consumption. Which material should be specified and what is the economic advantage of the selection per unit?
Solution (a) Cost of using aluminium metal for the jet engine part:
Weight of aluminium casting/unit = 1.2 kg
Cost of making aluminium casting = Rs. 80.00 per kg
Cost of machining aluminium casting per unit = Rs. 150.00
Total cost of jet engine part made of aluminium/unit
= Cost of making aluminium casting/unit + Cost of machining aluminium casting/unit
= 80 1.2 + 150 = 96 + 150 = Rs. 246
(b) Cost of jet engine part made of steel/unit: Weight of steel casting/unit = 1.35 kg
Cost of making steel casting = Rs. 35.00 per kg
Cost of machining steel casting per unit = Rs. 170.00 Penalty of excess weight of steel casting = Rs. 1,300 per kg
Total cost of jet engine part made of steel/unit
= Cost of making steel casting/unit
+ Cost of machining steel casting/unit
+ Penalty for excess weight of steel casting
= 35 1.35 + 170 + 1,300(1.35 – 1.2)
= Rs. 412.25
DECISION The total cost/unit of a jet engine part made of aluminium is less than that for an engine made of steel. Hence, aluminium is suggested for making the jet engine part. The economic advantage of using aluminium over steel/unit is Rs.
412.25 – Rs. 246 = Rs. 166.25
i.Design Selection for a Product
a. The design modification of a product may result in reduced raw material requirements, increased machinability of the materials and reduced labour.
b. Design is an important factor which decides the cost of the product for a specified level of performance of that product.
(Design selection for a process industry). The chief engineer of refinery operations is not satisfied with the preliminary design for storage tanks to be used as part of a plant expansion programme. The engineer who submitted
the design was called in and asked to reconsider the overall dimensions in the light of an article in the Chemical Engineer, entitled “How to size future
The original design submitted called for 4 tanks 5.2 m in diameter and 7 m in height. From a graph of the article, the engineer found that the present ratio of height to diameter of 1.35 is 111% of the minimum cost and that the minimum cost for a tank was when the ratio of height to diameter was 4 : 1. The cost for the tank design as originally submitted was estimated to be Rs. 9,00,000. What are the optimum tank dimensions if the volume remains the same as for the original design? What total savings may be expected through the redesign?
(a) Original design
Number of tanks = 4 Diameter of the tank = 5.2 m Radius of the tank = 2.6 m Height of the tank = 7 m
Ratio of height to diameter = 7/5.2 = 1.35 Volume/tank = (22/7)r 2h = (22/7)(2.6)2 7
= 148.72 m3
(b) New design
Cost of the old design = 111% of the cost of the new design (optimal design)
Optimal ratio of the height to diameter = 4:1
h : d = 4 : 1
4d = h
d = h/4
r = h/8
Volume = (22/7)r2h = 148.72 (since, the volume remains the same)
(22/7)(h/8)2h = 148.72
h = 14.47 m
r = h/8 = 14.47/8 = 1.81 m
Diameter of the new design = 1.81 2
= 3.62 m
Cost of the new design = 9,00,000 (100/111)
= Rs. 8,10,810.81
Expected savings by the redesign = Rs. 9,00,000 – Rs. 8,10,810.81 = Rs. 89,189.19