First law for closed and open systems

Introduction to the First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle that governs all energy interactions in engineering systems. Simply put, it states that energy can neither be created nor destroyed; it can only be transformed from one form to another. This principle is also known as the conservation of energy.

In mechanical engineering, especially in thermodynamics and heat transfer, understanding how energy flows and changes within systems is crucial. This knowledge helps design engines, refrigerators, turbines, compressors, and many other devices that power our industries and daily lives.

Before applying the first law, it is essential to understand the concept of a system. A system is a specific portion of the universe chosen for analysis, separated by boundaries from its surroundings. Systems are broadly classified into two types:

Closed System (Control Mass): A fixed amount of mass enclosed by rigid or movable boundaries. No mass crosses the system boundary, but energy in the form of heat or work can cross. Example: A piston-cylinder device containing gas.
Open System (Control Volume): A region in space through which mass and energy can flow in and out. Example: A steam turbine or compressor where fluid continuously flows through.

Understanding these system types is vital because the first law is applied differently depending on whether the system is closed or open.

In this chapter, we will explore the first law for both closed and open systems, using practical examples relevant to Indian engineering contexts. By mastering these concepts, you will be well-prepared for competitive exams and real-world engineering problems.

First Law of Thermodynamics for Closed Systems

For a closed system, the first law relates the change in the system's internal energy to the heat added to the system and the work done by the system on its surroundings.

Internal energy (\(U\)) is the total energy stored within the system due to the microscopic motion and interactions of molecules. It depends on the system's temperature, pressure, and phase.

The energy balance for a closed system undergoing a process from state 1 to state 2 is expressed as:

First Law for Closed Systems

\[\Delta U = Q - W\]

Change in internal energy equals heat added minus work done by the system

\(\Delta U\) = Change in internal energy (kJ)

Q = Heat added to system (kJ)

W = Work done by system (kJ)

Here:

\(Q\) is positive if heat is added to the system, negative if heat is lost.
\(W\) is positive if work is done by the system on the surroundings, negative if work is done on the system.

This sign convention is standard in thermodynamics and helps maintain consistency.

Work and Heat Transfer in Closed Systems

Work done by a closed system often involves boundary movement, such as a piston moving in a cylinder. The work done during a quasi-static (slow) process is given by:

Work Done by Gas in Quasi-Static Process

\[W = \int P dV\]

Work done during volume change at pressure P

W = Work done (kJ)

P = Pressure (kPa)

V = Volume (m³)

Heat transfer (\(Q\)) can occur via conduction, convection, or radiation, depending on the system and surroundings.

Diagram: Energy Transfer in a Closed System

Figure: A piston-cylinder device where heat is added to the gas, causing it to expand and do work on the piston.

First Law of Thermodynamics for Open Systems

Unlike closed systems, open systems allow mass to cross their boundaries. Examples include turbines, compressors, nozzles, and heat exchangers.

For open systems, the first law is applied to a control volume, a fixed region in space through which fluid flows steadily or unsteadily.

In many engineering applications, the flow is steady, meaning properties at any point do not change with time. The steady flow energy equation is the first law applied to such systems.

The general energy balance for a steady flow process is:

First Law for Open Systems (Steady Flow)

\[\dot{Q} - \dot{W} = \dot{m} \left(h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1) \right)\]

Energy balance accounting for heat, work, enthalpy, kinetic and potential energy changes

\(\dot{Q}\) = Heat transfer rate (kW)

\(\dot{W}\) = Work transfer rate (kW)

\(\dot{m}\) = Mass flow rate (kg/s)

h = Enthalpy (kJ/kg)

V = Velocity (m/s)

g = Gravitational acceleration (9.81 m/s²)

z = Elevation (m)

Here, \(h\) is the enthalpy, defined as:

Enthalpy Definition

h = u + Pv

Sum of internal energy and flow work per unit mass

h = Enthalpy (kJ/kg)

u = Internal energy (kJ/kg)

P = Pressure (kPa)

v = Specific volume (m³/kg)

Enthalpy represents the total energy content of the fluid including internal energy and the energy required to push the fluid into or out of the control volume (flow work).

Diagram: Energy Flow in an Open System (Turbine)

Figure: Control volume of a turbine showing inlet and outlet streams with enthalpy, velocity, and elevation changes, plus heat and work interactions.

