The first law for a closed system undergoing a process states that
Answer: A
Energy entering as heat either raises the internal energy or leaves as work, which is the content of the law for a fixed mass. The enthalpy statement holds only at constant pressure and the entropy statement belongs to the second law.
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State whether the following is true or false.
The first law by itself tells you which direction a process will go.
Answer: B
It says only that energy is conserved, and a process and its exact reverse both satisfy it equally well. Direction comes from the second law, which is why the two are always used together. Conservation alone permits heat to flow from cold to hot.
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Enthalpy rather than internal energy is the natural variable for flow processes because
Answer: C
Pushing a unit of fluid into the control volume requires work equal to pressure times volume, and grouping that term with internal energy defines enthalpy. The steady flow equation is therefore written in enthalpy rather than in internal energy.
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Assertion (A): The steady-state flow energy balance is written per unit mass flowing.
Reason (R): In steady state the mass in the control volume does not change, so every term can be referred to the flowing stream.
Answer: D
Correct answer: Both A and R are true, and R is the correct explanation of A
Both are true and the reason gives the justification. At steady state nothing accumulates in the control volume, so the energy entering with the stream plus the heat and work crossing the boundary must equal the energy leaving with the stream. Dividing throughout by the mass flow gives a balance per kilogram, which is the form in which turbine work, nozzle velocity and exchanger duty are all quoted.
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Match each device with the simplification usually made to the steady flow energy equation.
List I
A. Adiabatic turbine
B. Throttling valve
C. Adiabatic nozzle
D. Heat exchanger
List II
1. kinetic energy gain equals the enthalpy drop
2. shaft work equals the drop in enthalpy
3. heat transferred equals the enthalpy change
4. enthalpy is unchanged across the device
Answer: A
Correct answer: A-2, B-4, C-1, D-3
Each device is characterised by which terms of the steady flow energy equation survive. An adiabatic turbine has no heat and negligible velocity change, so the shaft work equals the enthalpy drop. A throttling valve has neither heat nor work, so the enthalpy is unchanged. An adiabatic nozzle has no work, so the enthalpy drop appears as kinetic energy. And a heat exchanger has no work, so the heat transferred equals the enthalpy change.
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Across an ideal throttling valve the property that stays constant is
Answer: B
Correct answer: enthalpy
Throttling is adiabatic, involves no shaft work and produces negligible change in velocity or elevation, so the steady flow energy equation collapses to \( h_1 = h_2 \). Note what is *not* constant: the entropy rises, since the process is thoroughly irreversible, and the temperature generally changes - the Joule–Thomson effect, which is the basis of the Linde liquefaction process.
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State whether the following is true or false.
Throttling an ideal gas leaves its temperature unchanged.
Answer: C
The enthalpy of an ideal gas depends on temperature alone, so constant enthalpy means constant temperature and the Joule-Thomson coefficient is zero. Real gases cool or warm depending on which side of the inversion curve they lie.
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The Joule-Thomson coefficient is defined as
Answer: D
It measures how much a fluid cools per unit fall in pressure when it is throttled, which is exactly a constant enthalpy path. Its sign decides whether throttling can be used for liquefaction. Its value is zero everywhere for an ideal gas.
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Consider the following statements about the Joule-Thomson inversion temperature.
1. Below it, throttling cools the gas.
2. Above it, throttling warms the gas.
3. Hydrogen and helium have inversion temperatures above room temperature.
Answer: A
The first two are correct. The third is wrong and the error matters: both gases have inversion temperatures far below ambient, so they must be precooled before throttling can liquefy them. Liquid nitrogen is the usual precooling medium.
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Steam enters an adiabatic turbine at 3200 kilojoules per kilogram and leaves at 2400. Neglecting kinetic and potential terms, what is the work output per kilogram?
Answer: B
Correct answer: 800 kilojoules
With no heat transfer and the kinetic and potential terms neglected, the shaft work is the enthalpy drop: \[ w = h_1 - h_2 = 3200 - 2400 = 800\ \mathrm{kJ/kg}. \] Adding the two enthalpies gives the 5600 offered, which corresponds to nothing physical, and the sign convention is worth stating: a positive result means work delivered by the fluid.
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