Energy Balances

If you want to find the secrets of the universe, think in terms of energy, frequency and vibration.

—Nikola Tesla


Rationale

When we learned material balances, we were able to track the movement of chemical species throughout a system or process.

Another important aspect of process analysis is the determination of the energy requirements and temperatures around systems. For instance, how can we maintain a consistent temperature in a reactor if the reaction inside is exothermic? How much energy do we need to add to a process stream to move it to some new temperature?


Important nomenclature

A closed system is one in which there is a fixed volume or space and no streams entering or leaving the system.

An open system is one in which there are streams entering and leaving. These streams can add or remove material and energy from a system.

An adiabatic process is one in which there is no heat added or removed from the system.

An isothermal process is one in which the system stays at a constant temperature. Heat may need to be added or removed to maintain this condition.


First law of thermodynamics for closed systems

\[\Delta E = Q + W\]

where

\(E\)

=

total energy of the system (units of energy)

\(Q\)

=

heat transferred from the environment to the system through the boundaries of the volume (units of energy) over the interval of time during which E changes

\(W\)

=

work done on the system by the environment (units of energy) over the interval of time during which E changes

Typical units for these energy terms are

  • erg and calorie (\(\si{cal}\)) in the cgs system

  • Joule (\(\si{J}\)) in the SI system

  • British thermal unit (\(\si{Btu}\)) in the American system

‘Direction’ of heat and work

If work is done on a system, energy is utilized to provide that work, and the result is an increased energy within the system.

Similarly, if heat passes across a system boundary into a system, it likewise shows up as increased energy within the system.

Consistent with this idea is the convention in this course that \(Q\) and \(W\) have a positive sign if they add energy to the system.


More symbols we need to know

Recall that a ‘dot’ over a symbol means a rate (quantity per unit time).

For example, \(\dot m\) means a mass per unit time in units such as \(\si{g/s}\).

We now introduce a ‘hat’ over symbols. This represents a quantity per unit mass. This is often called a specific quantity.

For instance \(\hat E\) is the specific energy (or energy per unit mass). Units in this case are something like \(\si{J/g}\).

First law of thermodynamics for open systems

For a steady-state open system, our conservation law becomes

\[\begin{split}\sum_{\substack{\text{output}\\\text{streams}}} (\dot m \, \hat E)_{out} - \sum_{\substack{\text{input}\\\text{streams}}} (\dot m \, \hat E)_{in} = \dot Q + \dot W\end{split}\]

where

\(\dot m\)

=

mass flow rate of a stream (units of mass per time)

\(\hat E\)

=

energy per mass of a stream of flowing material

\(\dot Q\)

=

rate of transfer of energy across the boundaries of a system into that system and/or generated within the system (units of energy per time)

\(\dot W\)

=

rate that work is done on a system (units of energy per time)

Typical units for these rate of energy transfer terms are

  • erg per second (\(\si{erg/s}\)) and calorie per second (\(\si{cal/s}\)) in the cgs system

  • Joule per second (i.e., Watts) (\(\si{J/s}=\si{W}\)) in the SI system

  • British thermal unit per hour (\(\si{Btu/hr}\)) in the American system

There are three forms of energy that we will consider (expressed per mass of material):

  • internal energy

  • kinetic energy

  • potential energy

Just as a flowing stream has kinetic and potential energy, individual molecules also have kinetic energy (from their individual motion) and potential energy (from the attraction and repulsion between molecules).

The sum of these molecular energies is expressed as the internal energy of the material, which is a strong function of temperature.

Now, the total energy per mass of material is

\[\begin{split}\begin{align*} \hat E_{\text{total}} & = \hat E_{\text{internal}} + \hat E_{\text{kinetic}} + \hat E_{\text{potential}} \\ & = \hat U + \frac{1}{2} \, \alpha \, v^{2} + g \, z \end{align*}\end{split}\]

What is \(\alpha\) in the equation above and those to follow?

In many cases of practical interest, the kinetic energy term is more appropriately written

\[\hat E_{\text{kinetic}} = \frac{1}{2} \, (v^{2})_{\text{avg}}\]

We can relatively easily measure the average velocity from the volumetric flow rate and cross sectional area, \(v_{\text{avg}} = \dot V / A_{c}\).

Thus, we also easily estimate the average velocity squared, \((v_{\text{avg}})^{2}\). However, our kinetic energy term, \((v^{2})_{\text{avg}}\), is not equal to \((v_{\text{avg}})^{2}\).

To make terms that are equivalent, we define a correction factor, \(\alpha\), such that

\[(v^{2})_{\text{avg}} = \alpha \, (v_{\text{avg}})^{2}\]

Though not strictly correct, in most cases in this course, we will assume that \(\alpha = 1\).