Why Use Enthalpy in Open Systems?

In open systems, fluids flow continuously, carrying energy as internal energy and flow work. Using enthalpy simplifies calculations by combining these energy forms into one property, making it easier to apply the first law.

Summary of Key Concepts

{"points": [ "First law expresses energy conservation: energy change equals heat added minus work done.", "Closed systems have fixed mass; energy transfer via heat and work only.", "Open systems allow mass flow; energy includes enthalpy, kinetic, and potential energy.", "Sign conventions: heat added (+), work done by system (+).", "Enthalpy = internal energy + flow work; essential for open system analysis." ], "conclusion": "Mastering these principles is essential for solving thermodynamics problems in engineering and exams."}

Formula Bank

First Law for Closed Systems

\[\Delta U = Q - W\]

where: \(\Delta U\) = change in internal energy (kJ), \(Q\) = heat added to system (kJ), \(W\) = work done by system (kJ)

First Law for Open Systems (Steady Flow)

\[\dot{Q} - \dot{W} = \dot{m} \left(h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)\right)\]

where: \(\dot{Q}\) = heat transfer rate (kW), \(\dot{W}\) = work transfer rate (kW), \(\dot{m}\) = mass flow rate (kg/s), \(h\) = enthalpy (kJ/kg), \(V\) = velocity (m/s), \(g\) = gravitational acceleration (9.81 m/s²), \(z\) = elevation (m)

Work Done by a Gas in a Quasi-Static Process

\[W = \int P dV\]

where: \(W\) = work done (kJ), \(P\) = pressure (kPa), \(V\) = volume (m³)

Change in Internal Energy (Ideal Gas)

\[\Delta U = m C_v \Delta T\]

where: \(m\) = mass (kg), \(C_v\) = specific heat at constant volume (kJ/kg·K), \(\Delta T\) = temperature change (K)

Change in Enthalpy (Ideal Gas)

\[\Delta h = C_p \Delta T\]

where: \(C_p\) = specific heat at constant pressure (kJ/kg·K), \(\Delta T\) = temperature change (K)

Worked Examples

Example 1: Energy Balance on a Piston-Cylinder Device Medium

A 2 kg gas is contained in a piston-cylinder device. The gas is heated, causing it to expand and push the piston outward. During the process, 50 kJ of heat is added to the gas, and the gas does 30 kJ of work on the piston. Calculate the change in internal energy of the gas.

Step 1: Identify the system type and given data.

Closed system (fixed mass of gas)
Mass, \(m = 2\) kg (given but not directly needed here)
Heat added, \(Q = +50\) kJ (positive since heat is added)
Work done by system, \(W = +30\) kJ (positive since work is done by gas)

Step 2: Apply the first law for closed systems:

\[\Delta U = Q - W\]

Step 3: Substitute values:

\[\Delta U = 50 - 30 = 20 \text{ kJ}\]

Answer: The internal energy of the gas increases by 20 kJ during the process.

Example 2: Steady Flow Analysis of a Steam Turbine Medium

Steam enters a turbine at 3 MPa, 400°C with a velocity of 50 m/s and leaves at 0.1 MPa, 150°C with a velocity of 150 m/s. The elevation change is negligible. The mass flow rate is 5 kg/s, and the turbine loses heat to the surroundings at a rate of 10 kW. Calculate the power output of the turbine. Use steam tables for enthalpy values: \(h_1 = 3210\) kJ/kg, \(h_2 = 2700\) kJ/kg.

Step 1: Identify system and data.

Open system (turbine control volume)
Mass flow rate, \(\dot{m} = 5\) kg/s
Heat loss, \(\dot{Q} = -10\) kW (negative since heat is lost)
Velocities: \(V_1 = 50\) m/s, \(V_2 = 150\) m/s
Elevation change negligible: \(z_2 - z_1 = 0\)
Enthalpies: \(h_1 = 3210\) kJ/kg, \(h_2 = 2700\) kJ/kg

Step 2: Write steady flow energy equation:

\[\dot{Q} - \dot{W} = \dot{m} \left(h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)\right)\]

Step 3: Calculate kinetic energy change per unit mass:

\[\frac{V_2^2 - V_1^2}{2} = \frac{150^2 - 50^2}{2} = \frac{22500 - 2500}{2} = \frac{20000}{2} = 10000 \text{ m}^2/\text{s}^2\]

Convert to kJ/kg (1 m²/s² = 0.001 kJ/kg):

\[10000 \times 0.001 = 10 \text{ kJ/kg}\]

Step 4: Substitute values:

\[-10 - \dot{W} = 5 \times (2700 - 3210 + 10 + 0)\]

\[-10 - \dot{W} = 5 \times (-500)\]

\[-10 - \dot{W} = -2500\]

Step 5: Solve for \(\dot{W}\):

\[\dot{W} = -10 + 2500 = 2490 \text{ kW}\]

Answer: The turbine power output is 2490 kW.

