EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 9780100257054
Author: CENGEL
Publisher: YUZU
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 12.6, Problem 71P
To determine

The change in enthalpy (h2h1) of water vapor using generalized chart relation.

The change in entropy (s2s1) of water vapor using generalized chart relation.

The change in enthalpy (h2h1) of water vapor using table.

The change in entropy (s2s1) of water vapor table.

Expert Solution & Answer
Check Mark

Answer to Problem 71P

The change in enthalpy (h2h1) of water vapor using generalized chart relation is 423.9kJ/kg.

The change in entropy (s2s1) of water vapor using generalized chart relation is 0.2329kJ/kgK.

The change in enthalpy (h2h1) of water vapor using table is 426.2kJ/kg.

The change in entropy (s2s1) of water vapor table is 0.2355kJ/kgK.

Explanation of Solution

Write the mean change in enthalpy (h¯2h¯1)ideal of water vapor.

(h¯2h¯1)ideal=(h¯2)ideal(h¯1)ideal (I)

Here, enthalpy of water vapor at temperature of 647.1 K is (h¯2)ideal and enthalpy of water vapor at temperature of 647.1 K  is (h¯1)ideal.

Write the change in enthalpy ((h2h1)ideal) of water vapor table.

(h2h1)ideal=(h¯2h¯1)idealMH2O (II)

Here, molar mass of water vapor is MH2O.

Write the mean change in entropy (s¯2s¯1)ideal for water vapor.

(s¯2s¯1)ideal=s¯2s¯1RlnP2P1 (III)

Here, gas constant is R, initial pressure is P1, final pressure is P2, entropy at final state is s¯2,and entropy at initial state is s¯1.

Write the change in entropy (s2s1)ideal for water vapor table.

(s2s1)ideal=(s¯2s¯1)idealMH2O (IV)

Write the reduced temperature (TR1) at initial state.

TR1=T1Tcr (V)

Here, critical temperature is Tcr and initial temperature is T1.

Write the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VI)

Here, critical pressure is Pcr and initial pressure is P1.

Write the reduced temperature (TR2) at final state.

TR2=T2Tcr (VII)

Here, critical temperature is Tcr and final temperature is T2.

Write the reduced pressure (PR2) at initial state.

PR2=P2Pcr (VIII)

Here, critical pressure is Pcr and initial pressure is P1.

Write the change in enthalpy (h2h1) of water vapor using generalized chart relation.

h2h1=RTcr(Zh1Zh2)+(h2h1)ideal (IX)

Here, change in enthalpy of water vapor is (h2h1)ideal and gas constant is R.

Write the change in enthalpy (s2s1) of water vapor using generalized chart relation.

(s2s1)=R(Zs1Zs2)+(s2s1)ideal (X)

Here, change in entropy of water vapor is (s2s1)ideal.

Conclusion:

Convert the unit of initial temperature °C to K.

T1=600°C=(600+273)K=873 K

Convert the unit of initial temperature °C to K.

T2=400°C=(400+273)K=673 K

Refer table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at temperature of 873 K and 673 K as 30754kJ/kmol and 23082kJ/kmol.

(h¯2)ideal=23082kJ/kmol(h¯1)ideal=30754kJ/kmol

Substitute 23082kJ/kmol for (h2)ideal and 30754kJ/kmol for (h1)ideal in Equation (I).

(h¯2h¯1)ideal=23082kJ/kmol30754kJ/kmol=7672kJ/kmol

Refer table A-1, “Molar mass properties table”, obtain the molar mass MH2O of water vapor as 18.015kg/kmol.

Refer the table A-20,”Water vapor properties of water vapor table”, select the entropy of water vapor at temperature of 873 K and 673 K as 217.141kJ/kmolK and 227.109kJ/kmolK.

s¯2=217.141kJ/kmolKs¯1=227.109kJ/kmolK

Substitute 217.141kJ/kmolK for s¯2, 227.109kJ/kmolK for s¯1, 8.314kJ/kgK for R, 500kPa for P2 and 1000kPa for P1 in Equation (III).

