GATE — Agricultural Engineering (AG), for dairy engineering aspirants · India · IITs and IISc on rotation
Original practice questions written to the published syllabus · not past papersMeasuring 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)
How our questions for GATE spread across the units of Instrumentation, Process Control & Automation. Difficulty: 0 easy, 18 medium, 4 hard.
| Temperature measurement: thermocouples, RTDs and recorders | 5 | |
| Generalised measurement system and instrument characteristics | 4 | |
| Pressure, flow and level measurement | 4 | |
| Control modes: on-off, P, PI and PID | 3 | |
| Final control elements: valves and drives | 3 | |
| Programmable logic controllers | 2 | |
| Automation of pasteurisers, CIP and milk reception | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A first-order temperature probe has a time constant of 5 s. About how long after a step change does it reach 95% of the final change?
D. About 15 s
1 − e^(−t/τ) = 0.95 gives t ≈ 3τ = 15 s. 5 s is one time constant (63.2%), 10 s is two (86.5%) and 25 s is five (about 99.3%).
A thermocouple with an assumed constant Seebeck coefficient of 41 µV/°C has its hot junction at 120 °C and its reference junction at 25 °C. What emf does it produce?
C. About 3.90 mV
emf = 41 µV/°C × (120 − 25) = 3,895 µV ≈ 3.90 mV. 4.92 mV ignores the reference junction; 1.03 mV uses only the reference temperature; 5.95 mV adds the two temperatures.
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?
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.
An orifice meter shows 40 kPa differential at 20,000 L/h. If the flow falls to 10,000 L/h, what differential will it show?
B. 10 kPa
Flow ∝ √ΔP, so ΔP ∝ flow²: halving flow gives (½)² × 40 = 10 kPa. 20 kPa assumes a linear relation; 5 kPa cubes; 28 kPa takes 40 × 1/√2.
The differential across a venturi falls to one quarter of its earlier value. The flow is now:
C. Half its earlier value
Flow ∝ √ΔP, so √(1/4) = 1/2. A quarter assumes linearity; a sixteenth squares instead of rooting; flow cannot rise when differential falls.
A hydrostatic level transmitter in a silo was set up for water (1,000 kg/m³). The silo now holds milk of 1,032 kg/m³. If the true milk depth is 5.0 m, what depth does the transmitter show?
B. About 5.16 m
Pressure = 1,032 × 9.81 × 5.0; read as water depth: 5.0 × 1,032/1,000 = 5.16 m. 4.84 m divides the wrong way; 5.00 m ignores the density change; 5.32 m doubles the error.
A valve with Kv = 8 passes water at a 1.44 bar pressure drop. Using Q = Kv √ΔP, what is the flow?
C. 9.6 m³/h
Q = 8 × √1.44 = 8 × 1.2 = 9.6 m³/h. 11.5 omits the square root; 6.7 divides by √ΔP; 8.0 ignores the pressure drop.
A centrifugal CIP pump's speed is reduced by 20% with a variable-frequency drive. By the affinity laws, roughly what saving in shaft power results?
A. About 49%
Power ∝ speed³: 0.8³ = 0.512, a saving of about 49%. 20% assumes power ∝ speed; 36% uses the square law for head; 64% is the head ratio, not a saving.
A 4–20 mA level transmitter is ranged 0–5 m. What current corresponds to 3.5 m?
A. 15.2 mA
I = 4 + 16 × (3.5 ÷ 5) = 15.2 mA. 14.0 mA uses 20 × 0.7; 11.2 mA omits the 4 mA offset; 17.5 mA uses 25 × 0.7.
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?
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%.
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