Example 3: Heat Transfer and Work in a Compressor Medium

Air at 100 kPa and 300 K enters a compressor at 10 m/s and leaves at 500 kPa and 450 K with velocity 20 m/s. The mass flow rate is 2 kg/s. Assuming negligible heat loss, calculate the power input to the compressor. Take \(C_p = 1.005\) kJ/kg·K.

Step 1: Identify system and data.

Open system (compressor control volume)
Mass flow rate, \(\dot{m} = 2\) kg/s
Inlet velocity, \(V_1 = 10\) m/s; outlet velocity, \(V_2 = 20\) m/s
Inlet temperature, \(T_1 = 300\) K; outlet temperature, \(T_2 = 450\) K
Negligible heat loss: \(\dot{Q} = 0\)

Step 2: Calculate enthalpy change using \(C_p\):

\[\Delta h = C_p (T_2 - T_1) = 1.005 \times (450 - 300) = 1.005 \times 150 = 150.75 \text{ kJ/kg}\]

Step 3: Calculate kinetic energy change per unit mass:

\[\frac{V_2^2 - V_1^2}{2} = \frac{20^2 - 10^2}{2} = \frac{400 - 100}{2} = \frac{300}{2} = 150 \text{ m}^2/\text{s}^2\]

Convert to kJ/kg:

\[150 \times 0.001 = 0.15 \text{ kJ/kg}\]

Step 4: Apply steady flow energy equation:

\[\dot{Q} - \dot{W} = \dot{m} \left(\Delta h + \Delta KE + \Delta PE\right)\]

Elevation change negligible, so \(\Delta PE = 0\), and \(\dot{Q} = 0\).

\[- \dot{W} = 2 \times (150.75 + 0.15) = 2 \times 150.9 = 301.8 \text{ kW}\]

Step 5: Solve for power input:

\[\dot{W} = -301.8 \text{ kW}\]

Negative sign indicates work is done on the system (power input).

Answer: Power input to the compressor is 301.8 kW.

Example 4: Energy Balance in a Throttling Valve Easy

Steam at 2 MPa and 300°C passes through a throttling valve and exits at 0.5 MPa. Assuming no heat transfer or work, and negligible changes in kinetic and potential energy, determine the enthalpy change across the valve.

Step 1: Identify system and assumptions.

Open system (valve control volume)
No heat transfer: \(\dot{Q} = 0\)
No work done: \(\dot{W} = 0\)
Negligible kinetic and potential energy changes

Step 2: Apply steady flow energy equation:

\[\dot{Q} - \dot{W} = \dot{m} (h_2 - h_1)\]

Since \(\dot{Q} = 0\) and \(\dot{W} = 0\),

\[0 = \dot{m} (h_2 - h_1) \Rightarrow h_2 = h_1\]

Answer: The enthalpy remains constant during throttling (\(\Delta h = 0\)).

Example 5: Energy Analysis of a Mixing Chamber Hard

Two steam streams mix in a chamber. Stream 1 has mass flow rate 3 kg/s, enthalpy 2800 kJ/kg, velocity 40 m/s. Stream 2 has mass flow rate 2 kg/s, enthalpy 3200 kJ/kg, velocity 60 m/s. The mixed stream leaves at velocity 50 m/s. Assuming no heat transfer or work, and negligible potential energy changes, find the enthalpy of the mixed stream.

Step 1: Identify data and assumptions.

