In an adiabatic nozzle the enthalpy drop appears as
Answer: C
A nozzle has no moving parts, so no shaft work is possible and the energy released by falling enthalpy can only accelerate the fluid. The pressure falls rather than rises, which is what makes the acceleration possible.
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A gas leaves a nozzle with an enthalpy drop of 20 kilojoules per kilogram and negligible inlet velocity. What exit velocity does that correspond to?
Answer: D
Setting the enthalpy drop equal to one half of the velocity squared gives a velocity of the square root of forty thousand, which is two hundred metres per second. Forgetting to convert kilojoules to joules is the usual source of an answer wrong by a factor of about thirty.
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Assertion (A): A diffuser is analysed with the same equation as a nozzle.
Reason (R): Both exchange enthalpy for kinetic energy with no shaft work and no heat transfer.
Answer: A
Both are true and the reason gives the reason they pair: the diffuser simply runs the exchange in the other direction, converting velocity back into pressure and enthalpy as the passage widens. The same reduced equation covers both cases.
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All of the following terms are normally dropped when analysing a large industrial heat exchanger EXCEPT
Answer: B
The enthalpy change is the whole point of the device, so it is the one term that must be kept. The other three are genuinely negligible in a shell and tube unit operating at ordinary velocities. Fluid velocities there are a few metres per second at most.
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For a steady flow compressor with heat loss, the work required is
Answer: C
Work put in must cover both the rise in enthalpy of the gas and the energy that escapes as heat, so cooling a compressor does not reduce the work unless the compression path itself is changed. Sign conventions cause most of the errors here.
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State whether the following is true or false.
Intercooling between compressor stages reduces the total work required.
Answer: D
Cooling the gas between stages lowers its specific volume before the next compression, and the work for a given pressure ratio scales with that volume. The saving is real and is why multistage machines are built with coolers.
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Arrange the steps of applying the steady flow energy equation to a device in the usual order.
1. delete the terms that are negligible
2. draw the control volume around the device
3. solve for the unknown work or heat
4. identify the streams crossing the boundary
Answer: A
The control volume must be fixed before anything can be said to cross it, and terms cannot be assessed as negligible until the streams are known. Solving comes last because the equation is only usable once it has been reduced.
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The energy balance term that is most often forgotten in a flow problem is
Answer: B
Flow work is invisible in a closed system and appears only when mass crosses a boundary, so it is the term newcomers omit. Using enthalpy instead of internal energy builds it in automatically. That is precisely why enthalpy was defined in the first place.
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Consider the following statements about the general energy balance for an open system.
1. It reduces to the closed system first law when no mass crosses the boundary.
2. It contains an accumulation term that vanishes at steady state.
3. It applies only to ideal gases.
Answer: C
Correct answer: 1 and 2 only
Statements 1 and 2 are correct: with no mass crossing the boundary the flow terms vanish and the balance becomes the closed system first law, and the accumulation term is exactly what distinguishes unsteady from steady operation. Statement 3 is wrong - the balance is a statement of energy conservation and holds for any substance in any phase. It is the *property data* that depend on the substance, not the balance.
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Water is pumped from 100 to 3000 kilopascals. Taking its specific volume as 0.001 cubic metres per kilogram, what is the ideal pump work?
Answer: D
For an incompressible liquid the reversible work is the specific volume times the pressure rise, which is two thousand nine hundred kilopascals times one thousandth of a cubic metre. That is why pump work is small beside turbine work in a steam cycle.
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