Explanation Write the expression for the energy balance equation for closed system. E i n − E o u t = Δ E s y s t e m (I) Here, energy transfer into the control volume is E i n , energy transfer exit from the control volume is E o u t and change in internal energy of system is Δ E s y s t e m . Write the expression to calculate the mass of the water. m w a t e r = ρ ν (II) Here, density of the water is ρ and volume of the container is ν . Write the expression for the entropy balance equation of the system. S i n − S o u t + S g e n = Δ S s y s t e m (III) Here, rate of net entropy in is S i n , rate of net entropy out is S o u t , rate of entropy generation is S g e n and change of entropy of the system is Δ S s y s t e m Conclusion: Substitute 0 for E i n , Q o u t for E o u t and Δ U c o n t a i n e r + Δ U w a t e r for Δ E s y s t e m in Equation (I). 0 − Q o u t = Δ U c o n t a i n e r + Δ U w a t e r − Q o u t = { [ m c o n t a i n e r c p , i r o n ( T 2 − T 1 ) ] c o n t a i n e r + [ m w a t e r c p , w a t e r ( T 2 − T 1 ) ] w a t e r } Q o u t = { [ m c o n t a i n e r c p , i r o n ( T 1 − T 2 ) ] c o n t a i n e r + [ m w a t e r c p , w a t e r ( T 1 − T 2 ) ] w a t e r } (IV) Here, change in total internal energy for container is Δ U c o n t a i n e r , change in total internal energy for container for water is Δ U w a t e r , heat loss to the surrounding air is Q o u t , mass of the container is m c o n t a i n e r , mass of the water is m w a t e r , initial temperature is T 1 , final temperature is T 2 , specific heat at constant pressure for iron is c p , i r o n and specific heat at constant pressure for water is c p , w a t e r . From the Table A-3E, “the properties of common liquids, solids and foods table”, select the density of water at room temperature as 62.1 lbm / ft 3 . Substitute 62.1 lbm / ft 3 for ρ and 15 ft 3 for ν in Equation (II). m w a t e r = ( 62.1 lbm / ft 3 ) ( 15 ft 3 ) = 931.5 lbm From the Table A-3E, “the properties of common liquids, solids and foods table”, select the specific heat at constant pressure at room temperature for water and iron as 1.00 Btu / lbm ⋅ °F and 0.107 Btu / lbm ⋅ °F respectively. Substitute 0.107 Btu / lbm ⋅ °F for c p , i r o n , 120 ° F for T 1 , 75 lbm for m c o n t a i n e r , 931 .5 lbm for m w a t e r , 1.00 Btu / lbm ⋅ °F for c p , w a t e r , and 70 ° F for T 2 in Equation (IV). Q o u t = { [ ( 75 lbm ) ( 0

8th Edition

ISBN: 9781307434316

Author: CENGEL

Publisher: INTER MCG

See similar textbooks

Concept explainers

Entropy

In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its m…

Videos

Question

Chapter 7.13, Problem 203RP

To determine

The total entropy generation during the process

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.13, Problem 202RP

Chapter 7.13, Problem 204RP

Students have asked these similar questions

Seven kilograms of air is contained in a constant volume tank at 327 degrees C and 120 kPa, then is cooled to the ambient temperature of 27 degrees C. Determine the entropy change of the air and the universe(surroundings) due to the process and sketch the process on a T-s diagram.

A 15-ft3 steel container that has a mass of 75 lbm when empty is filled with liquid water. Initially, both the steel tank and the water are at 120°F. Now heat is transferred, and the entire system cools to the surrounding air temperature of 70°F. Determine the total entropy generated during this process.

In an experiment, a certain amount of air is heated at constant pressure from 1.7 m3, 27 0C and 97 kPa to 414 0C. It is then cooled at constant volume back to its initial temperature. Calculate the change specific entropy in the cooling process in the correct unit in SI and 1 decimal place. Take Specific heat at constant pressure and constant volume for air as 1.005 kJ/kg K and 0.717 kJ/kg K respectively. add units.

Chapter 7 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

Ch. 7.13 - Prob. 1P Ch. 7.13 - Does the cyclic integral of heat have to be zero...Ch. 7.13 - Is a quantity whose cyclic integral is zero...Ch. 7.13 - Prob. 4P Ch. 7.13 - Prob. 5P Ch. 7.13 - How do the values of the integral 12Q/T compare...Ch. 7.13 - The entropy of a hot baked potato decreases as it...Ch. 7.13 - When a system is adiabatic, what can be said about...Ch. 7.13 - Prob. 9P Ch. 7.13 - A pistoncylinder device contains helium gas....

Knowledge Booster

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.

The total entropy generation during the process

Solution Summary: The author explains the total entropy generation during the process. Write the expression for the energy balance equation for closed system.

Concept explainers

Videos

The total entropy generation during the process

Want to see the full answer?

Chapter 7 Solutions