---
title: "Case study: revamping MBI system for CEO"
subject: "Instrumentation II"
chapter: 10
key: "semester_5/instrumentation_ii:cs-ceo-revamp"
source: https://www.bctengineer.com/q/semester_5/instrumentation_ii/cs-ceo-revamp
site: BCT Engineering
---

# Case study: revamping MBI system for CEO

Instrumentation II · Chapter 10: Case Study

- Asked in: 2079 Baisakh, 12 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/instrumentation_ii/cs-ceo-revamp
- Whole chapter: https://www.bctengineer.com/md/semester_5/instrumentation_ii/chapter-10

## Question

Suppose, the CEO of the company where you performed your case study is impressed with your case study report, and decides to hire you as a consulting engineer to oversee their existing MBI system. You are assigned the task of revamping their existing MBI system with the blueprint that you have designed. Show a well labeled, clear and detailed sketch of your design that you will be presenting to the board of directors to convince them to implement your idea. Your block diagram and supporting documents should include a minimum of the following items: the hardware solution, the software requirements, the advantages and disadvantages of your own strategy, the gain in efficiency of the plant after employing your plan, and a cost breakdown of realizing your project.

## Short answer

Microcontroller-based system with PT100/amplifier, 12-bit ADC, SSR-driven steam valve, RS-485 Modbus SCADA. Resolution 0.037°C, control band 72.5–74°C. Advantages: ±0.5°C accuracy, 8% steam saving, labour reduction; investment Rs 12,00,000, payback 10.4 months.

## Answer

**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.)

#### Proposed design (blueprint for the board)
```text
 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
```

#### Hardware solution
A microcontroller board (12-bit ADC, UART, timers, watchdog), PT100 sensors with an amplifier, level and flow sensors, optocoupler plus SSR/relay drivers for the pumps and steam valve, a 16x2 LCD and keypad, an RS-485 link to a PC, a 24 V/5 V power supply with UPS, and an enclosure (IP54) with grounding and surge protection.

#### Software requirements
Firmware in C: ADC acquisition and filtering, PID/hysteresis control, alarm and interlock logic, keypad/LCD menu, Modbus RTU communication, data logging and watchdog. PC side: SCADA/HMI screens with trends, alarms and reports (for example an open-source or vendor package).

**Key figures (calculated):** a PT100 amplifier scaled 0-150 C to 0-5 V and a 12-bit ADC give a resolution of 150/4096 = 0.037 C per count (1.22 mV per count at 5 V), enough for the ±0.5 C pasteurisation limit. An 8-byte Modbus-style frame at 9600 baud takes 8 × 10 bits / 9600 = 8.3 ms, so every point can be updated several times per second.

#### Advantages
Accurate and stable temperature, quality and traceability, savings in steam, labour and rejects, remote monitoring, and easy expansion.

#### Disadvantages
Upfront cost, the need for training, sensitivity to electrical noise and power quality, software maintenance, and dependence on spares.

#### Gain in efficiency (assumed)
The temperature swing reduces from ±3 C to ±0.5 C; steam saving is about 8 percent; rejects fall from 1.5 percent to 0.5 percent; and labour is reduced from 4 to 2 operators per shift.

#### Cost breakdown (assumed, Rs)
| Item | Cost |
|---|---|
| Controller panel and electronics | 3,00,000 |
| Sensors (temperature, level, flow) | 2,00,000 |
| Valves and actuators | 3,50,000 |
| SCADA PC and software | 1,50,000 |
| Installation and wiring | 1,50,000 |
| Training and commissioning | 50,000 |
| **Total** | **12,00,000** |

**Cost-benefit (assumed figures):** 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which set correctly lists the main hardware of the proposed blueprint?
   A. Gauges, bimetal thermostat and manual valve
   B. Microcontroller board, PT100 with amplifier, level/flow sensors, opto plus SSR/relay drivers, LCD/keypad, RS-485 link, UPS
   C. Only a PC and a printer
   D. Only a PLC with no sensors
   Answer: B. The blueprint is a microcontroller-based controller with sensors, isolated drivers, display, communication and backup power. (Key point: Hardware solution)

2. Which gains are assumed after the revamp?
   A. Swing +/-0.5 C to +/-3 C and rejects 0.5% to 1.5%
   B. No change in rejects but 4 more operators
   C. Swing +/-3 C to +/-0.5 C, rejects 1.5% to 0.5%, operators 4 to 2 per shift
   D. Steam use rises by 8%
   Answer: C. Tighter control cuts rejects and steam use (about 8% saved) and halves the labour. (Key point: Efficiency gains)

3. The cost items are 3,00,000 + 2,00,000 + 2,50,000 + 1,50,000 + 1,50,000 + 50,000. What is the total?
   A. Rs 10,00,000
   B. Rs 11,00,000
   C. Rs 13,00,000
   D. Rs 12,00,000
   Answer: B. The sum is Rs 12,00,000, matching the stated investment. (Key point: Cost breakdown total)

4. How long does an 8-byte frame take at 9600 baud with 10 bits per byte?
   A. 8.3 ms
   B. 0.83 ms
   C. 83 ms
   D. 1.2 ms
   Answer: A. 8 x 10 bits / 9600 = about 8.3 ms. (Key point: Modbus frame time)

## Related in Instrumentation II

- [Case study: new system vs existing (cost, manpower)](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acs-cost-manpower-automation) — Instrumentation II, chapter 10, asked 4×
- [Case study: existing vs proposed system](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acs-2074asoj) — Instrumentation II, chapter 10
- [Case study: feasibility, drawbacks, convincing](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acs-2072kartik-feasibility) — Instrumentation II, chapter 10
- [Case study: recommended changes and convincing](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acs-recommend-changes) — Instrumentation II, chapter 10, asked 2×
- [Case study: workflow, factors, proposed design, benefits](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acs-2076chaitra) — Instrumentation II, chapter 10
- [Case study: critical factors in process control](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Acs-critical-factors) — Instrumentation II, chapter 10, asked 2×

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