The Gaussian Integral

The Gaussian or Probability Integral is an essential concept in mathematics particularly in the fields of probability theory, statistics and quantum mechanics.

The Gaussian integral, closely related to the erf function, is the integral of the one-dimensional Gaussian function over (-∞, ∞).

The Gaussian Integral can also be defined as the integral of the exponential of -x² over the entire real line.

It can be calculated using the trick of combining two one-dimensional Gaussians:

In this case, Since the variable in the integral is a dummy variable i.e, it integrates out in the end, we can rename from x to y.

When switching to polar coordinates we get:

Example

Prove the following:

Solution

Convert the integral into the polar coordinates (r, θ) where x² + y² = r², 

and dxdy = rdrdθ:

Evaluating the integrals:

Therefore:

CONTROL VOLUME APPROACH

After studying the various type of flows, we need to address how to determine the velocity in the flow field. Two approaches as used for this:

The Lagrangian Approach

• This is used in solid mechanics and involves describing particle’s motion by position as a function of time.
• Can be used to describe the motion of an object falling due to gravity: s = ½gt². At any time, the distance from the body’s original position is known.
• This approach is difficult to use in fluid mechanics because a fluid is a continuous medium i.e., a single fluid volume changes shape, and different fluid particles within the fluid volume are traveling at different velocities. Thus, due to the nature of fluids, the Lagrangian Approach is generally not a desirable analysis method.

The Eulerian (or control volume) Approach

• This is preferred in fluid mechanics.
• In this method, a region in the flow field is chosen for study. For example, consider flow draining from a sink, as shown below. A control volume is chosen around the region of study and is bounded by the dashed line called the control surface while everything outside is called the surroundings. The control volume or shape is chosen for convenience in solving the problem.

• Generally, the control volume shape is selected so that fluid and flow properties can be evaluated at locations where mass crosses the control surface or, if no mass enters or exits, where energy crosses the control surface. Furthermore, the control surface can move or change shape with time as illustrated above.
• The control volume is to fluids as the free-body diagram is to solids. 

The Eulerian approach is suitable in solving fluid mechanical problems. The aim is to develop equations of fluid dynamics that are conservation equations each developed from a general conservation equation:

• the continuity equation (conservation of mass),
• the momentum equation (conservation of linear momentum),
• the energy equation (conservation of energy).

Let N be defined as a flow quantity (mass, momentum, or energy) associated with a fluid volume or system of particles and n represents the flow quantity per unit mass. Thus:

Using the following image consider a system of particles at two different times: V₁ at t₁ and V₂ at t₂ where V₁ is bounded by the solid line and V₂ is bounded by the double line. V₁ consists of V_A and V_B while V₂ consists of V_B and V_C. The control volume is bounded by the dashed line.

The amount of flow quantity N contained in V₁ is the amount in V_A and the amount in V_B at t₁ which is N_A1 + N_B1. The amount of flow quantity contained in V₂ is the amount in V_B and V_C at t₂, which is N_B2 + N_C2. During the time interval, the change in N is, therefore:

To obtain a specific limiting expression for this last term, let us consider a differential area dA through which fluid particles flow:

The fluid velocity at the differential area, dA is Ṽ, which has components normal and tangential to dA: V_n and V_t. The tangential velocity carries no fluid out of the control volume with it as all fluid leaving dA is in the V_n direction. During the time interval Δt, the mass of fluid crossing dA is:

By substituting this into Equation 3, we get the general conservation equation:

Equation 4 gives us a relationship between the various quantities associated with a system of particles i.e., equation 4 simply means:

KINEMATICS OF FLOW

Types of flow and how to characterize them:

• Closed-conduit flows: Type of flow that is completely enclosed by restraining solid surfaces. Examples include flow through a pipe.
• Open-channel flows: In this flow type there is one surface exposed to atmospheric pressure. Examples include flow in a river and flow in a spillway.
• Unbounded flows: In this flow the fluid is not in contact with any solid surface. Examples are the jet that issues from a tap and the jet from a can of spray paint.

