Sensitivity analysis of fertilizer plant

Introduction

-A number of researchers have analyzed availability parameters of various industrial system using different methods. In this research paper Behavioral Analysis of Urea Fertilizer Industry has been discussed using regenerative point graphical technique. Kumar, J. & Malik, S. C. [1] have discussed the concept of preventive maintenance for a single unit system. Others like, Nakagawa, T. and Osaki, S. [2] have discussed reliability analysis of a one unit system with un-repairable spare units and its applications. Some interconnected studies were also found such as Goel, P. & Singh J. [3] , Gupta, P., Singh, J. & Singh, I.P. [4] , Gupta, V. K. [5] , Chaudhary, Goel& Kumar [6] Sharma &Goel [7] , Gupta, R., Sharma, S. & Bhardwaj, P. [8] , Ms. Rachita and Garg, D. [9] and Garg, D. and Yadav, R. [10] have discussed behavior with perfect and imperfect switch-over of systems using various techniques. A urea fertilizer industry consists of six sub-units Ammonia Making Section (A), Medium Pressure Section (B), Low Pressure Section (C), Pre-vacuum Section (D), Vacuum Section (E) & Periling Section (F) and three pressure units low pressure, medium pressure and high pressure units (P1, P2& P3), whenever low or high pressure units fail, then low and medium pressure steams can be obtained from high pressure unit and whenever high pressure unit fail then high pressure steam cannot be obtained, thus the whole system is failed. The failed pressure units are repaired only when the whole system is in failed state. System works in full capacity when all units are good and fails when any one of the six sub-systems fails. Common cause failures may also lead the system to complete failure. All the units have sub units in series, so any one of the sub unit fails then the unit fails and the system fails when the number of failed units is more than two. Fuzzy concept is used to determine failure/working state of a unit. For failure / repair rates constant, independent and considering transition probability, a transition diagram of the system is drawn in Figure 1 to find Primary circuits, Secondary circuits & Tertiary circuits and Base state. Problem is solved using RPGT to determine system parameters. System behavior is discussed with the help of graphs and tables. Particular cases are also taken for different repair and failure rates of the system.

Assumptions, Notations & Model Description:

-The following assumptions and notations are taken: -1. There is single repair facility which is always available.

2. The distributions of failure and repair times of units are constant and also different. 3. Failures and repairs are statistically independent. 4. Repair is perfect and repaired system is as good as new one. 5. Nothing can fail when the system is in failed state. 6. The system is discussed for steady-state conditions. 7. Failure of any of the subsystems leads to failure of the system. 7. Low pressure and medium pressure can be obtained from high pressure unit by scientific logic. 8. The failed pressure units will be repaired only when the system is failed. Ri (t): reliability of the industry at general time t Ai (t): Probability of the system in up time at time't', given that the system entered Regenerative state 'i' at t = 0. Bi (t): The server is busy in repair at time't' Vi (t): The expected no. of server visits for doing a job in (o,t] given that the system Entered regenerative state 'i' at t = 0. ',': denote derivative μi: Mean sojourn time spent in state i, before visiting any other states; 𝜇 𝑖: = ∫ 𝑅 𝑖 (𝑡)𝑑𝑡 ∞ 0 ni: Expected waiting time spent while doing a given job, given that the system entered regenerative state 'i' at t=0; 𝑛 𝑖 𝑊 𝑖 * (0). ξ: Base state of the system. fj: Fuzziness measure of the j-state.

: Full Capacity Working State: Failed State Pn(t): Probability that the system is in state Snat time t.(n = 0,1,2,...,29) αi: Constant failure rates of subsystem A,B,C,D,E and F respectively for i = 1,2,..,6. α7, α8, α9: Constant failure rate of pressure unit P1,P2 and P3 respectively. α0: Constant failure rate of entire system from any of its operative state. βi, gi(t): Constant repair rate of subsystem A,B,C,D,E and F respectively for i = 1,2,..,6 and corresponding p. d. f., such that) =βi (t) exp [-∫ 𝛽 𝑖 (𝑢)𝑑𝑢] 𝑡 0 h, gh(t): Constant repair rate of system failed due to pressure unit P3 and corresponding p.d.f. gh(t) = h(t)exp[-∫ ℎ 𝑖 (𝑢)𝑑𝑢] 𝑡 0 c, gc(t): Constant repair rate of system failed due to common cause failure and corresponding p. d. f. gc(t) = c(t)exp [-∫ 𝑐 𝑖 (𝑢)𝑑𝑢] 𝑡 0 a: Indicates that subsystem A is failed, similarly for other units. S0: Initial operative state of the system where all the subsystems and units are in good condition. S21: Failed state of the system due to the failure of pressure unit P3. S22: Failed state of the system due to the common cause failure. Si, S7+i, S14+i, S23+i: Failed states of the system due to the failure of any of the subsystem A,B,C,D,E and F respectively for i = 1,2,3,4,5 and 6 when all the three pressure units are good. Taking into consideration the above assumptions and notations the Transition Diagram of the system is given in Figure 1.

