First Law Applied to Flow Processes Questions and Answers

Answer: A

Both are true and the reason gives the magnitude: work for a given pressure change scales with specific volume, so the pump handling liquid needs a tiny fraction of what the turbine delivers. Neglecting it costs about one per cent in cycle efficiency.

Answer: B

The flow work done pushing gas into the tank ends up as internal energy inside it, so the internal energy of the contents equals the enthalpy of the line gas. For an ideal gas that raises the temperature by the ratio of the specific heats.

Answer: C

Filling is by definition unsteady, since the mass inside is changing, so the accumulation term must be retained. Dropping it is what leads students to predict no temperature rise at all. The tank contents are not a steady stream but a growing mass.

Answer: D

Steady state means no accumulation of mass, which fixes the mass flow rates but says nothing about volume, since density changes freely with pressure and temperature. Confusing the two is a common source of error in compressible flow.

Answer: A

Correct answer: a centrifugal compressor

A compressor has a rotating shaft and absorbs work - it is the one device here that does. A throttling valve has no moving part at all, a nozzle accelerates the fluid by shaping the passage rather than by driving anything, and a heat exchanger transfers heat through a wall. Identifying which terms are zero is the whole method of applying the steady flow energy equation.

Answer: B

Only for an ideal gas does the pressure dependence vanish, and the residual enthalpy measures exactly that departure. The Joule-Thomson effect exists because the pressure dependence is not zero. Residual enthalpy is the quantity that measures it.

Answer: C

Both are true and the reason gives the construction: the ideal gas part is easy from heat capacity data, so isolating the departure lets an equation of state supply only the correction needed. Heat capacity data and an equation of state together suffice.

Answer: D

Isentropic efficiency for a turbine is actual work divided by ideal work, so the actual output is eighty-five per cent of five hundred. For a compressor the ratio is inverted, and mixing the two up is the standard trap.

Answer: A

The first two are correct and reflect that the ideal is the best case in each direction. The third is impossible, because the isentropic path is the limit that no real adiabatic machine can beat. Values above ninety per cent are considered very good.

Answer: B

These are the devices designed specifically to change velocity, so the term they exist to alter cannot be dropped. In the others the velocities are low enough that the kinetic term is a rounding error on the enthalpy change.