Modern Engineering Thermodynamics

6.7 Throttling Calorimeter 183
The unknowns are the quality of the wet steam in the pipe and its Joule-Thomson coefficient. The system is open and the
material is wet steam.
A throttling calorimeter is clearly a steady state, steady flow, single-inlet, single-outlet open system. The material flowing is
steam, and the unknown is the quality of the inlet flow stream ðx in = x 1 = ?Þ and the steam’s Joule-Thomson coefficient μ J .
The system and its properties follow:
Station 1 ðinletÞ
! Throttling ! Station 2 ðoutletÞ
process
p 1 = p sat = 200: MPa
x 1 = ?
p 2 = 0:100 MPa
T 2 = 150:°C
h 2 = 2776:4 kJ/kg
This is an example of a problem where the process path plus three of the four state properties needed to fix the two states
are given, and the unknown is a property in the undetermined state. This is a common problem format.
A quick check of the steam tables shows that the outlet state is in the superheated region, and therefore all the outlet
properties are easily found from the superheated steam table. Since we are given no information on mass flow rates
or velocities, we assume that the changes in flow stream specific kinetic and potential energies are negligible. The
calorimeter was stated to be insulated, so it will have no heat transfer, and we acknowledge that this device can
neither do work nor have work done on it. Under these conditions, the modified energy rate balance reduces to
Eq. (6.24), or
h in = h 1 = h out = h 2
From the superheated steam table (Table C.3b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics),
we find that
h 2 = hð0:100 MPa, 150:°CÞ = 2776:4kJ=kg
and this value is listed previously as part of the data for station 2. Therefore, from the modified energy rate balance, we
have
h 1 = hð2:00 MPa, ?Þ = 2776:4kJ=kg
and the pair of independent properties p 1 =2.00MPaandh 1 = 2776.4 kJ/kg now fix the inlet state. From the saturation
tables for water (Table C.2b), we find that, at 2.00 MPa,
h f1 = h f ð2:00 MPaÞ = 908:8 kJ/kg
h fg1 = h fg ð2:00 MPaÞ =1890:7 kJ/kg
h g1 = h g ð2:00 MPaÞ = 2799:5 kJ/kg
Since h f 1 < h 1 < h g1 , we can now use the auxiliary formula for quality x to determine its value at station 1 as
So x 1 = 98.8%, which is the quality of the steam in the pipe.
x 1 = ðh 1 − h f1 Þ/h fg1 = ð2776:4 − 908:8Þ/1890:7 = 0:9878
A rough estimate for the Joule-Thomson coefficient for this process is given by Eq. (6.26) as
μ J ≈ ðΔT/ΔpÞ throttling
process
where, from the saturation tables, T 1 = T sat ð2:00 MPaÞ = 212:4 o C:
μ J ≈ ð212:4 − 150:°CÞ/ð2:00 − 0:100 MPaÞ = 32:8°C/MPa
Note that this is not a particularly accurate value, since μ J is a point function and consequently the values of ΔT and Δp used
in its calculation should really be much smaller than those used in the preceding calculation. This calculation does, however,
provide a reasonable average value of μ J for this throttling process.
Exercises
7. What would be the quality of the steam in the pipe in Example 6.3 if the pressure in the pipe were 3.00 MPa instead of
2.00 MPa and everything else remained the same? Answer: x = 0.985 = 98.5%.
8. What would be the quality of the steam in the pipe in Example 6.3 if the temperature in the calorimeter was 100.°C
instead of 150.°C and everything else remained the same? Answer: x = 0.935 = 93.5%.