Open system (mixing chamber control volume)
No heat transfer: \(\dot{Q} = 0\)
No work done: \(\dot{W} = 0\)
Negligible potential energy changes
Mass flow rates: \(\dot{m}_1 = 3\) kg/s, \(\dot{m}_2 = 2\) kg/s
Enthalpies: \(h_1 = 2800\) kJ/kg, \(h_2 = 3200\) kJ/kg
Velocities: \(V_1 = 40\) m/s, \(V_2 = 60\) m/s, \(V_3 = 50\) m/s (mixed stream)

Step 2: Apply steady flow energy equation for mixing chamber:

\[\sum \dot{m}_{in} \left(h + \frac{V^2}{2}\right) = \dot{m}_{out} \left(h + \frac{V^2}{2}\right)\]

Calculate kinetic energy terms (in kJ/kg):

\[\frac{V_1^2}{2} = \frac{40^2}{2} = 800 \text{ m}^2/\text{s}^2 = 0.8 \text{ kJ/kg}\]

\[\frac{V_2^2}{2} = \frac{60^2}{2} = 1800 \text{ m}^2/\text{s}^2 = 1.8 \text{ kJ/kg}\]

\[\frac{V_3^2}{2} = \frac{50^2}{2} = 1250 \text{ m}^2/\text{s}^2 = 1.25 \text{ kJ/kg}\]

Step 3: Calculate total mass flow rate out:

\[\dot{m}_3 = \dot{m}_1 + \dot{m}_2 = 3 + 2 = 5 \text{ kg/s}\]

Step 4: Write energy balance:

\[3 \times (2800 + 0.8) + 2 \times (3200 + 1.8) = 5 \times (h_3 + 1.25)\]

\[3 \times 2800.8 + 2 \times 3201.8 = 5 \times (h_3 + 1.25)\]

\[8402.4 + 6403.6 = 5h_3 + 6.25\]

\[14806 = 5h_3 + 6.25\]

Step 5: Solve for \(h_3\):

\[5h_3 = 14806 - 6.25 = 14799.75\]

\[h_3 = \frac{14799.75}{5} = 2959.95 \text{ kJ/kg}\]

Answer: The enthalpy of the mixed stream is approximately 2960 kJ/kg.

Tips & Tricks

Tip: Always define system boundaries clearly before applying the first law.

When to use: At the start of any thermodynamics problem to avoid confusion and errors.

Tip: Use consistent sign conventions: heat added to system (+), work done by system (+).

When to use: While writing energy balance equations to avoid sign errors.

Tip: Convert all units to metric SI before calculations to maintain consistency.

When to use: Before starting numerical problems, especially in exams.

Tip: For open systems, focus on enthalpy changes rather than internal energy.

When to use: When analyzing turbines, compressors, nozzles, and other flow devices.

Tip: Remember throttling processes are isenthalpic (\(\Delta h=0\)), simplifying calculations.

When to use: When analyzing valves and expansion devices in thermodynamics problems.

Common Mistakes to Avoid

❌ Confusing system types and applying closed system equations to open systems.

✓ Identify the system as closed or open and use the appropriate first law form.

Why: Different energy terms and variables apply to each system type, leading to incorrect results if mixed.

❌ Ignoring kinetic and potential energy changes in open system analysis.

✓ Include velocity and elevation terms when they are significant.

Why: Neglecting these can cause inaccurate energy balances and wrong power or heat transfer calculations.

❌ Incorrect sign convention for work and heat transfer.

✓ Adopt and stick to standard sign conventions throughout the problem.

Why: Inconsistent signs cause errors in energy calculations and confusion in interpretation.

❌ Mixing units, e.g., using kJ with kgf or non-metric units.

✓ Convert all quantities to SI units before calculation.

Why: Unit inconsistency leads to wrong numerical answers and failed exams.

❌ Assuming zero heat transfer in all open system problems.

✓ Evaluate problem conditions carefully; heat transfer may be present.

Why: Incorrect assumptions reduce accuracy and may lead to wrong conclusions.

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First law for closed and open systems

Introduction to the First Law of Thermodynamics

First Law of Thermodynamics for Closed Systems

First Law for Closed Systems

Work and Heat Transfer in Closed Systems

Work Done by Gas in Quasi-Static Process

Diagram: Energy Transfer in a Closed System

First Law of Thermodynamics for Open Systems

First Law for Open Systems (Steady Flow)

Enthalpy Definition

Diagram: Energy Flow in an Open System (Turbine)

Why Use Enthalpy in Open Systems?

Summary of Key Concepts

Formula Bank

Formula Bank

Worked Examples

Tips & Tricks

Common Mistakes to Avoid

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