Technology & Engineering
TdS Equation
The TdS equation is a fundamental concept in thermodynamics, representing the change in entropy of a system due to heat transfer. It is expressed as TdS = δQ, where T is the temperature, dS is the change in entropy, and δQ is the heat transfer. This equation is used to analyze and quantify the relationship between heat and entropy in various engineering and scientific applications.
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4 Key excerpts on "TdS Equation"
- Allan D. Kraus, James R. Welty, Abdul Aziz(Authors)
- 2011(Publication Date)
- CRC Press(Publisher)
The Gibbs property relations, or TdS Equations, are TdS = dU + PdV (7.14) Tds = du + Pdv (7.15) TdS = dH − VdP (7.18) and Tds = dh − vdP (7.19) For solids and liquids, the specific energy change is given by s 2 − s 1 = ¯ c ln T 2 T 1 (7.21) where ¯ c is the average specific heat. For ideal gases s 2 − s 1 = ¯ c v ln T 2 T 1 + R ln v 2 v 1 (7.23) Entropy 211 and s 2 − s 1 = ¯ c p ln T 2 T 1 − R ln P 2 P 1 (7.24) For an isentropic process of an ideal gas, the P -V -T relations are T 2 T 1 = V 1 V 2 k − 1 T 2 T 1 = P 2 P 1 ( k − 1) / k P 2 P 1 = V 1 V 2 k and the work will be W = V 2 V 1 PdV = P 2 V 2 − P 1 V 1 1 − k = mR ( T 2 − T 1 ) 1 − k (7.31) Isentropic efficiencies of steady flow devices include the turbine s = h 1 − h 2 h 1 − h 2 s (7.32) the compressor or pump s = h 2 s − h 1 h 2 − h 1 (7.33) and the nozzle s = ˆ V 2 act ˆ V 2 S = C (7.39) The entropy balance equation for the closed system is S sys = Q T + S gen (7.40) and for the open system or control volume i ˙ m i s i + j ˙ Q j T j + ˙ S gen = e ˙ m e s e (7.45) 7.11 Problems The Clausius Inequality 7.1: The following heat engines operate between a hot reservoir at 1000 K and a cold reservoir at 300 K. Verify the validity of the Clausius inequality in (a) Q H = 500 kJ and Q L = 200 kJ, (b) Q H = 600 kJ and Q L = 180 kJ, and (c) Q H = 900 kJ and Q L = 250 kJ. 7.2: A heat engine draws heat from a flame at 800 K at the rate of 20 W. The engine rejects heat to an ocean at 290 K and the engine efficiency is 32%. Determine the cyclic integral ˙ Q / T and comment on whether the engine satisfies the Clausius inequality. 7.3: A hot reservoir at 1500 K supplies 120 kJ to heat engine 1, which has a thermal efficiency of 28%. The heat rejected by engine 1 is supplied to engine 2, which operates with a thermal efficiency of 35%. Heat engine 2 rejects heat to a sink at 400 K and another 212 Introduction to Thermal and Fluid Engineering 12 kJ to a sink at 300 K.
eBook - PDF- R. Prasad(Author)
- 2016(Publication Date)
- Cambridge University Press(Publisher)
TdS Equations and their Applications 6.0 Introduction The first and the second laws of thermodynamics may be written, respectively, in the following differential forms; d ¢ Q = dU + d ¢ W 6.1 and d ¢ Q r = T dS 6.2 In Eq. 6.1, that represents the first law of thermodynamics, d ¢ Q represents the path dependent heat flow in the system, dU the increase in the internal energy, and d ¢ W the path dependent work done by the system. Equation 6.2, which is the mathematical representation of the second law, is applicable only for reversible processes and tells that (for reversible processes) heat flow to the system at constant temperature d ¢ Q r , which is path dependent, is equal to the multiplication of the constant temperature T and the change in the entropy of the system dS. Substituting the value of d ¢ Q from Eq. 6.2 in Eq. 6.1, one gets for reversible processes, T dS = dU + d¢W 6.3 However, for reversible processes d ¢ W = P dV and so Eq. 6.3 reduces to T dS = dU + P dV 6.4 Equation 6.4 that contains both the first and the second laws is in principle applicable for reversible processes. However, a closer look to this equation reveals that no quantity that depends on the process or the path is involved in the equation. The equation relates the small variations of state parameters, S, U and V . As a matter of fact Eq. 6.4 gives a relation between the state functions of two nearby equilibrium states of the system. Since state functions do not depend on path, Eq. 6.4 is applicable for any two nearby equilibrium states of a system, irrespective of whether the other state is reached from the first state by an irreversible or a reversible process. Sometimes it may also happen that the final state with entropy ( S + dS), internal energy ( U+ dU) and volume (V + dV) may not be reached from the initial state of entropy S, internal energy U and volume V , by any process and vice-versa.
