9-Entropy calculation and ideal work [相容模å&f

9-Entropy calculation and ideal work [相容模å&f

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: BioE BioE Entropy calculation and ideal work ideal work Entropy calculation for ideal gases in a simple closed system Ideal work done by simple closed systems work done by simple closed systems Helmholtz free energy Entropy balance over a flow system balance over flow system Ideal work done by a flow system Gibbs free energy free energy Molecular BioEngineering Laboratory 1. Calculation of ΔS for ideal gases under an Calculation of for ideal gases under an isothermal process dU = TdS − PdV , dH = TdS + VdP dS = ( dU + PdV ) / T = ( dH − VdP ) / T dS = nCV , m dT / T + nRdV / V = nC P , m dT / T − nRdP / P 1. Isothermal process, dS = nRdV / V = − nRdP / P Isothermal Δ S = nR ln V2 P = − nR ln 2 V1 P1 BioE 2. Calculation of 2. Calculation of ΔS for ideal gases under a constant for ideal gases under constant volume process dU = TdS − PdV , dH = TdS + VdP dS = (dU + PdV ) / T = (dH − VdP) / T dS = nCV ,m dT / T + nRdV / V = nCP ,m dT / T − nRdP / P 2. Constant volume process, dS = nCV ,m dT / T T2 ΔS = nCV ,m ln , if CV ,m is a constant. if T1 BioE 3. Calculation of ΔS for ideal gases under an isobar Calculation of for ideal gases under an isobar process dU = TdS − PdV dS = ( dU + PdV ) / T = ( dH − VdP ) / T dS = nC P , m dT / T + nRdP / P 3. Constant pressure process, dS = nC P , m dT / T process ΔS = nC P , m ln T2 , if C P , m is a constant. if is T1 BioE Th thi The third law of thermodynamics th Tf ( S (T ) − S (0) = ∫ C Ps ) dT + 0 ΔH f Tf Tb ΔH vap Tf Tb ( + ∫ C Pl ) dT + T ( + ∫ C Pg ) dT Tb CP/T S ( 0) = 0 BioE 4. Calculation of ΔS for ideal gases under an Calculation of for ideal gases under an adiabatic process dU = TdS − PdV dS = (dU + PdV ) / T = (dH − VdP) / T dS = nCP,m dT / T − nRdP / P dS = nCV ,m dT / T + nRdV / V 4 - 1. Adiabatic reversible, dS = 0 Adiabatic 4 - 2. Adiabatic irreversible, dS = nCP ,m dT / T V2 T2 ΔS = nCV ,m ln + nR ln , if CV ,m is a constant. V1 T1 ΔS = nCP ,m ln T2 P − nR ln 2 , if CP,m is a constant. T1 P 1 BioE A system is divided into two chambers by a movable diathermic piston and is placed in a constant temperature water bath at T1. i l Chamber A contains gas A at a pressure of PA1 and chamber B contains gas B at a pressure of PB1. Initially chambers A and B ll are kept at equal size by a mechanical lock and the initial temperature of the system is T1. Estimate the changes after the mechanical lock is released. 1. Gas A is an ideal gas. PB1 = 0 The system is completely insulated. system is completely insulated. 2. Gas A is a van der Waals gas. PB1 = 0 The wall of the system is diathermic. wall of the system is diathermic B A BioE Two gas chambers are connected as the following diagram. Chamber A is filled with ideal gas A and chamber is filled with ideal gas B. The initial temperatures and pressures in both chambers are the same, T0 and P0. Please calculate the entropy change after the valve is opened. Assume the outside wall of the system is (1) adiabatic, (2) diathermal. A B BioE Maximum obtainable work 1 海面溫度為TH = 330K, 深 海洋流溫度為TL=280K, 請問利用此洋流做功, 每 小時最多可做多少功? TH E W TL BioE Maximum obtainable work 2 Assuming that there is no heat released to the environment, calculate released to the environment, calculate the