Section 1

Thermodynamics Comprehensive Exam - Cheatsheet

STUDY GUIDE

🎓 Thermodynamics Comprehensive Exam - Study Guide

📖 Chapter 1: Introduction to Thermodynamics and Fundamental Concepts

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Units
System	Region in space chosen for study	Defining the scope of analysis	N/A
Surroundings	Mass or region outside the system	Identifying external interactions	N/A
Closed System	Fixed mass, no mass transfer	Analyzing processes with constant mass	kg
Open System	Mass can cross the boundary	Analyzing flow processes	kg/s
Intensive Property	Independent of mass (e.g., temperature, pressure)	Characterizing state without mass dependence	K, Pa
Extensive Property	Dependent on mass (e.g., volume, energy)	Characterizing state with mass dependence	$m^3$ , J
Specific Volume	Volume per unit mass	$v = V/m$	$m^3/kg$
Density	Mass per unit volume	$\rho = m/V$	$kg/m^3$

🛠️ Problem Types

Type A: Identifying System Type

Setup: "Given a description of a system and its interactions with the surroundings."

Method: Determine if mass crosses the boundary. If yes, it's open; if no, it's closed. If no mass or energy crosses, it's isolated.

Example: A piston-cylinder device with a gas being heated is a closed system.

Type B: Unit Conversion

Setup: "Given a value in one unit system, convert it to another."

Method: Use conversion factors. For example, 1 atm = 101.325 kPa.

Example: Convert 100 psi to kPa. 100 psi * (101.325 kPa / 14.696 psi) = 689.47 kPa.

🧮 Solved Example

Problem: A rigid tank contains 5 kg of air at 200 kPa and 300 K. Determine the specific volume of the air. Steps:

Calculate the total volume: Assume ideal gas behavior: $V = \frac{mRT}{P} = \frac{5 kg * 0.287 kJ/kg.K * 300 K}{200 kPa} = 2.1525 m^3$
Calculate the specific volume: $v = \frac{V}{m} = \frac{2.1525 m^3}{5 kg} = 0.4305 m^3/kg$

"

✅
Answer: $0.4305 m^3/kg$

Section 2

📖 Chapter 2: Energy, Energy Transfer, and the First Law of Thermodynamics

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Units
Kinetic Energy (KE)	Energy due to motion	$KE = \frac{1}{2}mv^2$	J
Potential Energy (PE)	Energy due to elevation	$PE = mgh$	J
Internal Energy (U)	Energy due to molecular activity	State function, depends on T	J
Heat (Q)	Energy transfer due to temperature difference	Calculating energy transfer	J
Work (W)	Energy transfer due to force acting over a distance	Calculating energy transfer	J
First Law (Closed System)	Energy conservation	$\Delta U = Q - W$	J
First Law (Open System)	Energy conservation with flow	$h_1 + \frac{V_1^2}{2} + gz_1 + q = h_2 + \frac{V_2^2}{2} + gz_2 + w$	J/kg

🛠️ Problem Types

Type A: Calculating Kinetic and Potential Energy

Setup: "Given mass, velocity, and height."

Method: Use $KE = \frac{1}{2}mv^2$ and $PE = mgh$ .

Example: A 2 kg object moving at 5 m/s at a height of 10 m has KE = 25 J and PE = 196.2 J.

Type B: Applying the First Law to a Closed System

Setup: "Given heat transfer, work done, and initial/final states."

Method: Use $\Delta U = Q - W$ .

Example: A system receives 100 J of heat and does 50 J of work; the change in internal energy is 50 J.

🧮 Solved Example

Problem: A piston-cylinder device contains 0.5 kg of air initially at 300 K. 150 kJ of heat is added, and the air expands, doing 30 kJ of work. Find the final internal energy. Steps:

Apply the First Law: $\Delta U = Q - W = 150 kJ - 30 kJ = 120 kJ$
Calculate the initial internal energy (not needed for this problem, but generally good practice).
Calculate the final internal energy: $U_2 = U_1 + \Delta U$ . Since we only need the change, the answer is 120 kJ more than the initial.

"

✅
Answer: The internal energy increases by 120 kJ.

