SKH Thermodynamics Labs

Lab 07 · Adiabatic flash

How much liquid flashes when no heat is added?

The Lab 01 mixture, z₁ = 0.40, enters as a compressed liquid at 370 K and flashes to 90 kPa through a valve. No heat is added or removed.

Find the outlet temperature and vapor amount, then explain where the energy to vaporize comes from.

2027 prototype
Review the student workflow

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Core: predict, calculate, read the energy balance and explain the result. Steps 1–5.

Graduate: the core steps plus a feed-temperature sweep, a liquid-model comparison and the heater duty in Step 4.

Graduate route. Complete Steps 1–3 to verify the energy balance. In Step 4, find where the flash begins, compare liquid models and calculate the heater duty that replaces Q = 0. Go to the graduate investigation.

Choose a feedAll named cases use the Lab 01 synthetic ideal mixture with z₁ = 0.40.

Picture an adiabatic flash

Energy, not temperature, is specified
A hot compressed liquid passes a valve into an insulated flash drum The liquid feed at 370 kelvin passes a pressure-reducing valve into an insulated drum at 90 kilopascals with no heat transfer. Vapor leaves at the top and liquid at the bottom, both at the outlet temperature. FeedVaporLiquid 370 K · liquidvalveQ = 0 · 90 kPaz₁ = 0.40

Lab 01 specified T and P. Here P and the energy balance (Q = 0) are specified, and the outlet temperature is unknown.

Enthalpy of the feed and of the outlet relative to liquid at 350 K The feed carries 2240 joules per mole of sensible enthalpy above the 350 kelvin liquid reference. At the outlet, the sensible enthalpy is minus 172 joules per mole because the outlet is colder than 350 kelvin, and the latent enthalpy of the vapor formed is 2412 joules per mole. Both totals equal 2240 joules per mole. sensible+2240latent+2412sensible −172 Feed · 370 KOutlet · 348.5 K 02240 Total h = 2240 J mol⁻¹ on both sides

Calculated for the starting case. Enthalpy is relative to liquid at 350 K. Cooling the whole 1 mol stream by 21.5 K releases the 2412 J needed to vaporize 0.0755 mol.

Because liquid and vapor have the same heat capacities here, the energy balance reduces to Cp(TF − T) = β Δhvap(y): sensible heat lost equals latent heat gained.

1. Set the feed and outlet pressure

Basis · 1.00 mol feed · Q = 0
K
kPa
mol/mol

The feed is a compressed liquid upstream of the valve. Property range 300–390 K.

Optional. Your choice does not change the calculation.

Add a calculated result to the comparison before changing a setting.

Caloric model and assumptions

Enthalpies are relative to the pure liquids at 350 K. Cp,1 = 100 and Cp,2 = 120 J mol⁻¹ K⁻¹ in both phases; Δhvap,1 = 30 000 and Δhvap,2 = 35 000 J mol⁻¹, the same constants as the Lab 01 vapor-pressure laws. The vapor is ideal. Activity parameters are held constant, so the consistent excess enthalpy is zero. Pressure effects on liquid enthalpy, heat loss and shaft work are neglected.

Open the current Lab 07 for a specified-enthalpy feed and the solver iterations.

2. Read the outlet state and the energy balance

Calculate the flash to see the outlet temperature, the vapor amount and the energy balance.

4. Check your understanding

Write first, then compare

Use Steps 1–3 to answer each question in your own words. Then open the explanation and compare. Your answers are saved with your study.

A. Why is the outlet colder than the feed?

Calculate the starting case and compare T with TF.

Compare with the explanation

The outlet is at 348.47 K, 21.5 K below the 370 K feed. Vaporizing 0.0755 mol needs 0.0755 × 31 951 = 2412 J. With no heat supplied, that energy comes from the stream itself: cooling 1 mol by 21.5 K releases 112 × 21.5 = 2412 J.

The outlet settles just above the feed's bubble temperature at 90 kPa (347.95 K), where the remaining liquid is saturated.

B. Why does so little of the feed vaporize?

Compare the starting case with “Hotter feed · 390 K”. How much more vapor does 20 K of extra feed temperature buy?

Compare with the explanation

β rises from 0.0755 to 0.1436 mol. From Cp(TF − T) = β Δhvap, each kelvin of superheat vaporizes only about Cp/Δhvap = 112/32 000 ≈ 0.0035 mol per mol of feed.

Latent heats are roughly 300 times larger than heat capacities, so an adiabatic flash of a liquid usually vaporizes only a small fraction.

C. What would it take to reach the Lab 01 state?

Lab 01 found β = 0.30 at 350 K and 90 kPa for the same feed. Use the energy balance to estimate the feed temperature that would give this state with Q = 0.

Compare with the explanation

At 350 K the vapor has y₁ = 0.5556, so Δhvap(y) = 32 222 J mol⁻¹. Then TF = 350 + 0.30 × 32 222/112 = 436 K, outside the 300–390 K property range.

In practice the same state is reached from a 370 K feed by adding heat: Q = hout − hF = 9667 − 2240 = 7427 J per mol of feed. Specifying T (Lab 01), Q (here) or β (Lab 06) are interchangeable ways to close the problem.

4. Investigate what the energy balance decides

Master's core work

Calculate a state in Step 1 before running an investigation.

A. Find where the flash begins

  1. Run the sweep of feed temperatures from 330 to 390 K at the current pressure and feed.
  2. Identify the feed temperature below which no vapor forms, and compare it with the bubble temperature of the feed.
  3. Explain why β rises almost linearly with TF while the outlet temperature barely moves.

B. Compare liquid models

  1. Run the same feed and pressure with an ideal liquid and with Margules A = −1 and +1.
  2. Compare the bubble temperature, the outlet temperature and β.
  3. Explain the direction of each change from the activity coefficients, and why hE = 0 here.

C. Replace Q = 0 with a heater duty

  1. Choose a target vapor amount and calculate the outlet temperature and the heat duty Q needed from the current feed.
  2. Check that β = 0.30 reproduces the Lab 01 state at 90 kPa.
  3. Explain why specifying T, Q or β each closes the same problem.
mol

D. Defend a limited conclusion

The record check only confirms that the sections and calculations are present. Scientific reasoning still needs instructor review.

5. Compare and save

  1. Calculate and save the four named cases.
  2. Compare the outlet temperature and β across feed temperatures and pressures.
  3. Answer questions A–C in Step 4 using the saved comparison.
  4. Download the complete study, reopen it, and reproduce a saved case.
  1. Calculate and save the four named cases.
  2. Complete investigations A–C and write the graduate claims.
  3. Download the complete study, reopen it, and reproduce a saved case and its investigation record.

Your calculated cases will appear here after you add them.

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Your inputs, comparisons and explanations are kept in this open page. Download one study file to keep the complete record and reopen it later. Results are recalculated on restore.