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02.07 The Closed-System Energy Balance
Book navigation
- Chapter 1 - Basic concepts
-
Chapter 2 - The energy balance
- 02.01 Expansion/Contraction Work
- 02.03 Work Associated with Flow
- 02.04 Lost Work Versus Reversibility
- 02.06 Path Properties and State Properties
- 02.07 The Closed-System Energy Balance
- 02.08 The Open-System, Steady-State Energy Balance
- 02.09 The Complete Energy Balance
- 02.10 Internal Energy, Enthalpy, and Heat Capacities
- 02.11 Reference States
- 02.13 Energy Balances for Process Equipment
- 02.15 Closed and Steady-State Open Systems
- 02.16 Unsteady State Open Systems
- 02.18 Chapter 2 Summary
- Chapter 3 - Energy balances for composite systems.
- Chapter 4 - Entropy
- Chapter 5 - Thermodynamics of Processes
- Chapter 6 - Classical Thermodynamics - Generalization to any Fluid
- Chapter 7 - Engineering Equations of State for PVT Properties
- Chapter 8 - Departure functions
- Chapter 9 - Phase Equlibrium in a Pure Fluid
- Chapter 10 - Introduction to Multicomponent Systems
- Chapter 11 - An Introduction to Activity Models
- Chapter 12 - Van der Waals Activity Models
- Chapter 13 - Local Composition Activity Models
- Chapter 14 - Liquid-liquid and solid-liquid equilibria
- Chapter 16 - Advanced Phase Diagrams
- Chapter 15 - Phase Equilibria in Mixtures by an Equation of State
- Chapter 17 - Reaction Equilibria
- Chapter 18 - Electrolyte Solutions
Constant Pressure Process with Steam
Constant pressure process using steam (LearnChemE.com, 5min). 2000 kJ of heat is isobarically added to steam piston/cylinder starting at 0.45MPa with 0.9kg as vapor and 0.1kg as liquid. Compute the final temperature, work, and state of the steam. Once we have our general energy balance defined, we can straigntforwardly reduce it to its simplest applicable form to solve problems. The energy balance is the same regardless of whether the process uses an ideal gas, steam, or some other working fluid. But the method of solving the problem changes quite a bit depending on the working fluid.
Hint: "steam" and H2O are the same thing. So "liquid steam" is also known as "water."
Comprehension questions:
1. Describe how you would solve this problem if the H2O was replaced with a monatomic ideal gas (MW=40). Use the same starting pressure and temperature as the steam, but obviously the entire 1.0 kg will be gas, with no liquid.
2. Describe how you would write the energy balance if the cylinder was 5m3 total, open to the atmosphere, and the pressure was suddenly reduced to 1 bar. Assume the piston has a mass of 0.1kg.
Adiabatic, Reversible Compression of an Ideal Gas (Bang!)
Adiabatic, Reversible Compression of an Ideal Gas in a Piston/Cylinder (LearnChemE.com, 5min). The standard formula for an adiabatic, reversible, ideal gas is derived here in the T,V form. You should be able to rearrange the given equation into the usual form:
(T2/T1) = (P2/P1)^(R/Cp) to show they are equivalent. (Hint: PV=RT and Cp=Cv+R). The interesting part of this video is during the last 10 seconds. Watch what happens!
Comprehension Questions:
1. My bicycle pump is about 50cm tall and 2.0cm diameter. When I pump it down, the pressure goes to 100psig (after pumping once or twice). What is the temperature of the air that goes into the tire at that point?
2. What is the length (cm) remaining in the pump?