- 2076 Chaitra · 12 marks
Case study: workflow, factors, proposed design, benefits
Short answer
Existing manual pasteurization with analogue gauges suffers from temperature swings, power quality, and operator errors. Proposed microcontroller system with PT100, 12-bit ADC, PID/hysteresis control, Modbus SCADA, watchdog achieves ±0.5°C accuracy, 0.037°C resolution. Investment Rs 12 lakh, monthly savings Rs 1,15,000, payback 10.4 months.
Answer the following questions with regard to your case study. (a) Describe the existing work flow mechanism of the industrial instrumentation system. (b) What are the critical factors affecting the production of existing system and what measures you can recommend for mitigating those factors? (c) Design a proposed system using microprocessor/microcontroller, input/output devices, interfacing process, communication protocols, data converters and handshake signals with neatly labeled block diagram. (d) List out the different advantages of the proposed plan in terms of technology, production rate, quality assurance, cost-benefit and return on investment (ROI).
Assumed case study: a milk pasteurisation and packaging section of a dairy plant (HTST process: raw milk tank, heat exchanger, steam-heated holding tube, cooling, filling). The existing system uses analog panel indicators, bimetal thermostats, a manual steam valve and operator logging. (If your own visit was to another plant, replace the process names; the structure of the answer stays the same.)
(a) Existing workflow
Raw milk tank --> pump --> heat exch. --> holding tube
(level sw) (motor (steam valve (72 C, 15 s)
starter) manual) |
cooling --> filling
Panel meters / thermostats <-- operator reads, adjusts
Milk is received, tested, chilled and pumped to the pasteuriser. The operator watches the thermometer and adjusts the steam valve by hand. The milk goes through the holding tube, is cooled, stored and then filled into pouches. Records are written manually.
(b) Critical factors and mitigation
| Critical factor | Mitigation |
|---|---|
| Manual temperature control | Closed-loop PID/hysteresis control |
| Power cuts and voltage swings | UPS, stabiliser, generator |
| Steam pressure variation | Pressure sensor and control valve |
| Sensor drift | Scheduled calibration, redundant sensor |
| Operator error and missing records | Automatic logging and alarms |
(c) Proposed system
PT100 --> Amp/SigCond --+
Level sensor ----------+--> MUX/ADC (12-bit)
Flow sensor ---(pulse)-+ |
Keypad ----------------+---> [Microcontroller] <--> LCD 16x2
| ARM/AVR + flash |
Opto + SSR/relay <----------+ watchdog, timers +--> RS-485
| | (Modbus)
Pump / steam valve / alarm +--> EEPROM log PC / SCADA
Power supply: 24 V + 5 V + UPS
- Controller: a microcontroller (on-chip ADC, UART, timers).
- Input devices: PT100 temperature sensor, level switch, flow sensor, keypad.
- Output devices: steam valve and pump (via optocoupler and SSR/relay), LCD, buzzer.
- Data converters: 12-bit ADC (0.037 C/count); optional DAC (4-20 mA) for a proportional valve.
- Communication: RS-485 Modbus RTU, 9600 baud, 8N1.
- Handshake signals: ADC start/end of conversion, LCD enable strobe, RS-485 DE/RE and master-slave polling.
/* Assumed: 8-bit MCU, ADC on channel 0 (PT100 amplifier), relay on PORTB.0 */
unsigned int read_adc(void){ /* 12-bit conversion */
ADC_START(); while(!ADC_DONE()); /* start, wait for EOC */
return ADC_DATA; /* 0..4095 */
}
void control(void){
float t = read_adc() * 150.0 / 4096; /* temperature in C */
if (t < 72.5) STEAM_ON(); /* open steam valve */
else if (t > 74.0) STEAM_OFF(); /* hysteresis control */
wdt_reset(); /* watchdog */
}
The control band (72.5 to 74 C) stays above the 72 C minimum holding temperature.
(d) Advantages of the proposed plan
- Technology: digital, programmable, networked and easy to upgrade.
- Production rate: continuous automatic operation and less downtime (assume +10 percent).
- Quality assurance: tight temperature control and records.
- Cost-benefit and ROI: investment Rs 12,00,000 (controller panel, sensors, valves, SCADA PC, installation, training). Monthly savings: manpower Rs 60,000, steam/energy Rs 35,000, reduced rejection Rs 20,000 = Rs 1,15,000. Payback = 12,00,000 / 1,15,000 ≈ 10.4 months; yearly return = 13.8 lakh on 12 lakh, about 115 percent.
✓ Answer checked by Claude Sonnet 5.5
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