Modern Engineering Thermodynamics

392 CHAPTER 11: More Thermodynamic Relations
To find the change in specific enthalpy between states 1 and 2 using this figure, values of p R and T R are calculated
and values of h* − h /T c are then read from the figure. The change in specific enthalpy between states 1 and 2 is
then determined from
h 2 − h 1 = ðh 2 * − h 1 *Þ −
h* − h
T c
−
2
h* − h
T c
1
Tc
M
(11.41)
where h 2 * − h 1 * is determined from the gas tables (Table C.16 in Thermodynamic Tables to accompany Modern Engineering
Thermodynamics) orbyassumingconstantspecificheats over the temperature range from T 1 to T 2 and
using h 2 * − h 1 * = c p ðT 2 − T 1 Þ:
EXAMPLE 11.15
As chief engineer at Precision Throttles Inc., you are responsible for designing an adiabatic, aergonic, steady state, steady flow
device to throttle carbon dioxide (CO 2 ) from 20.0 MPa, 150.°C to 0.101 MPa. As part of this design, you must accurately
predict the exit temperature of the throttle. Neglecting all kinetic and potential energies, use the generalized charts under the
assumption of constant specific heats to determine the exit temperature of the throttle.
Solution
First, draw a sketch of the system (Figure 11.10).
CO 2
p 1 =20.0 MPa
p 2 = 0.101 MPa
T 1 = 150.°C
T 2 = ?
FIGURE 11.10
Example 11.15.
The unknown is the exit temperature of the throttle. The energy rate balance for an adiabatic, aergonic, steady state, steady
flow, single-inlet, single-outlet throttle is
_Q − _W + _m ðh 1 − h 2 Þ = 0
and since _Q = _W = 0 here, we get h 2 – h 1 = 0. From Table C.12b, we find the critical temperature and pressure for CO 2 are
T c = 304:2 K and p c = 7:39 MPa
and that the molecular mass of CO 2 is 44.01 kg/kgmole. From Table C.13b, we now find the value of the constant pressure
specific heat of CO 2 as
Then,
and
c p = 0:845 kJ/kg .K
p R1 = 20/7:39 = 2:71
T R1 = ð150: + 273:15 KÞ/ð304:2KÞ = ð423:15 KÞ/ð304:2KÞ = 1:39:
Then, Figure 11.9 gives [(h* – h)/T c ] 1 ≈ 14.0 kJ/kgmole · K. At the exit of the throttle, we have
p R2 = ð0:101 MPaÞ/7:39 ð MPaÞ = 0:0135
and Figure 11.9 gives [(h* – h)/T c ] 2 ≈ 0. Then, using Eq. (11.41), we have
h 2 − h 1 = 0
= 0:845 kJ
kJ
ðT 2 − 423:15 KÞ − 0 − 14:0
304:2K
kg .K
kgmole.K 44:01 kg/kgmole