KINEMATICS OF FLOW

In fluid flow situations velocity often needs to be determined. However, velocity generally varies in the flow field. The flow can thus be classified according to how the velocity varies.

One-dimensional flow occurs if the parameters of both the fluid and the flow are constant at any cross section normal to the flow. It can also occur when represented by average values over the cross section. Although the flow velocity can change from point to point it remains constant at each location.

The figure below illustrates velocity distributions for a one-dimensional flow:

Two-dimensional flow occurs when the fluid or flow parameters have a gradient in two directions. For flow in a pipe, the velocity at any cross section is parabolic thus the velocity is a function of the radial coordinate. Additionally, a pressure gradient exists in the axial direction that maintains the flow i.e., a pressure difference between inlet and outlet that causes the fluid to flow. Even though the fluid flows in one direction, due to the pressure and velocity gradient, the flow is classified as two-dimensional. This is illustrated as follows:

Another example of two-dimensional flow occurs when at the constant area section velocity is a function of one variable but at the convergent section, velocity is a function of two space variables. Furthermore, a pressure gradient exists that maintain the flow. This is illustrated as follows:

Another example of a one-directional, two-dimensional flow where gradients exist in two dimensions is illustrated below:

Three-dimensional flow occurs when the fluid velocity or flow parameters vary with respect to all three space variables. Thus, a gradient exists in three directions.

Steady Flow occurs when conditions do not vary with time or when fluctuating variations are small with respect to mean flow values, and the mean flow values do not vary with time.

Unsteady Flow occurs when flow conditions change with time.

Quasi-steady Flow occurs in some unsteady flows, where it is permissible or even necessary to assume that steady flow exists to obtain a solution.

Ideal Reactors - Solved Examples

 Example 1

The following reaction is irreversible and first order:

The reaction is carried out in a PFR with 80 tubes. Each tube has a diameter of 5cm and a height of 1m. The feed consists of reactant A and 30% inerts flowing at 200kg/h with a pressure of 50bar and temperature of 600°C (873K). The output conversion is 90%. What is the average residence time? (The molecular weight of A is 60g/mol & R = 0.0821 L • atm/ mol • K)

Solution

The space time for a first order reaction in a PFR is given by the following equation:

The mass flow in each tube is: (200/80) = 1.25kg/h

In molar basis:

Thus, the initial concentration of A is given by:

We can now calculate the initial volumetric flow:

The volume of each tube can be calculated as follows:

The residence time can be calculated as follows:

The expansion parameter, ε_A can be determined as follows:

The rate constant, k, can now be solved using equation 1:

Thus, the average residence time can now be solved using:

Example 2

The following second order irreversible reaction is carried out in gas phase in a PFR reactor:

The PFR feed consists of 50% by weight reactant A (MW = 40) and the rest is inert (MW_inert = 20). The reaction occurs at a constant temperature of 70C and pressure of 5.25atm. The rate constant is 400m³/(kmol.ks).

For a production rate of R set at 30kmol/h with a 40% conversion, what should be the volume of the reactor?

Solution

For a second-order reaction, the rate can be expressed as:

Substituting the PFR design equation into the above equation and integrating we obtain:

With a mass basis of 1.0g, we can calculate ε_A. The molar values for each feed components are 0.5/40 = 0.0125 mol of A and 0.5/20 = 0.025 mol of inert. Thus, the volume balance is as shown below:

The expansion parameter, ε_A is = (0.05 - 0.0373)/0.0373 = 0.340. The initial concentration of A is:

Equation 1 can now be solved: 

Given that the output molar flow F_R = 30kmol/h, the volumetric flow rate, v_o is:

Thus, the reactor volume is:

Solved Example for Thermodynamics of Energy Conversion

For the following examples, Appendix B for steam values that I have referred to in these questions was obtained from:

M.D Koretsky, Engineering and Chemical Thermodynamics, Wiley, 2004.

