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ICAR AICE-JRF/SRF Instrumentation, Process Control & Automation: practice questions

ICAR AICE-JRF/SRF (PhD): Dairy Chemistry, Dairy Microbiology, Dairy Technology, Dairy Engineering · India · NTA for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
108written to this exam in this subject
10syllabus units

What this subject covers

Measuring and controlling temperature, flow, pressure and level in a milk plant, from sensors to PLC and SCADA.

In the exams: ICAR AICE subject 38 Unit 7; ICAR AIEEA-PG group 17 Unit I (computerisation and automatic process control in milk processing)

Where the 108 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Instrumentation, Process Control & Automation. Difficulty: 33 easy, 54 medium, 21 hard.

Temperature measurement: thermocouples, RTDs and recorders12
Control modes: on-off, P, PI and PID12
Generalised measurement system and instrument characteristics11
Pressure, flow and level measurement11
pH, conductivity and inline composition sensors11
Final control elements: valves and drives11
Automation of pasteurisers, CIP and milk reception11
Programmable logic controllers10
Data integrity and electronic records10
SCADA and distributed control systems9

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Generalised measurement system and instrument characteristics · hard

A 12-bit analogue-to-digital converter digitises a 0–150 °C range. What is the approximate temperature resolution?

  1. About 0.15 °C
  2. About 0.037 °C
  3. About 0.59 °C
  4. About 0.012 °C
Answer and working

B. About 0.037 °C

A 12-bit converter has 2^12 = 4,096 steps: 150 ÷ 4,096 ≈ 0.037 °C. 0.59 °C is the 8-bit value (150 ÷ 256); 0.15 °C divides by 1,000; 0.012 °C divides by 12,288, which is not a converter size.

2 · Temperature measurement: thermocouples, RTDs and recorders · hard

A Pt100 (α = 0.00385 per °C) is wired with two copper leads of 0.5 Ω each in a two-wire circuit. Roughly what error do the leads cause?

  1. About 2.6 °C too low
  2. About 2.6 °C too high
  3. About 0.4 °C too high
  4. About 1.3 °C too high
Answer and working

B. About 2.6 °C too high

Two leads add 1.0 Ω in series; 1.0 ÷ 0.385 Ω/°C ≈ 2.6 °C, and extra resistance reads as higher temperature. 1.3 °C counts one lead only; the error is not low; 0.4 °C uses 0.385 as the answer.

3 · pH, conductivity and inline composition sensors · easy

In CIP, return-line conductivity is mainly used to:

  1. Measure the flow velocity in the return pipe
  2. Measure the residual protein film on surfaces
  3. Count the number of cleaning cycles
  4. Detect the change between rinse water and detergent
Answer and working

D. Detect the change between rinse water and detergent

Detergents conduct far more than water, so the return conductivity shows when full-strength detergent arrives and when rinse water replaces it. It does not measure soil films, cycle counts or velocity.

4 · Control modes: on-off, P, PI and PID · easy

A controller's proportional band is 25%. What is its gain?

  1. 0.25
  2. 4
  3. 400
  4. 25
Answer and working

B. 4

Gain = 100 ÷ PB = 100 ÷ 25 = 4. 0.25 is PB as a fraction; 25 is PB itself; 400 multiplies instead of dividing.

5 · Control modes: on-off, P, PI and PID · easy

Why does a proportional-only temperature controller leave an offset after a change in milk flow?

  1. It responds only to the rate of change of temperature
  2. Its gain falls automatically as the load increases
  3. A non-zero error is needed to hold the valve at its new opening
  4. Its derivative term cancels the integral term at steady state
Answer and working

C. A non-zero error is needed to hold the valve at its new opening

With P-only control, output changes only when there is error, so a new steady load needs a steady error. P-only has no integral or derivative terms, its gain is fixed, and it responds to error, not rate.

6 · Programmable logic controllers · easy

A PLC analogue input on a 4–20 mA loop reads 0.2 mA. This most likely means:

  1. The process is above its top range value
  2. A broken wire or failed transmitter
  3. The loop is at half scale
  4. The process is at the bottom of its range
Answer and working

B. A broken wire or failed transmitter

A live zero of 4 mA means any reading well below it signals a fault such as an open circuit. The bottom of the range is 4 mA, not near zero.

7 · SCADA and distributed control systems · easy

What distinguishes a distributed control system (DCS) from a PLC-plus-SCADA system?

  1. A DCS cannot handle analogue signals such as temperature or pressure at all
  2. A DCS works only for batch processes and never for continuous ones like UHT
  3. One integrated platform supplies controllers, I/O, stations and tools
  4. A DCS uses no programmable controllers at any level of the system hierarchy
Answer and working

C. One integrated platform supplies controllers, I/O, stations and tools

A DCS integrates its control, I/O and operator layers in one system, traditionally for large continuous plants. It uses controllers, handles analogue signals and suits continuous processes.

8 · Automation of pasteurisers, CIP and milk reception · medium

An HTST pasteuriser gives an average holding time of 16 s at 9,000 L/h. If flow rose to 10,000 L/h with the same holding tube, what would the average holding time be?

  1. 17.8 s
  2. 16.0 s
  3. 12.8 s
  4. 14.4 s
Answer and working

D. 14.4 s

Holding time ∝ 1/flow: 16 × 9,000 ÷ 10,000 = 14.4 s. 17.8 s scales the wrong way; 16.0 s ignores the change; 12.8 s subtracts 20%.

9 · Automation of pasteurisers, CIP and milk reception · medium

A CIP detergent tank holds 2,000 L of water to be made up to 1.5% w/v sodium hydroxide using a 50% w/w caustic solution. Ignoring the volume added, how much 50% solution is needed?

  1. 120 kg
  2. 30 kg
  3. 15 kg
  4. 60 kg
Answer and working

D. 60 kg

NaOH needed = 2,000 L × 15 g/L = 30 kg; as 50% solution: 30 ÷ 0.5 = 60 kg. 30 kg is the pure NaOH; 15 kg halves again; 120 kg divides by 0.25.

10 · Automation of pasteurisers, CIP and milk reception · easy

Why is an air eliminator fitted ahead of the reception flowmeter?

  1. So entrained air is not measured and paid for as milk
  2. So the milk is cooled before it reaches the silo
  3. So the tanker empties faster through a wider pipe
  4. So foam is added to protect the milk from oxygen
Answer and working

A. So entrained air is not measured and paid for as milk

Air bubbles occupy volume in the meter and would be counted as milk, overpaying the supplier. The air eliminator does not cool, speed unloading or add foam.

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Read the subject

Instrumentation, Process Control & AutomationThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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