📖 Chapter 3: Properties of Pure Substances

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Units
Pure Substance	Fixed chemical composition	Identifying working fluids	N/A
Saturated Liquid	Liquid at boiling temperature	Determining state at saturation	N/A
Saturated Vapor	Vapor at condensation temperature	Determining state at saturation	N/A
Superheated Vapor	Vapor above saturation temperature	Determining state above saturation	N/A
Compressed Liquid	Liquid below saturation temperature	Determining state below saturation	N/A
Quality (x)	Mass fraction of vapor in a saturated mixture	$x = \frac{m_{vapor}}{m_{total}}$	N/A
Specific Volume (mixture)	$v = v_f + x(v_g - v_f)$	Calculating specific volume of a mixture	$m^3/kg$
Ideal Gas Law	$PV = mRT$	Approximating gas behavior at low pressure	N/A

🛠️ Problem Types

Type A: Determining Phase of a Substance

Setup: "Given temperature and pressure, or specific volume."

Method: Compare to saturation tables. If T > Tsat(P) or v > vg at given P, it's superheated vapor. If T < Tsat(P) or v < vf at given P, it's compressed liquid.

Example: Water at 100 kPa and 150°C is superheated vapor (T > Tsat(100 kPa)).

Type B: Using Property Tables

Setup: "Given two independent properties, find other properties."

Method: Use steam tables or other property tables to find the desired properties.

Example: For water at 1 MPa and 200°C, find h and v from superheated steam tables.

🧮 Solved Example

Problem: A rigid tank contains 2 kg of water at 100°C. 1 kg is liquid and 1 kg is vapor. Find the pressure and the volume of the tank. Steps:

Determine the saturation pressure at 100°C from steam tables: Psat = 101.42 kPa.
Calculate the quality: x = 1 kg / 2 kg = 0.5.
Find vf and vg at 100°C from steam tables.
Calculate the specific volume: $v = v_f + x(v_g - v_f)$ .
Calculate the total volume: $V = m * v = 2 kg * v$ .

"

✅
Answer: Pressure = 101.42 kPa, Volume = calculated V.

📖 Chapter 4: Energy Analysis of Closed Systems

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	When to Use	Units
Work (Constant Volume)	$W = 0$	Rigid tanks	J
Work (Constant Pressure)	$W = P(V_2 - V_1)$	Piston-cylinder with constant pressure	J
Work (Isothermal, Ideal Gas)	$W = mRT \ln(\frac{V_2}{V_1})$	Isothermal process for ideal gas	J
Work (Adiabatic, Ideal Gas)	$W = \frac{P_2V_2 - P_1V_1}{1-k}$	Adiabatic process for ideal gas	J
Polytropic Process	$PV^n = constant$	General process	N/A
Heat Transfer (Q)	Energy transfer due to temperature difference	Calculating energy transfer	J
Change in Internal Energy (ΔU)	$\Delta U = m c_v \Delta T$	Calculating change in internal energy	J

🛠️ Problem Types

Type A: Constant Volume Process

Setup: "Rigid tank with heat transfer."

Method: W = 0, so $\Delta U = Q$ .

Example: A rigid tank with air heated from 20°C to 100°C. Calculate Q using $\Delta U = m c_v \Delta T$ .

Type B: Constant Pressure Process

Setup: "Piston-cylinder with constant pressure heating/cooling."

Method: $W = P(V_2 - V_1)$ , $\Delta U = Q - W$ .

Example: Water heated in a piston-cylinder at 1 atm. Calculate W and Q.

🧮 Solved Example

Problem: A piston-cylinder device contains 0.2 kg of nitrogen at 100 kPa and 27°C. It is compressed adiabatically to 500 kPa. Find the final temperature. Steps:

Use the adiabatic relation: $P_1V_1^k = P_2V_2^k$ and $T_1P_1^{\frac{1-k}{k}} = T_2P_2^{\frac{1-k}{k}}$ .
Find k for nitrogen (approximately 1.4).
Solve for $T_2$ : $T_2 = T_1 (\frac{P_2}{P_1})^{\frac{k-1}{k}} = 300 K * (\frac{500 kPa}{100 kPa})^{\frac{1.4-1}{1.4}} = 475.4 K$ .

"

✅
Answer: 475.4 K.

8 more sections

Create a free account to import and read the full study notes — all 10 sections.

No credit card · 2 free imports included

Thermodynamics Comprehensive Exam - Cheatsheet

Study Notes Preview

Thermodynamics Comprehensive Exam - Cheatsheet

🎓 Thermodynamics Comprehensive Exam - Study Guide

📖 Chapter 1: Introduction to Thermodynamics and Fundamental Concepts

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

📖 Chapter 2: Energy, Energy Transfer, and the First Law of Thermodynamics

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

📖 Chapter 3: Properties of Pure Substances

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

📖 Chapter 4: Energy Analysis of Closed Systems

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

8 more sections