Example 

Consider a power plant operating on a Rankine cycle using steam as the working fluid. The boiler pressure is 2.5MPa and the steam leaving the boiler is superheated to a temperature 126°C above its saturation temperature. The condenser temperature is 50°C and it discharges saturated liquid. The efficiency of the turbine is 0.90 and of the pump 0.80 as compared to reversible and adiabatic machines operating at the same pressure ranges.

a) Sketch the cycle on a T-S diagram.

b) Calculate the thermal efficiency of the cycle.

c) Compare the thermal efficiency of this Rankine cycle with the thermal efficiency of a Carnot cycle receiving heat at the saturation temperature of steam at 2.5MPa and discarding it at 50°C.

Solution

a) 

b) In order to calculate the thermal efficiency, we need to calculate enthalpy values at states 1, 2, 3 and 4.

c)

Thermodynamics of Energy Conversion

1. Vapour Power Cycles

Power cycle is a process where devices are used to continuously produce power. A typical vapor power cycle is the steam power plant shown below:

In this system:

• the boiler receives heat from a high temperature source and converts liquid water into high pressure and high-temperature steam.
• This steam is then fed to a turbine that drives an electric generator. The steam expands through the turbine and exhausts at a low pressure. This expansion process occurs adiabatically and is nearly as reversible as possible.
• The exhaust from the turbine is sent to a condenser where cooling water is used. A pump is used to increase the pressure of the liquid condensate. A small fraction of the work obtained from the turbine is used to operate the pump.

The most efficient cycle that can operate between two constant temperature reservoirs, T_H and T_C is the Carnot engine and its thermal efficiency is given by:

                           

Let us consider a typical nuclear power plant that has a capacity of 750,000 kW. Steam is generated from a boiler at around 300°C and 7 MPa. The condenser operates at around 40°C and 7 kPa. The thermal efficiency of a Carnot engine operating between these two temperatures is:

This means that, in the best-case scenario the power plant can only convert 45% of heat received by the boiler into work while the other 55% is discarded to the surroundings. 

In practice, the efficiency of an actual power plant will be much less than that of a Carnot engine. If we assume that η = 0.3 then:

The heat of vaporization of water is approximately 2260 kJ/kg. Therefore, the steam circulation rate is:

The specific volume of steam at 300°C and 7 MPa is 0.02947 m³/kg and a reasonable velocity for high-pressure steam in a pipe is 25 m/s. Hence, the diameter of the pipe is:

The representation of a Carnot heat engine on a T-S diagram is:

Shaft Work in a Reversible Steady-State Process

In the analysis of a power cycle, it is necessary to calculate the work required for the pump to increase the pressure of the exit stream from the condenser to the boiler pressure, i.e., process 1 → 2. In a reversible steady-state process, the equation to calculate the shaft work is derived as follows.

The first law of thermodynamics for a closed system is given by:

Note that equations 3 and 6 consist of only properties and their differential changes.

These properties and their changes are state functions and are not dependent on the path or process involved. Therefore, both equations hold for all reversible and irreversible processes and for a change of state in either a steady-state flow system or a closed system.

Now consider the first law of thermodynamics for a steady-state flow system

Thus, the reversible shaft work in a steady-flow process with negligible changes in kinetic and potential energies is given by:

The power required can be obtained by multiplying equation 12 by the mass flow rate, ṁ = Q̇ρ. The result is:

2. Rankine Cycle

Named after William J.M. Rankine (1820-1872), the Rankine cycle is the ideal cycle for a simple power plant. As was shown in the T-S diagram of a Carnot cycle from the previous section, the fluid at the exit of a condenser i.e., state 1 is a mixture of liquid and vapour. Practically, it is much easier to pump a pure liquid rather than a two-phase mixture. Thus, for a Rankine cycle, state 1 is a saturated liquid.

The Rankine cycle can be represented on a T-S diagram as follows:

In a Rankine cycle with superheat, the steam at the exit of the boiler is superheated as shown:

In analyzing the Rankine cycle, it is helpful to think of the efficiency depending on the average temperature at which heat is supplied and the average temperature at which heat is rejected. 

Thus, if the boiler and condenser pressures are the same, the efficiency of a Rankine cycle with superheat is greater than the efficiency of a Rankine cycle.