We consider the First Law of Thermodynamics applied to stationary closed systems as a conservation of energy principle. Thus energy is transferred between the system and the surroundings in the form of heat and work, resulting in a change of internal energy of the system. Internal energy change can be considered as a measure of molecular activity associated with change of phase or temperature of the system and the energy equation is represented as follows:

Energy transferred across the boundary of a system in the form of heat always results from a difference in temperature between the system and its immediate surroundings. We will not consider the mode of heat transfer, whether by conduction, convection or radiation, thus the quantity of heat transferred during any process will either be specified or evaluated as the unknown of the energy equation. By convention, positive heat is that transferred from the surroundings to the system, resulting in an increase in internal energy of the system

In this course we consider three modes of work transfer across the boundary of a system, as shown in the following diagram:

In this course we are primarily concerned with
**Boundary Work** due to compression or expansion of a system in a piston-cylinder
device as shown above. In all cases we assume a perfect seal (no
mass flow in or out of the system), no loss due to friction, and
quasi-equilibrium processes in that for each incremental movement
of the piston equilibrium conditions are maintained. By convention
positive work is that done by the system on the surroundings,
and negative work is that done by the surroundings on the system,
Thus since negative work results in an increase in internal energy
of the system, this explains the negative sign in the above energy
equation.

Boundary work is evaluated by integrating the
force F multiplied by the incremental distance moved **d**x
between an initial state (1) to a final state (2). We normally
deal with a piston-cylinder device, thus the force can be replaced
by the piston area A multiplied by the pressure P, allowing us
to replace A.**d**x by the change in volume **d**V, as follows:

This is shown in the following schematic diagram, where we recall that integration can be represented by the area under the curve.

Note that work done is a **Path Function** and not a
property, thus it is dependent on the process path between the
initial and final states. Recall in **Chapter
1** that we introduced some typical process paths of interest:

**Isothermal**(constant temperature process)**Isochoric**or**Isometric**(constant volume process)**Isobaric**(constant pressure process)**Adiabatic**(no heat flow to or from the system during the process)

It is sometimes convenient to evaluate the
specific work done which can be represented by a *P-v* diagram
thus if the mass of the system is m [kg] we have finally:

We note that work done by the system on the surroundings (expansion process) is positive, and that done on the system by the surroundings (compression process) is negative.

Finally for a closed system **Shaft Work**
(due to a paddle wheel) and **Electrical
Work** (due to a voltage applied to an
electrical resistor or motor driving a paddle wheel) will always
be negative (work done on the system). Positive forms of shaft
work, such as that due to a turbine, will be considered in Chapter
4 when we discuss open systems.

The third component of our Closed System Energy
Equation is the change of internal energy resulting from the transfer
of heat or work. Since specific internal energy is a property
of the system, it is usually presented in the Property Tables
such as in the **Steam
Tables**. Consider for example the following solved problem.

**Solved Problem 3.1 -**
Recall the Solved Problem 2.2 in **Chapter
2a** in which we presented a constant pressure process.
We wish to extend the problem to include the energy interactions
of the process, hence we restate it as follows:

Two kilograms of water at 25°C are placed in a piston cylinder device under 3.2 MPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the temperature of the steam reaches 350°C (State (2)). Determine the work done by the fluid (W) and heat transferred to the fluid (Q) during this process.

**Solution Approach:**

We first draw the diagram of the process including all the relevant data as follows:

Notice the four questions to the right of the
diagram, which we should always ask before attempting to solve
any thermodynamic problem. What are we dealing with - liquid?
pure fluid, such as steam or refrigerant? ideal gas? In this case
it is steam, thus we will use the steam tables to determine the
various properties at the various states. Is the mass or volume
given? If so we will specify and evaluate the energy equation
in kiloJoules rather than specific quantities (kJ/kg). What about
entropy? Not so fast - we have not yet considered enthalpy (below)
- wait patiently until **Chapter 6**.

Since work involves the integral of P.**d**v
we find it convenient to sketch the *P-v* diagram of the
problem as follows:

Notice on the *P-v* diagram how we determine
the specific work done as the area under the process curve. We
also notice that in the Compressed Liquid region the constant
temperature line is essentially vertical. Thus all the property
values at State (1) (compressed liquid at 25°C) can be determined
from the saturated liquid table values at 25°C.

**Enthalpy (h) - a New Property**

In the case studies that follow we find that
one of the major applications of the closed system energy equation
is in heat engine processes in which the system is approximated
by an ideal gas, thus we will develop relations to determine the
internal energy for an ideal gas. We will find also that a new
property called **Enthalpy** will be useful both for Closed Systems and in particular
for Open Systems, such as the components of steam power plants
or refrigeration systems. Enthalpy is not a fundamental property,
however is a combination of properties and is defined as follows:

As an example of its usage in closed systems, consider the following constant pressure process:

Applying the energy equation we obtain:

However, since the pressure is constant throughout the process:

Substituting in the energy equation and simplifying:

Values for specific internal energy (u) and
specific enthalpy (h) are available from the **Steam
Tables**, however for ideal gasses it is necessary to develop
equations for Δu and Δh in terms of Specific Heat
Capacities. We develop these equations in terms of the differential
form of the energy equation in the following web page:

We have provided property values for various ideal gases, including the gas constant and specific heat capacities in the following web page:

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Engineering Thermodynamics by Israel Urieli is licensed under a
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