Explanation Write the formula for volume expansivity ( β ) . β = 1 v ( ∂ v ∂ T ) P (I) Write the formula for isothermal compressibility ( α ) . α = − 1 v ( ∂ v ∂ P ) T (II) Here, the specific volume is v , the pressure is P , the temperature is T and the symbol ∂ indicates the partial derivative of variables. Refer Table A-13, “Superheated refrigerant-134a”. Consider the two temperatures 20 ° C and 40 ° C (at constant pressure 200 kPa ), these temperatures can give the average temperature of 30 ° C . Also consider the two pressures 180 kPa and 240 kPa (at constant temperature 30 ° C ), these pressures can give the average pressure of 210 kPa (close to the value of 200 kPa ). Rewrite the Equations (I) and (II) by approximating the partial differentials about the specified state. β ≅ 1 v ( Δ v Δ T ) P = 200 kPa = 1 v ( v 40 ° C − v 20 ° C 40 ° C − 20 °C ) P = 200 kPa (III) α ≅ − 1 v ( Δ v Δ P ) T = 30 ° C = − 1 v ( v 240 kPa − v 180 kPa 240 kPa − 180 kPa ) T = 30 ° C (IV) Refer Table A-13, “Superheated refrigerant-134a”. The specific volume ( v ) corresponding to the pressure of 200 kPa ( 0.2 MPa ) and the temperature of 30 ° C is 0.11874 m 3 /kg . And also obtain the specific volumes corresponding pressure of 200 kPa ( 0.2 MPa ) and the temperatures 20 ° C and 40 ° C . v 40 ° C = 0.12322 m 3 /kg v 20 ° C = 0.11418 m 3 /kg Refer Table A-13, “Superheated refrigerant-134a”. Obtain the specific volumes corresponding to the temperature of 30 ° C at the pressures 180 kPa ( 0.18 MPa ) and 240 kPa ( 0

9th Edition

ISBN: 9781259822674

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

See similar textbooks

Concept explainers

Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

Videos

Question

Chapter 12.6, Problem 40P

To determine

The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30 °C.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.6, Problem 39P

Chapter 12.6, Problem 41P

Students have asked these similar questions

A 1.40 kg sample of water at 15.0°C is in a calorimeter. You drop a piece of steel with a mass of 0.330 kg at 230°C into it. After the sizzling subsides, what is the final equilibrium temperature (in °C)? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don't affect the outcome.) °C

Derive a relation for the volume expansivity β and the isothermal compressibility κ (a) for an ideal gas and (b) for a gas whose equation of state is P(v - a) = RT and (c) calculate β and κ for nitrogen at T=175K, v = 0.00375 m3 /kg, and a = 0.00138 m3 /kg and compare the results obtained from ideal gas law and from P(v - a) = RT.

There are 1.5 kgm of gas for which R=0.377kJ/kg-K and k=1.3 that undergo an isometric process from an initial state of 550kPa and 60C. During the process there are 1500kJ of heat removed from the gas. Find the change in enthalphy (kJ)

Chapter 12 Solutions

Thermodynamics: An Engineering Approach

Ch. 12.6 - What is the difference between partial...Ch. 12.6 - Consider the function z(x, y). Plot a differential...Ch. 12.6 - Consider a function z(x, y) and its partial...Ch. 12.6 - Prob. 4P Ch. 12.6 - Prob. 5P Ch. 12.6 - Consider a function f(x) and its derivative df/dx....Ch. 12.6 - Conside the function z(x, y), its partial...Ch. 12.6 - Consider air at 350 K and 0.75 m3/kg. Using Eq....Ch. 12.6 - Consider air at 350 K and 0.75 m3/kg. Using Eq....Ch. 12.6 - Nitrogen gas at 800 R and 50 psia behaves as an...

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 volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30 ° C .

Solution Summary: The author explains the volume expansivity and isothermal compressibility of refrigerant-134a at 200kPa, and 30°

Concept explainers

Videos

The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30 °C.

Want to see the full answer?

Chapter 12 Solutions