22-Agric-B8 Food Process Engineering (Part 1) · December 2013
Question 5 of 10: Vapour Recompression and Steam-Jet Ejection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-B8 Food Process Engineering (Part 1), National
Exams December 2013 — a three-hour open-book exam (any non-communicating
calculator permitted). Ten questions are set in four sections (I–IV), each with a
"choose N of M" instruction; candidates who follow the choice rule answer six questions for a
100-mark paper. All ten are worked here so the set is a complete study resource.
Reference texts. R.T. Toledo, Fundamentals of Food Process Engineering,
3rd ed. (thermal-process lethality, D and z values, Ball/Stumbo process calculation, aseptic
holding-tube residence time — this is the exam's own appendix source); C.J. Geankoplis,
Transport Processes and Separation Process Principles, 4th ed. (evaporator heat and
mass balances, multiple-effect steam economy, vapour recompression); R.P. Singh and
D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation,
modified Plank equation, unsteady-state heat transfer in canned foods); A.C. Cleland,
Food Refrigeration Processes: Analysis, Design and Simulation (Plank/Cleland-Earle
freezing-time correlations); F.P. Incropera and D.P. DeWitt, Fundamentals of Heat and Mass
Transfer (transient conduction, Heisler charts, composite-wall resistance).
Question 5: Vapour Recompression and Steam-Jet Ejection (20 marks)
Vapour-recompression stream data (from the source figure)
Stream
Condition
Vapour leaving the evaporating liquid
70°C, saturated, 31.19 kPa
Compressed vapour to heating coil
100°C, 50 kPa
Condensate leaving the coil
80°C (≈ sat. at 50 kPa)
Auxiliary live steam
120°C, saturated
Find. Whether (and how much) auxiliary live steam is required beyond the
recompressed vapour.
Mechanical vapour recompression (MVR): evaporator vapour is compressed and
returned to the tank's heating coil, condensing and giving up heat to re-boil the liquid.
Check: the source figure's own label places "Compressed Vapour 100°C
50 kPa" on the stream entering the compressor, which is thermodynamically backwards for
a compressor (its inlet must be the lower-pressure 70°C/31.19 kPa vapour). This is read as
an extraction/labelling artifact and the stream is treated as the compressor's discharge,
feeding the tank's internal heating coil — the only reading consistent with vapour
recompression and with the coil's condensate leaving at 80°C (near the 50 kPa saturation
temperature of 81.3°C).
Approach. Per kg of vapour recompressed (1:1 with the vapour generated, since
all of it is recycled), compare the heat released as the compressed vapour desuperheats and
condenses in the coil against the heat required to evaporate 1 kg of liquid at 70°C.
Heat released by 1 kg of compressed vapour condensing to 80°C liquid.
From the superheated-steam table at 50 kPa, 100°C: \(h=2682.5\ \text{kJ/kg}\). Saturated
liquid at 80°C: \(h_f=334.9\ \text{kJ/kg}\).
\(q_{released} = 2682.5-334.9 = \boxed{2347.6\ \text{kJ/kg}}\).
Heat required to evaporate 1 kg at 70°C.
\(h_{fg,70^\circ C}=\boxed{2333.8\ \text{kJ/kg}}\).
Balance. Neglecting sensible heat (as instructed), the recompressed vapour
alone supplies \(2347.6-2333.8=\boxed{+13.8\ \text{kJ per kg vapour generated}}\), a small
surplus. The recompressed vapour therefore meets — and very slightly exceeds — the
evaporator's heat requirement, so no auxiliary steam is required at steady
state; the 120°C live-steam line exists for start-up (before enough vapour exists to
recompress) rather than for continuous makeup.
Find. The mass of vapour evaporated from the food (i.e. entrained through the
ejector) per kg of motive steam consumed.
Approach. A steam-jet ejector uses high-pressure motive steam expanding
through a nozzle to entrain and re-pressurize a low-pressure vapour stream; the ratio of
entrained-to-motive mass is the "vapour evaporated from the food per kg steam consumed"
that the question asks for, since the entrained stream is precisely the vapour being pulled off
the food evaporator. A mixed-stream enthalpy check confirms the two streams are being combined
consistently.
Read off the entrainment ratio. The question states directly that 0.4 kg of
evaporator vapour is entrained per kg of motive steam, i.e.
\(\dfrac{\dot m_{vapour}}{\dot m_{steam}} = \boxed{0.40\ \text{kg food-vapour/kg steam}}\).
Consistency check via the mixed-stream enthalpy. Motive steam
\(h_g(135^\circ C)=2727.3\ \text{kJ/kg}\); entrained vapour \(h_g(90^\circ C)=2660.1\ \text{kJ/kg}\).
For 1.4 kg combined,
\(\bar h = \dfrac{(1)(2727.3)+(0.4)(2660.1)}{1.4} = \boxed{2708.1\ \text{kJ/kg}}\), a value
between the two saturated-vapour enthalpies as required by a simple adiabatic mix —
confirming the entrained stream is genuinely a saturated low-pressure vapour and not, e.g.,
liquid condensate being mislabeled.
Final results
Quantity
Value
Food-vapour evaporated per kg steam consumed
0.40 kg/kg
Mixed-stream enthalpy (check)
2708.1 kJ/kg
Check: the source figure also shows a downstream barometric-condenser
stage (an additional "20 kPa, 60°C" water inlet, a further "0.4 kg vapour to condense"
outlet, and "condensate at 90°C") describing how the combined 1.4 kg motive+entrained stream
is finally condensed with contact water. That stage answers a different question (the
condenser's cooling-water requirement) and is not needed for the entrainment ratio asked here;
it is not solved as part of this sub-question.
