Economics of chem. plants

1. OPTIMUM DESIGN
ENG. KAREEM HOSSAM

2. WHAT ARE WE GOING TO DEAL WITH
• 1- one variable
• 2- two variables
• 3- optimum production rates

3. HOW TO KNOW THE OPTIMUM DESIGN
PARAMETERS

4. HOW TO KNOW THE OPTIMUM DESIGN
PARAMETERS

5. QUESTION 1
• Find the optimum thickness to achieve minimum total cost knowing that the fixed
cost of the insulation is 3x+20 egp where x is the insulation in cm, and the cost of
heat lost is 20/x +50.
• Solution
Total Cost = fixed + variable = 3x+20 + 20/x +50.
dcost/ dx = 3-20/x2  x= 2.58 cm
D2cost/dx2= 40/x3  at x = 2.58 the value will be positive

6. QUESTION 2
• if you know that the temperature and pressure are two variables affecting the
industry, on the other hand both of them contribute in the total cost if you know
that the effect of temperature is 2.33T + (3000/PT) +3 and the effect of pressure
is 5P+ (5000/PT) +5
• 1- Determine the optimum values for P and T which gives you the minimum cost
knowing that T in C and P in Psi.
• 2- Determine the total cost at the optimum conditions

7. QUESTION 2
• Total cost = 2.33T + 3000/PT +3 + 5P+ 5000/PT +5 =2.33T+8000/PT +5P+ 8
• Derivative w.r.t T = 2.33 – 8000/PT2 = 0
• Derivative w.r.t P = 5 – 8000/TP2 = 0
• Solving both equations
• P = 9 psi and T = 19.5 C
• Second derivatives are:
• 2*8000/PT3 and 2*8000/TP3 and both of them has + values after substitution
• The total cost at optimum conditions is 144 EGP

8. QUESTION 3
• A plant produces at a rate of P units per day. The variable costs per unit was 30
+ 0.1P, The total daily fixed charges are 1000, and all other expenses are constant
at 7000 per month. If the selling price per unit is 100 determine:
• (a)The daily profit at production giving the minimum cost per unit.
• (b)The daily profit at production giving the maximum daily profit.

9. QUESTION 3 – CONTINUE (A)
a) Total Cost per unit = 30+0.1P + 1000/P + (7000/30)/P = 30 + 0.1 P + 1233.33/P
d(cost)/P = 0.1 -1233.33/P2  P= 111 Unit
• Second derivative of the previous equation is 2* 1233.33/P3 (+ve at all cases)
• Daily profit is = selling revenue per day – total cost per day
• Total cost per unit at p =111 = 52.2 $
• Total cost per day = 52.2 * 111 = 5794.2$
• Selling revenue per day = 100*111 = 11100$
• Daily profit is 5305.8 $

10. QUESTION 3 – CONTINUE (B)
• B) daily profit = selling price per unit * no. of units per day - Total cost per unit * no. of units
per day
• daily profit = 100P – [30 + 0.1 P + 1233.33/P]P
• d profit/dP = 100-30 - 0.2P  P = 350 unit
• Second derivative = - 0.2 always negative which indicates a maximum value
• Daily profit at a production giving the maximum daily profit = 11016.67 $

11. QUESTION 4
• An organic chemical is being produced by a batch operation. Each cycle consists of
the operating time necessary to complete the reaction plus a discharging time of 0.7
hour and charging time of 0.3 hour. The operating time per cycle is equal to 2P0.5 h,
where P is the kilograms of product produced per batch. The operating costs during
the operating period are $20 per hour, and the costs during the discharge-charge
period are $10 per hour. The annual fixed costs for the equipment vary with the size of
the batch as the fixed cost is equal to 300P1.2 dollars per year. the plant can be
operated 24 h per day for 300 days per year. The annual production is 1 million kg of
product. At this capacity, the other costs rather than those already mentioned is
$200,000 per year. Determine the cycle time for conditions of minimum total cost per
year.

12. QUESTION 4 SOLUTION
• Givens:
• P production per cycle (kg)
• discharging time of 0.7 hour/cycle
charging time of 0.3 hour/cycle
operating time per cycle = 2P0.5 h
• operating costs
during the operating= $20/h
during the discharge-charge are
$10/h
• fixed cost= 300P1.2 $/year
• Operation time
24 h per day for 300 days per year
• annual production is 106 kg of
product
• other costs= 200000$/year
• Determine the cycle time for conditions of minimum total cost
per year.
• Total cost/year = fixed cost/year + variable cost/year
• Cycle time= operating + charge + discharge = 2P0.5+1
• Total cost/year = 300P1.2 + 200000 + variable cost/year
• Variable cost /year =
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡
𝑏𝑎𝑡𝑐ℎ
*
𝑁𝑜. 𝑜𝑓𝑏𝑎𝑡𝑐ℎ
𝑦𝑒𝑎𝑟
•
𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑡𝑐ℎ
𝑦𝑒𝑎𝑟
=
𝑎𝑛𝑛𝑢𝑎𝑙 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒
=
106
𝑃
•
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡
𝑏𝑎𝑡𝑐ℎ
= cost of operation + cost of charge and discharge
• =
20
ℎ𝑜𝑢𝑟
* time of operation +
10 $
ℎ𝑜𝑢𝑟
* time of charge and Discharge

13. QUESTION 4 SOLUTION
• Givens:
• P production per cycle (kg)
• discharging time of 0.7 hour/cycle
charging time of 0.3 hour/cycle
operating time per cycle = 2P0.5 h
• operating costs
during the operating= $20/h
during the discharge-charge are
$10/h
• fixed cost= 300P1.2 $/year
• Operation time
24 h per day for 300 days per year
• annual production is 106 kg of
product
• other costs= 200000$/year
•
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡
𝑏𝑎𝑡𝑐ℎ
= cost of operation + cost of charge and discharge
• =
20
ℎ𝑜𝑢𝑟
* time of operation +
10 $
ℎ𝑜𝑢𝑟
* time of charge and Discharge
• =
20
ℎ𝑜𝑢𝑟
* 2P0.5+
10 $
ℎ𝑜𝑢𝑟
* (0.7+0.3)
• Total cost/year = 300P1.2 + 200000 +[
20
ℎ𝑜𝑢𝑟
* 2P0.5+
10 $
ℎ𝑜𝑢𝑟
* (0.7+0.3)]*
106
𝑃
• From dTotal annual Cost/d(p)= zero  get = 625 kg/bath
• Substitute in Cycle time= 2P0.5+1  get the minimum time