(s¯2s¯1)ideal=(217.141227.109)kJ/kmolK(8.314kJ/kgK)ln(1000kPa500kPa)=4.2052kJ/kmolK

Substitute 873 K  for T1 and 647.1 K for Tcr in Equation (V).

TR1=873K647.1K=1.349

Substitute 1000kPa for P1 and 22060kPa for Pcr in Equation (VI).

PR1=1000kPa22060kPa=0.0453

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs1 and Zh1 at reduced pressure and temperature of 1.349 and 0.0453 as 0.0157 and 0.0288.

Zs1=0.0157Zh1=0.0288

Substitute 673 K for T2 and 647.1 K for Tcr in Equation (VII).

TR2=673K647.1K=1.040

Substitute 500kPa for P2 and 22060kPa for Pcr in Equation (VIII).

PR2=500kPa22060kPa=0.0227

Refer the table A-15, “Nelson-Obert generalized compressibility chart”, select the initial state of compressibility factor Zs2 and Zh2 at reduced pressure and temperature of 1.040 and 0.0227 as 0.0146 and 0.0223.

Zs2=0.0146Zh2=0.0223

From the gas constant properties table A-1, select the gas constant of water vapor as 0.1889kJ/kgK.

Substitute 0.0223 for Zh2, 0.0288 for Zh1, 7672kJ/kg for (h2h1)ideal, 8.314kJ/kgK for R and 647.1K for Tcr in Equation (IX).

h2h1=(8.314kJ/kgK)(647.1K)(0.02230.0288)+24.32kJ/kg=7637kJ/kg

Substitute 7637kJ/kmol for (h¯2h¯1)ideal and 18.015kg/kmol for MH2O in Equation (II).

(h2h1)ideal=7637kJ/kmol18.015kg/kmol=423.9kJ/kg

Thus, the change in enthalpy (h2h1) of water vapor using generalized chart relation is 423.9kJ/kg.

Substitute 4.2052kJ/kgK for (s2s1)ideal, 0.0146 for Zs2, 0.0157 for Zs1 and 8.314kJ/kgK for R in Equation (X).

s2s1=(8.314kJ/kgK)(0.0150.0146)+4.2052kJ/kgK=4.1961kJ/kgK

Substitute 4.1961kJ/kgK for (s¯2s¯1)ideal and 18.015kg/kmol for MH2O in Equation (IV).

(s2s1)ideal=4.1961kJ/kgK18.015kg/kmol=0.2329kJ/kgK

Thus, the change in entropy (s2s1) of water vapor using generalized chart relation is 0.2329kJ/kgK.

Refer table A-6, “Superheated water table”, select the inlet enthalpy and exit enthalpy at pressure of 1000kPa and temperature of 600°C as 3698.6kJ/kg and 3272.4kJ/kg.

h1=3698.6kJ/kgh2=3272.4kJ/kg

h2h1=3272.4kJ/kg3698.6kJ/kg=426.2kJ/kg

Thus, the change in enthalpy (h2h1) of water vapor using table is 426.2kJ/kg

Refer table A-6, “Superheated water table”, select the inlet entropy and exit entropy at pressure of 500kPa and temperature of 400°C as 8.0311kJ/kgK and 7.7956kJ/kgK.