Fig 1

From the state transition Figure 1, we see that there are maximum number of primary circuits and minimum number of secondary circuits; hence vertex '0' is the base state. Primary, Secondary, Tertiary Circuits w. r. t. the Simple Paths (Base-State '0') qi,j(t) Pij = q*i,j(0) 𝑞 0,𝑖 (𝑡) = 𝛼 𝑖 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 9 +𝛼 8 +𝛼 0 )𝑡 𝑞 0,14 (𝑡) = 𝛼 8 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 8 +𝛼 9 +𝛼 0 )𝑡 𝑞 0,21 (𝑡) = 𝛼 9 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 8 +𝛼 9 +𝛼 0 )𝑡 𝑞 0,22 (𝑡) = 𝛼 0 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 8 +𝛼 9 +𝛼 0 )𝑡 Where i = 1 to 7 𝑝 0,𝑖 = αi/(α1+α2+α3+α4+α5+α6+α7+α9+α8+α0) 𝑝 0,14 = α8/(α1+α2+α3+α4+α5+α6+α7+α8+α9+α0) 𝑝 0,21 = α9/(α1+α2+α3+α4+α5+α6+α7+α8+α9+α0) 𝑝 0,22 = α0/(α1+α2+α3+α4+α5+α6+α7+α8+α9+α0)

𝑝 𝑖,0 = 1, where 1≤ i ≤ 6 𝑞 7,7+𝑖 (𝑡) = 𝛼 𝑖 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 8 +𝛼 9 )𝑡 𝑞 7,23 (𝑡) = 𝛼 8 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 8 +𝛼 9 )𝑡 𝑞 7,21 (𝑡) = 𝛼 9 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 8 +𝛼 9 )𝑡 𝑞 7,22 (𝑡) = 𝛼 0 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 8 +𝛼 9 )𝑡 𝑝 7,7+𝑖 = αi/(α1+α2+α3+α4+α5+α6+α0+α8+α9) 𝑝 7,23 = α8/(α1+α2+α3+α4+α5+α6+α0+α8+α9) 𝑝 7,21 = α9/(α1+α2+α3+α4+α5+α6+α0+α8+α9) 𝑝 7,22 = α0/(α1+α2+α3+α4+α5+α6+α0+α8+α9)

𝑝 7+𝑖,7 = 1, where 1≤ i ≤ 6 𝑞 14,14+𝑖 (𝑡) = 𝛼 𝑖 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 9 +𝛼 0 )𝑡 𝑞 14,22 (𝑡) = 𝛼 0 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 9 +𝛼 0 )𝑡 𝑞 14,23 (𝑡) = 𝛼 7 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 9 +𝛼 0 )𝑡 𝑞 14,21 (𝑡) = 𝛼 9 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 7 +𝛼 9 +𝛼 0 )𝑡 𝑝 14,14+𝑖 = αi/(α1+α2+α3+α4+α5+α6+α7+α9+α0) 𝑝 14,22 = α0/(α1+α2+α3+α4+α5+α6+α7+α9+α0) 𝑝 14,23 = α7/(α1+α2+α3+α4+α5+α6+α7+α9+α0) 𝑝 14,21 = α9/(α1+α2+α3+α4+α5+α6+α7+α9+α0) 𝑞 14+𝑖,14 (𝑡)= 𝛽 1 𝑒 -𝛽 1 𝑡 , 𝑞 23+𝑖,23 (𝑡)= 𝛽 𝑖 𝑒 -𝛽 𝑖 𝑡 𝑝 14+𝑖,14 = 1, 𝑝 23+𝑖,23 = 1, where 1≤ i ≤ 6 𝑞 21,0 (𝑡)= h𝑒 -ℎ𝑡 , 𝑞 22,0 (𝑡)= c𝑒 -𝑐𝑡 𝑝 21,0 = 1, 𝑝 22,0 = 1 𝑞 23,22 (𝑡) = 𝛼 0 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 9 )𝑡 𝑞 23,23+𝑖 (𝑡) = 𝛼 𝑖 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 9 )𝑡 𝑞 23,21 (𝑡) = 𝛼 9 𝑒 -(𝛼 1 +𝛼 2 +𝛼 3 +𝛼 4 +𝛼 5 +𝛼 6 +𝛼 0 +𝛼 9 )𝑡 𝑝 23,22 = α0/(α1+α2+α3+α4+α5+α6+α0+α9) 𝑝 23,23+𝑖 = αi/(α1+α2+α3+α4+α5+α6+α0+α9) 𝑝 23,21 = α9/(α1+α2+α3+α4+α5+α6+α0+α9)

Mean Sojourn Times Ri(t): Reliability of the system at time t, given that the system in regenerative state i.: Mean sojourn time spent in state i, before visiting any other states; From the above table and graph we conclude that increase in repair rates do not have significant increase in the value of availability of the system. However, for maximum of availability repair rate of unit 'A' kept maximum.

Busy Period of the Server's Visits (B0) Busy Period of the Server's Visits(B0) Table Table 6 From the table 6 we see that busy period of the server maximum when repair rate of unit 'F' minimum in comparison to other units, hence repairman should be efficient in repairing the unit 'F' to have lower value of busy period of the server. Busy period is minimum when repair rate of pressure unit P3 and that from common cause failure is maximum. From the table and graph we see that MTSF is maximum when failure rate of higher pressure unit P3 is minimum and MTSF is minimum when common cause failure rate is maximum. Hence, to have optimum value of MTSF failure rate of high pressure unit and common cause should be minimum.

Availability of the System (A0)

Availability of the System (A0) Table Table From the table and graph we see that availability is maximum when failure rate of high pressures unit P3 and common cause failure rate are minimum and availability is minimum when failure rate of sub-units are maximum.

Busy Period of the Server's Visits (B0) Busy Period of the Server's Visits (B0) Table From the above table and graph for optimum value of V0 failure rate of sub-unit should be minimum in comparison to the failure rate of pressure units and common cause failure rate.

Conclusion

To have optimum value of system parameters management may control the failure and repair rates of units depending upon the availability of finances and market circumstances.

References

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