eBook - PDFRadial Flow Turbocompressors
Design, Analysis, and Applications
- Michael Casey, Chris Robinson(Authors)
- 2021(Publication Date)
- Cambridge University Press(Publisher)
It is a measure of the increase in entropy due to dissipative processes and, as it occurs in many equations, following Traupel (2000) and Baehr and Kabelac (2012), it is given its own symbol, j: dj ¼ Tds irr 0: (2.33) For an adiabatic process, dq ¼ 0, the change in entropy is always greater than zero. A common misconception of the second law is that it states that entropy always increases, but this is not the case for a system with heat extraction. The entropy of the flow falls across a turbine blade row with intensive cooling, but this reduction in entropy in the turbine flow is associated with an increase in entropy of the surround- ings of the turbine blade. For the turbine and the surroundings taken together, the entropy always increases. When applied to an adiabatic control volume in a turbo- machine, the equation states that the entropy between inlet and outlet increases and only for a fully reversible ideal adiabatic process does it remain constant. Another case where entropy would remain constant is a diabatic process with heat transfer in which the heat extraction exactly matches and cancels the effect of dissipation on the entropy change, and this becomes important when considering polytropic processes in Section 2.7.4. The product Tds irr is the sum of all the dissipation losses related to internal frictional dissipation in the flow and is called specific dissipation here. Alternative terms are the frictional work, the dissipation work, lost work or simply the dissipation. It represents the lost opportunity to obtain work during a frictional dissipative process. As it appears together in an equation with the term dq, it is also sometimes confusingly 67 2.5 The Second Law of Thermodynamics
eBook - PDF- Kalyan Annamalai, Ishwar K. Puri, Milind A. Jog(Authors)
- 2011(Publication Date)
- CRC Press(Publisher)
dS/dt Q/T b m i s i σ m e s e ˙ ˙ ˙ ˙ Figure 3.28 Illustration of the entropy band diagram. Second Law of Thermodynamics and Entropy ◾ 171 For a rigid closed system, Equation 3.90 reduces to Equation 3.31. If the process is reversible, ( σ ˙ cv = 0) Equation 3.87 becomes dS c.v. /dt = Σ m ˙ i s i – Σ m ˙ e s e + Σ j Q ˙ j /T b,j . (3.91) If it becomes difficult to evaluate Σ j Q ˙ j /T b,j , the system boundary may be drawn so that T b,j = T 0 , ambient temperature that is, all irreversibilities are contained inside the selected control volume. For instance, if in Figure 3.27 the dashed boundary is selected just outside the control volume, then T b,j = T 0 . At steady state Equation 3.87 becomes Σ m ˙ i s i – Σ m ˙ e s e + Σ j Q ˙ j /T b,j + σ ˙ cv = 0. (3.92) For single inlet and exit at steady state, m ˙ i = m ˙ e = m ˙ s i – s e + Σ q j /T bj + σ m = 0 or s e – s i = Σ q j /T bj + σ m, (3.93) where σ m = dotnosp σ cv / dotnosp m , entropy generation per unit mass. Equation 3.93 is similar to entropy balance equation for a closed system except that one follows an unit mass in an open system. If process is reversible σ m = 0, T bj = T and hence s i – s e + Σ q j /T = 0 or ds = Σ ( δ q j ) rev /T, which is similar to the relation for a closed system. In case of a single inlet and exit, but for a substance containing multiple components, the relevant form of Equation 3.86 is dS /dt N s N s Q /T c.v. k k,i k,i k k,e k,e j j b, = ∑ -∑ + ∑ dotnosp dotnosp dotnosp j cv , + dotnosp σ (3.94) where s k denotes the entropy of the k-th component in the mixture. For ideal gas mixtures s T P s T PX P k k k ( , ) ( ) ln( ) = -0 0 / . Example 13 Water enters a boiler at 60 bar as a saturated liquid (state 1). The boiler is supplied with heat from a nuclear reactor maintained at 2000 K while the boiler interior walls are at 1200 K. The reactor transfers about 4526 kJ of heat to each kg of water.
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