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Section 9.pptx

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kareemhamad3

The document discusses optimum design parameters and provides examples of determining optimum values that minimize costs. Question 1 finds the optimum insulation thickness. Question 2 determines optimum temperature and pressure values that minimize total cost. Question 3 calculates the daily profit at the production rate giving minimum cost per unit and maximum daily profit. Question 4 involves determining the batch cycle time that gives the minimum total annual cost of producing 1 million kg of chemical product.

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OPTIMUM DESIGN ENG. KAREEM HOSSAM
WHAT ARE WE GOING TO DEAL WITH 1- one variable 2- two variables 3- optimum production rates
HOW TO KNOW THE OPTIMUM DESIGN PARAMETERS
HOW TO KNOW THE OPTIMUM DESIGN PARAMETERS
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
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
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
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.
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 $
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 $
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.
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 = ｐ ＄ * . ＄ . 巨＄ = ＄ ＄ = 106 ｐ ＄ = cost of operation + cost of charge and discharge = 20 * time of operation + 10 $ * time of charge and Discharge
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 ｐ ＄ = cost of operation + cost of charge and discharge = 20 * time of operation + 10 $ * time of charge and Discharge = 20 * 2P0.5+ 10 $ * (0.7+0.3) Total cost/year = 300P1.2 + 200000 +[ 20 * 2P0.5+ 10 $ * (0.7+0.3)]* 106 From dTotal annual Cost/d(p)= zero get = 625 kg/bath Substitute in Cycle time= 2P0.5+1 get the minimum time

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Section 9.pptx

  • 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 = ｐ ＄ * . ＄ . 巨＄ = ＄ ＄ = 106 ｐ ＄ = cost of operation + cost of charge and discharge = 20 * time of operation + 10 $ * 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 ｐ ＄ = cost of operation + cost of charge and discharge = 20 * time of operation + 10 $ * time of charge and Discharge = 20 * 2P0.5+ 10 $ * (0.7+0.3) Total cost/year = 300P1.2 + 200000 +[ 20 * 2P0.5+ 10 $ * (0.7+0.3)]* 106 From dTotal annual Cost/d(p)= zero get = 625 kg/bath Substitute in Cycle time= 2P0.5+1 get the minimum time

Editor's Notes

  • #5: Source: https://xaktly.com/DerivativesIII.html