s1=8.0311kJ/kgKs2=7.7956kJ/kgK

(s2s1)=7.7956kJ/kgK8.0311kJ/kgK=0.2355kJ/kgK

Thus, the change in entropy (s2s1) of water vapor table is 0.2355kJ/kgK.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 12 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 12.6 - Prob. 11PCh. 12.6 - Verify the validity of the last Maxwell relation...Ch. 12.6 - Prob. 14PCh. 12.6 - Prob. 15PCh. 12.6 - Prob. 16PCh. 12.6 - Prob. 17PCh. 12.6 - Prove that (PT)=kk1(PT)v.Ch. 12.6 - Prob. 19PCh. 12.6 - Prob. 20PCh. 12.6 - Using the Clapeyron equation, estimate the...Ch. 12.6 - Prob. 22PCh. 12.6 - Prob. 23PCh. 12.6 - Determine the hfg of refrigerant-134a at 10F on...Ch. 12.6 - Prob. 25PCh. 12.6 - Prob. 26PCh. 12.6 - Prob. 27PCh. 12.6 - Prob. 28PCh. 12.6 - Prob. 29PCh. 12.6 - 12–30 Show that = Ch. 12.6 - Prob. 31PCh. 12.6 - Prob. 32PCh. 12.6 - Prob. 33PCh. 12.6 - Prob. 34PCh. 12.6 - Prob. 35PCh. 12.6 - Prob. 36PCh. 12.6 - Determine the change in the internal energy of...Ch. 12.6 - Prob. 38PCh. 12.6 - Determine the change in the entropy of helium, in...Ch. 12.6 - Prob. 40PCh. 12.6 - Derive expressions for (a) u, (b) h, and (c) s for...Ch. 12.6 - Derive an expression for the specific heat...Ch. 12.6 - Show that cpcv=T(PT)V(VT)P.Ch. 12.6 - Prob. 44PCh. 12.6 - Prob. 45PCh. 12.6 - Derive an expression for the specific heat...Ch. 12.6 - Derive an expression for the isothermal...Ch. 12.6 - Show that = ( P/ T)v.Ch. 12.6 - Prob. 49PCh. 12.6 - Prob. 50PCh. 12.6 - Show that the enthalpy of an ideal gas is a...Ch. 12.6 - Prob. 52PCh. 12.6 - Prob. 53PCh. 12.6 - The pressure of a fluid always decreases during an...Ch. 12.6 - Does the Joule-Thomson coefficient of a substance...Ch. 12.6 - Will the temperature of helium change if it is...Ch. 12.6 - Prob. 59PCh. 12.6 - Prob. 60PCh. 12.6 - 12–61E Estimate the Joule-Thomson-coefficient of...Ch. 12.6 - Prob. 62PCh. 12.6 - Consider a gas whose equation of state is P(v a)...Ch. 12.6 - Prob. 64PCh. 12.6 - On the generalized enthalpy departure chart, the...Ch. 12.6 - Why is the generalized enthalpy departure chart...Ch. 12.6 - Prob. 67PCh. 12.6 - Prob. 68PCh. 12.6 - Prob. 69PCh. 12.6 - Prob. 70PCh. 12.6 - Prob. 71PCh. 12.6 - Prob. 72PCh. 12.6 - Prob. 73PCh. 12.6 - Prob. 75PCh. 12.6 - Propane is compressed isothermally by a...Ch. 12.6 - Prob. 78PCh. 12.6 - Prob. 80RPCh. 12.6 - Starting with the relation dh = T ds + vdP, show...Ch. 12.6 - Show that cv=T(vT)s(PT)vandcp=T(PT)s(vT)PCh. 12.6 - Temperature and pressure may be defined as...Ch. 12.6 - For ideal gases, the development of the...Ch. 12.6 - Prob. 85RPCh. 12.6 - For a homogeneous (single-phase) simple pure...Ch. 12.6 - For a homogeneous (single-phase) simple pure...Ch. 12.6 - Prob. 88RPCh. 12.6 - Estimate the cpof nitrogen at 300 kPa and 400 K,...Ch. 12.6 - Prob. 90RPCh. 12.6 - Prob. 91RPCh. 12.6 - An adiabatic 0.2-m3 storage tank that is initially...Ch. 12.6 - Prob. 93RPCh. 12.6 - Methane is to be adiabatically and reversibly...Ch. 12.6 - Prob. 96RPCh. 12.6 - Prob. 98RPCh. 12.6 - Prob. 99RPCh. 12.6 - Prob. 100FEPCh. 12.6 - Consider the liquidvapor saturation curve of a...Ch. 12.6 - Prob. 102FEPCh. 12.6 - For a gas whose equation of state is P(v b) = RT,...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
What is entropy? - Jeff Phillips; Author: TED-Ed;https://www.youtube.com/watch?v=YM-uykVfq_E;License: Standard youtube license