**work**performed by a system is the energy transferred by the system to its surroundings.

**Work is a form of energy**, but it is

**energy in transit**. A system contains no work, and work is a process done by or on a system. In general, work is defined for mechanical systems as the action of a force on an object through a distance.

*W = F . d*

where:

W = work (J)

F = force (N)

d = displacement (m)

## pΔV Work

Pressure-volume work (or *pΔV ***Work**) occurs when the volume *V* of a system changes. The *pΔV ***Work** is equal to the area under the process curve plotted on the pressure-volume diagram. It is also known as **boundary work**. ** ****Boundary work** occurs because the mass of the substance within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and move it. **Boundary work** (or **pΔV*** ***Work**) occurs when the **volume**** ****V**** ****of a system changes**. It is used for calculating piston displacement work in a **closed system**. This happens when **steam** or gas contained in a piston-cylinder device expands against the piston and forces the piston to move.

**Example:**

Consider a frictionless piston that is used to provide a constant pressure of **500 kPa** in a cylinder containing steam (superheated steam) of a volume of **2 m**** ^{3 }** at

**500 K**.

Calculate the final temperature if **3000 kJ** of** heat** is added.

**Solution:**

Using steam tables we know, that the **specific enthalpy** of such steam (500 kPa; 500 K) is about** 2912 kJ/kg**. Since the steam has a density of 2.2 kg/m^{3}, we know about **4.4 kg of steam** in the piston at an enthalpy of 2912 kJ/kg x 4.4 kg =** 12812 kJ**.

When we use simply **Q = H**_{2}** − H**** _{1}**, then the resulting enthalpy of steam will be:

H_{2} = H_{1} + Q = **15812 kJ**

From **steam tables**, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of **828 K (555°C)**. Since at this enthalpy, the steam has a density of 1.31 kg/m^{3}, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m^{3} x 1.67 = 3.34 m^{3} and ∆V = 3.34 m^{3} – 2 m^{3} = 1.34 m^{3}.

The **p∆V **part of enthalpy, i.e., the work done is:

**W = p∆V = 500 000 Pa x 1.34 m**^{3}** = 670 kJ**

———–

During the **volume change**, the **pressure** and **temperature** may also change. To calculate such processes, we would need to know how pressure varies with volume for the actual process by which the system changes **from state i to state f**. The **first law of thermodynamics** and the work can then be expressed as:

When a thermodynamic system changes from an** initial state** to a **final state**, it passes through a **series of intermediate states**. We call this series of states a **path**. There are always infinitely many different possibilities for these intermediate states. When they are all equilibrium states, the path can be plotted on a **pV-diagram**. One of the most important conclusions is that:

*The work done by the system depends not only on the initial and final states but also on the intermediate states—that is, on the path.*

**Q and W are path-dependent, whereas ΔE**_{int}** is path-independent. **As can be seen from the picture (p-V diagram), work is a path-dependent variable. The blue area represents the ** pΔV Work** done by a system from an initial state i to a final state f. Work W is positive because the system’s volume increases. The second process shows that work is greater, and that depends on the path of the process.

Moreover, we can take the system through a series of states forming a **closed loop**, such **i ⇒ f ⇒ i**. In this case, the **final state is the same as the initial state**, but the **total work done** by the system **is not zero**. A positive value for work indicates that work is done by the system in its surroundings. A negative value indicates that work is done on the system by its surroundings.

## Example: Turbine Specific Work

A **high-pressure stage** of **steam turbine** operates at a steady state with inlet conditions of **6 MPa**, **t = 275.6°C**, x = 1 (point C). Steam leaves this turbine stage at a pressure of **1.15 MPa**, **186°C,** and **x = 0.87** (point D). Calculate the enthalpy difference between these two states. Determine the specific work transfer.

The enthalpy for the state C can be picked directly from** steam tables**, whereas the enthalpy for the state D must be calculated using vapor quality:

*h*_{1, wet}* = ***2785 kJ/kg**

*h*_{2, wet}* = h*_{2,s}* x + (1 – x ) h***_{2,l}** = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 =

**2523 kJ/kg**

**Δh = 262 kJ/kg**

Since in adiabatic process **dh = dw**, **Δh = 262 kJ/kg is **the **turbine-specific work**.

## First Law in Terms of Enthalpy dH = dQ + Vdp

The** enthalpy** is defined to be the sum of the internal energy E plus the product of the pressure p and volume V. In many thermodynamic analyses, the sum of the internal energy U and the product of pressure p and volume V appears. Therefore it is convenient to give the combination a name, **enthalpy**, and a distinct symbol, H.

**H = U + pV**

See also: Enthalpy

The **first law of thermodynamics** in terms of **enthalpy** shows us why engineers use the enthalpy in thermodynamic cycles (e.g.,, **Brayton cycle** or **Rankine cycle**).

The classical form of the law is the following equation:

**dU = dQ – dW**

In this equation, **dW** is equal to **dW = pdV** and is known as the **boundary work**.

Since ** H = U + pV**, therefore

*and we substitute*

**dH = dU + pdV + Vdp***into the classical form of the law:*

**dU = dH – pdV – Vdp****dH – pdV – Vdp = dQ – pdV**

We obtain the law in terms of enthalpy:

*dH = dQ + Vdp*

or

*dH = TdS + Vdp*

In this equation, the term * Vdp* is a

**flow process work.**This work,

*, is used for*

**Vdp****open flow systems**like a

**turbine**or a

**pump**in which there is a

**“dp”**, i.e., change in pressure. There are no changes in the control volume. As can be seen, this form of the law

**simplifies the description of energy transfer**.

**At constant pressure**, the

**enthalpy change**equals the

**energy**transferred from the environment through heating:

**Isobaric process (Vdp = 0):**

**dH = dQ → Q = H**_{2}** – H**_{1}

**At constant entropy**, i.e., in isentropic process, the **enthalpy change** equals the **flow process work** done on or by the system:

**Isentropic process (dQ = 0):**

**dH = Vdp → W = H**_{2}** – H**_{1}

It is obvious, and it will be very useful in the analysis of both thermodynamic cycles used in power engineering, i.e., in the Brayton and Rankine cycles.

## Example: First Law of Thermodynamics and Brayton Cycle

Let assume the **ideal Brayton cycle** that describes the workings of a **constant pressure** **heat engine**. **Modern gas turbine** engines and **airbreathing jet engines** also follow the Brayton cycle. This cycle consist of four thermodynamic processes:

**Isentropic compression**– ambient air is drawn into the compressor, pressurized (1 → 2). The work required for the compressor is given by**W**_{C}**= H**_{2}**– H**_{1}**.****Isobaric heat addition**– the compressed air then runs through a combustion chamber, burning fuel, and air or another medium is heated (2 → 3). It is a constant-pressure process since the chamber is open to flow in and out. The net heat added is given by**Q**_{add}**= H**_{3 }**– H**_{2}**Isentropic expansion**– the heated, pressurized air then expands on the turbine, gives up its energy. The work done by the turbine is given by**W**_{T}**= H**_{4}**– H**_{3}**Isobaric heat rejection**– the residual heat must be rejected to close the cycle. The net heat rejected is given by**Q**_{re}**= H**_{4 }**– H**_{1}

As can be seen, we can describe and calculate (e.g.,, thermodynamic efficiency) such cycles (similarly for **Rankine cycle**) using enthalpies.