Our generalized conservation of energy equation is therefore

\[\begin{split}\sum_{\substack{\text{output}\\\text{streams}}} \left [\dot m \left(\hat U + \frac{1}{2}\, \alpha \, v^{2} + g \, z \right) \right ]_{out} - \sum_{\substack{\text{input}\\\text{streams}}} \left[\dot m \left(\hat U + \frac{1}{2}\, \alpha \, v^{2} + g \, z \right) \right]_{in} = \dot Q + \dot W\end{split}\]

Note that the internal energy is often combined with the flow work into a property called the enthalpy, which is represented by the symbol \(\hat H\) and defined as

\[\hat H = \hat U + P \, \hat V\]

To be consistent with other units in our equation, specific enthalpy has units of energy per unit mass, e.g., \(\si{Btu/lb_{m}}\) or \(\si{J/g}\).


Rate of Work

When external forces do work on a fluid, the energy of that fluid increases.

For example, when a pump does work on a fluid, that work increases fluid velocities, potential energy, and/or fluid temperature. Fluids can also do work on their environment, and thereby lose energy.

It is important to note that the rate of work \(\dot W\) consists of both ‘rate of flow’ work and shaft work: \(\dot W = \dot W_{PV} + \dot W_{S}\).

Rate of flow work, \(\dot W_{PV}\):

This work results from the displacement of fluid during flow and is similar to the pressure-volume work associated with the compression or expansion of a closed system. However, in the case of an open system (with inlet and outlet streams), the flow of fluid into and out of a system represents a continual performance of work as upstream fluid “pushes” fluid into the system entrance and the fluid in the system “pushes” downstream fluid out of the system exit.

Rate of shaft work, \(\dot W_{S}\):

When the flowing fluid in a system contacts moving parts, work is performed. Shaft work is positive when the net work is on the system (such as in a pump, where the moving parts are driven by external forces and thereby “push” the fluid). Conversely, the shaft work has a negative value when the net work is done by the fluid (such as in a turbine, where the fluid causes the parts to move).

Utilizing the above relationships and the fact that \(P \, \hat V = P/\rho\), we obtain the common form of the conservation of energy equation:

Conservation of energy equation

\[\begin{split}\sum_{\substack{\text{output}\\\text{streams}}} \left [\dot m \left(\hat H + \frac{1}{2}\, \alpha \, v^{2} + g \, z \right) \right ]_{out} - \sum_{\substack{\text{input}\\\text{streams}}} \left[\dot m \left(\hat H + \frac{1}{2}\, \alpha \, v^{2} + g \, z \right) \right]_{in} = \dot Q + \dot W_{S}\end{split}\]

Special cases

For now, we’ll focus on applications of the steady-state energy balance in which there is negligible change in kinetic and potential energies and no shaft work.

Common form of the conservation of energy equation

\[\begin{split}\sum_{\substack{\text{output}\\\text{streams}}} \left( \dot m \,\hat H \right )_{out} - \sum_{\substack{\text{input}\\\text{streams}}} \left( \dot m \,\hat H \right )_{in} = \dot Q\end{split}\]

Sensible heating or cooling

When a material is warmed or cooled without a phase change, we call this process sensible heating or cooling.

For sensible heating/cooling, the specific enthalpy, \(\hat H\), can be approximated as

\[\hat H \approx \bar C_{p} \, (T - T_{\text{ref}})\]

where \(\bar C_{p}\) is the heat capacity averaged from the reference temperature \(T_{\text{ref}}\) to the temperature of interest, \(T\).

The units of \(\bar C_{p}\) are energy per mass per temperature change, such as \(\si{cal/g.\degree C}\) or \(\si{Btu/lbm.\degree F}\).

Energy balance for sensible heating or cooling

\[\begin{split}\sum_{\substack{\text{output}\\\text{streams}}} \left[ \dot m \, \bar C_{p} \, (T - T_{\text{ref}}) \right ]_{out} - \sum_{\substack{\text{input}\\\text{streams}}} \left[ \dot m \, \bar C_{p} \, (T - T_{\text{ref}}) \right ]_{in} = \dot Q\end{split}\]

Exercise: Energy balance for sensible heating or cooling

Consider a process designed to process biomass for further processing into biofuels.

digraph simple_process { splines = ortho; bgcolor=transparent; rankdir=LR; node [shape=box]; "i1" [style=invis]; "i2" [style=invis]; "o1" [style=invis]; "o2" [style=invis]; "m" [label="processing device", color=black]; "i1" -> "m" [label="raw biomass"]; "i2" -> "m" [label="pure water"]; "m" -> "o1" [label="diluted biomass"]; "m" -> "o2" [label="contaminated water"]; }

Stream

Mass flow rate

Heat capacity

Temperature

(\(\si{kg/min}\))

(\(\si{J/g\,K}\))

(\(\si{\degree C}\))

raw biomass

\(15\)

\(0.9\)

\(90\)

pure water

\(120\)

\(4.2\)

?

diluted biomass

\(45\)

\(2.6\)

\(60\)

contaminated water

?

