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ADVANCED THERMODYNAMICS ENGINEERING phần 6

Ghi chú Đối với khí lý tưởng, v = RT / P để T (∂ v / ∂ T) = v, và do đó, μJT = 0. Không có nhiệt độ thay đổi do điều tiết đối với khí lý tưởng. Đối với chất lỏng không nén được ∂ v / ∂ T = βP = 0, và μJT = -v/cP có một giá trị âm. Do đó, chất lỏng nóng lên khi can thiệp. Nếu 0 v, và ngược lại. Tại các điểm đảo ngược, μJT = 0. Inversion xảy ra khi Tinv βP = | jj. Example 36 The experimentally determined value of u for N2 is 0.31 K bar-1 at-70 C and 5 MPa. If the gas is throttled from 6 MPa and -67 C to 4 MPa what is the final exit temperature Solution p.JT Ah cp A P A T A P Te -203 40 - 60 0.31 K bar-1. Therefore Te -73.2 C. a. Evaluation of U T Recall that dh CpdT v - T dv dT p dP. Since dh 0 during throttling we obtain the relation J dT dP h - v - T dv dT p cp T2 cp d v T dT or 156 Mil idl dl - 1 - Tpp cp v . 157 b. Remarks For ideal gases v RT P so that T dv dT v and hence U.IT 0. There is no temperature change due to throttling for ideal gases. For incompressible fluids dv dT pP 0 and U.IT -v cP has a negative value. Therefore liquids generally heat up upon throttling. If v T dv dT i or T pP 1 UIT 0 and vice versa. At the inversion point UIT 0. Inversion occurs when Tinv pP T v dv dT P 1. For a real gas Pv ZRT i.e. dv dT ZR P RT P dZ dT. If T v dv dT p 1 this implies that ZRT Pv RT2 Pv dZ dT 1 or dZ dT P 0. For cooling to occur the inlet pressure must be lower than the inversion pressure. Cooling of a gas can also be accomplished using isentropic expansion. We define us dT dP s that is related to the temperature decrease due to the work delivered during an isentropic process. Recall that ds cp dT T - dv dT p dP. For an isentropic process ds 0 and ls T dv dT p cp TvPp cp. 158 Dividing Eq. 157 by Eq. 158 qs 1 - 1 Tpp 159 Equation 159 provides a relation to indicate relative degree of heating of the isentropic to isenthalpic throttling process. For substances that expand upon heating pP 0 and hence us 0. If pPT 1 then UIT qs i.e. the isentropic expansion results in greater cooling than isenthalpic expansion for the same pressure ratio. The values of Tinv pP pT and Psat can be directly obtained from the state equations while u IT h u cv and other such properties depend both on the equations of state and Pi Pa Pb Pe Inlet Exit Thermometers a Rigid Porous Foam--------- Porous Plug P Figure 20 a Illustration of a .