Problem 1 The following data are available for unnamed fluid X: P = 10 bar, Tsat

Question

Problem 1 The following data are available for unnamed fluid X: P = 10 bar, Tsat-234°C saturated liq. saturated vapor enthalpy (kJ/kg) 318.19 520.6 In addition, the heat capacities of the liquid and the vapor are C%--3.3 J/kg K, and Cr-29 J/kg K, respectively, and can be assumed constant with temperature 1kg of fluid X is cooled under constant pressure P = 10 bar from 300 C until it forms a vapor-liquid mixture that contains 12% vapor by mass. The cooling takes place inside a temperature bath that is maintained at constant temperature Tbath -200°C. a) Calculate the amount of heat that is exchanged between the fluid and the bath. b) Calculate the entropy change of the fluid. c) Calculate the entropy generation. d) What are the irreversible features of this process?

Explanation / Answer

The change in entropy in the process involving phase change can be evaluated

using the specific heat and enthalpy values

?S = Sensible change due to change in temperature + latent values

Given Data: P 10 bar T in 300 °C T out 200 °C T sat 234 °C Vap. Fraction 0.12 Step a: Calculation of amount of heat exchanged between the fluid and the bath Since the fluid is cooled using a bath, the reduction in temperature would be 300°C to 200°C. There will be phase change, so we will consider the enthalpy also while calculating the final heat transfer. Heat exchanged = Sensible heat + Latent heat Therefore, Q = mCp?T + m? Q = Sensible heat of fluid cooling from 300°C to 234°C + latent heat of liquid + sensible heat of fluid from 234°C to 200°C Q = Q1 + Q2 + Q3 Q1 = mCp(300-234) Q2 = m? Q3 = mCp (234-200) Since the vapour fraction is 0.12, 0.88 fraction of fluid will condense. Q1 0.2 kJ Q2 280.0 kJ Q3 0.1 kJ Q 280.3 kJ So, the amount of heat exchanged between the fluid and bath is 280.3 kJ Step b: Calculation of change in entropy

The change in entropy in the process involving phase change can be evaluated

using the specific heat and enthalpy values

?S = Sensible change due to change in temperature + latent values

?S = mCv ln(300/234) + ?H/Tsat + mCp(avg) ln (234/200) ?S = ?S1 + ?S2 + ?S3 ?S1 0.001 kJ/kg K ?S2 1.364 kJ/kg K ?S3 0.001 kJ/kg K ?S 1.365 kJ/kg K So, the change in entropy for the system is 1.365 kJ/kg K Step c: Calculation of entropy generation Rate of entropy transfer due to heat & mass transfer + Entropy generation = change in entropy Since, the cooling is done in the bath the generation of entropy will be zero as the system is not insulated. 1.365 + Entropy generation = 1.365 Therefore, Entropy generation = 0 kJ/kg K Step d: Irreversible features of the process 1. Since the entropy being a state function depends only on the end states. 2. The entropy generation obtained is zero that means the entropy has been trasferred to the universe as the system is open.

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