\(3.5\)

\(80\)

1. If the process is adiabatic, what temperature of pure water is required to produce diluted biomass at a temperature of \(\SI{60}{\degree C}\)? Assume a reference temperature of \(\SI{25}{\degree C}\).

2. If the temperature of the pure water is the same as in part a, how much heat (\(\si{W}=\si{J/s}\)) must be added or removed from the process to lower the diluted biomass temperature an additional \(\SI{10}{\degree C}\)?

Answers


Latent (phase change) heating or cooling

When a material changes phase (water to steam, water to ice) without a change in temperature, we call this process latent heating or latent cooling.

Considering phase changes with only one inlet stream and one outlet stream and where the phase change occurs isothermally at the temperature for which we have a value for \(\Delta H_{\text{phase change}}\), we can write

\[\hat H_{\text{out}} - \hat H_{\text{in}} = \Delta \hat H_{\text{phase change}}\]

This relationship, when combined with our energy balances, leads to

Steady-state energy balance for phase change

\[\dot m_{\text{phase change}} \, \Delta \hat H_{\text{phase change}} = \dot Q_{\text{phase change}}\]

Chemical reactions

\[\begin{split}\begin{align*} \sum_{\substack{\text{output}\\\text{streams}}} \left( \dot m \, \hat H \right )_{out} - \sum_{\substack{\text{input}\\\text{streams}}} \left( \dot m \, \hat H \right )_{in} &= \left ( \frac{\text{moles A reacted}}{\text{time}} \right ) \Delta \tilde H_{\text{reaction, A}} \\ &= r_{\text{consumption, A}} \, \Delta \tilde H_{\text{reaction, A}} \end{align*}\end{split}\]

Steady-state energy balance for systems with chemical reactions

\[r_{\text{consumption, A}} \, \Delta \tilde H_{\text{reaction, A}} = \dot Q_{\text{reaction}}\]

Exercise: Energy balance for a system with a chemical reaction

Suppose the following reaction is carried out in a chemical reactor: \(\ce{A + B -> C}\).

The reactor has a single inlet and a single effluent (outlet) and the entire reactor system is at constant density (\(\rho = \SI{0.9}{kg/L}\)).

The desired conversion of \(A\) is \(0.8\).

Operating conditions and parameter values

  • feed conditions:
    • \(\dot V_{\text{feed}} = \SI{50}{L/hr}\)

    • \(c_{\text{A, feed}} = \SI{1}{M}\)

    • \(T_{\text{feed}} = \SI{50}{\degree C}\)

  • heat (enthalpy) of reaction:
    • \(\Delta \tilde H_{\text{reaction, A}} = \SI{−200}{kJ/\text{mol of A}}\)

  • heat capacities:
    • \(C_{\text{p, feed}} = \SI{1.7}{kJ/kg.K}\)

    • \(C_{\text{p, outlet}} = \SI{2.1}{kJ/kg.K}\)

1. Assuming that the reactor is perfectly insulated (adiabatic), what would be the temperature of the effluent stream?

2. If we want the system to act isothermally and have the temperature of the effluent stream equal to that of the inlet stream, how much energy per time (in \(\si{Watts}=\si{J/s}\)) would we have to add or remove from the system?

Solution

Procedure for using the energy balance

Similar to those used in material balances, here are the recommended steps in solving problems in which energy balances are relevant:

  1. Draw a diagram if one is not already available.

  2. Write all known quantities (flow rates, densities, etc.) in the appropriate locations on the diagram.

  3. Identify and assign symbols to all unknown quantities and write them in the appropriate locations on the diagram.

  4. Write the appropriate simplified energy balance depending on whether the problem involves sensible heating/cooling, phase change, or chemical reaction. Along with the balance equation, write down the given information associated with that equation, such as average heat capacities, enthalpy changes for a phase change. or enthalpy changes of reaction.

  5. Construct appropriate material balance equations to aid in determining unknown flow rates or other material-re lated information. Continue to seek such equations, as needed, until the total number of equations equals the number of unknowns.

  6. Solve the equations to determine the desired unknown quantities.