maximum work can be done by a 1000g metal blocks. The initial temperature of Metal A is 800K and the surrounding temperature is 300K. The constant volume heat capacity of metal A is 3 J K-1 g-1. A W E Surrounding BioE dS ≥ δQ Tsurr δQ ≤ Tsurr dS dU = δQ + δW − δW = δQ − dU ≤ Tsurr dS − dU − Wmax = Tsurr ΔS − ΔU = −(ΔU − Tsurr ΔS ) ΔS = ∫ Tsurr TA 0 ΔU = ∫ dS = ∫ Tsurr TA 0 Tsurr TA 0 dU = ∫ CV dT Tsurr = CV ln TA 0 T Tsurr TA 0 CV dT = CV (Tsurr − TA0 ) − Wmax = +CV (TA0 − Tsurr ) − CV Tsurr ln TA 0 Tsurr BioE Helmholtz free energy, A A ≡ U − TS at constant temperature T dA = dU − TdS − δWmax = TdS − dU = −dA BioE Entropy balance for an open system d & & & (mS )sys + dS surr = (mS )in − (mS )out + SG dt dt d & & (mS )sys + dS surr + Δ(mS ) = SG ≥ 0 dt dt & d Q & & (mS )sys − + Δ(mS ) = SG ≥ 0 Tsurr dt BioE Entropy balance for a steady state flow py system & d Q & & (mS )sys − + Δ (mS ) = S G ≥ 0 dt Tsurr d (mS )sys = 0 dt & Q & & + Δ (mS ) = S G ≥ 0 ∴− Tsurr & Q & Δ (mS ) ≥ Tsurr Q BioE Id Ideal work & Q & Δ (mS ) ≥ Tsurr if Tsurr & Q & ≈ Tsys , Δ (mS ) = Tsys 1 ⎡ ⎤&& & Δ ⎢ m ( H + u 2 + gz ) ⎥ = Q + W s 2 ⎣ ⎦ 1 ⎡ ⎤ & & & & Ws ( rev ) = Wideal = Δ ⎢ m ( H + u 2 + gz ) ⎥ − Tsys Δ (mS ) 2 ⎣ ⎦ & & & & W = Δ ( mH ) − T Δ (mS ) = m (ΔH − T ΔS ) ideal sys Wideal = ΔH − Tsys ΔS sys BioE Gibbs free energy G ≡ H − TS at constant temperature T dG = dH − TdS − δWideal = TdS − dH = −dG dG BioE Maximum obtainable work 3 Assuming that there is no heat released to the environment, calculate released to the environment, calculate the maximum work can be done by two 1000g metal blocks and the final temperature of these two blocks. Metal A is at 800K and metal B is at 200K. The constant volume heat capacity of metal A is 3 J K-1 g-1 and that of metal B is 5 J K-1 g-1. A E W B BioE Maximum obtainable work 4 E W TA0 PA0 A0 T0, P0 Calculate the maximum work the maximum work can be done by a N2 gas cylinder. The initial cylinder. The initial temperature and pressure of the N2 gas are 400 K and 10 bar. The work is done by connecting an ideal engine to the cylinder and exhaust the N2 gas into the outside environment at 1 bar and 300K. BioE HW#10 HW#10-1 1. Estimate the maximum work that can be done by the big bird. The bird has m grams ether in its body. Its head is always wet by keeping on dipping in the water The atmospheric water. The atmospheric temperature is Ta and the wet bulb temperature is Tw. The heat of evaporation per gram of ether is ΔHvap. BioE Answer#10-1 m ΔHvap x (1- Tw/ Ta) BioE HW#10 HW#10-2 2. Without contacting with environment please calculate the maximum work that can be obtained from the three metal blocks A, B, and C. The initial temperatures of metal blocks, A, B, and C are TA0, TB0, and TC0, respectively. The masses of , blocks A, B, and C are mA, mB, and mC, respectively. The specific heat capacities of the three blocks are CPgA, CPgB, and CPgC, respectively. A C B BioE Answer #10-2 #10 Tf A CP, g TA 0 T ΔS = ∫ m A Tf B CP, g TB 0 T dT + ∫ mB Tf C CP, g TC 0 T dT + ∫ mC dT Solve for Tf − W ≤ TΔS − ΔU = −ΔU Tf = ∫ m AC TA 0 A P, g Tf dT + ∫ mB C TB 0 B P,g Tf C dT + ∫ mC C P , g dT TC 0 BioE ...
View Full Document

This note was uploaded on 11/11/2010 for the course CHE CH2005 taught by Professor 曹恒光 during the Spring '10 term at National Central University.

Ask a homework question - tutors are online