Ammonia enters a valve as saturated liquid at 9 bar and undergoes a throttling p

Question

Ammonia enters a valve as saturated liquid at 9 bar and undergoes a throttling process to a pressure of 2 bar. Determine the rate of entropy production per unit mass of ammonia flowing, in kW/(kg·K). If the valve were replaced by a power-recovery turbine operating at steady state, determine the maximum theoretical power that could be developed per unit mass of ammonia flowing, in kJ/kg, and comment. In each case, ignore heat transfer with the surroundings and changes in kinetic and potential energy.

Explanation / Answer

The rate of entropy production per unit mass of ammonia:

ds = s2 - s1

Using Ammonia thermo properties table,
s1 = sf @ 9 bar gage = 6.0502 kJ/kg-K

s2 = sf2 + x2 (sg2 - sf2)
sf2 = sf @ 2 bar gage = 5.3708 kJ/kg-K
sg2 = sg @ 2 bar gage = 10.5826 kJ/kg-K

x2 = h2 - hf2 / hfg2 = h1 - hf2 / hfg2
note thant h1 = h2 in a throttling process
hf2 = hf @ 2 bar gage = -848.91 kJ/kg
hfg2 = hfg @ bar gage = 1325.3286 kJ/kg
h1 = h2 = hf @ 9 bar gage = -661.54 kJ/kg

x2 = (-661.54 - -848.91) / 1325.3286 = 0.1414

s2 = 5.3708 + 0.1414*(10.5826 - 5.3708)
s2 = 6.1077 kJ/kg-K

ds = 6.1077 - 6.0502
ds = 0.0575 kJ/kg-K

The maximum theoretical power of a power recovery turbine following a constant entropy expansion process:

dh = h1 - h2

h2 = hf2 + x2' * hfg2

x2' = (s1 - sf2) / (sg2 - sf2)
x2' = (6.0502 - 5.3708) / (10.5826 - 5.3708)
x2' = 0.1304

h2 = -848.91 + (0.1304)*(1325.3286)
h2 = -676.14 kJ/kg

dh = -661.54 - (-676.14)
dh = 14.6 kJ/kg

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