# Demand Calculations in PE Power Exam

Welcome to my blog, where we will explore and uncover demand calculations in the Power exam, focusing on their significance in the PE Power Exam and throughout your power engineering career. Demand calculations are critical to power engineering, providing valuable insights into load analysis, capacity planning, and system optimization.

Mastering demand calculations are essential for success in the PE Power Exam, as it tests your ability to accurately estimate electrical load requirements and make informed decisions regarding equipment sizing, energy management, and grid stability.

This blog will explore the intricacies of demand calculations, unravel their importance in the PE Power Exam, and uncover their relevance in shaping a successful power engineering career. Get ready to enhance your technical expertise and gain a competitive edge in power engineering.

## Importance of Demand Calculations in Power Engineering

Demand calculations in the PE power exam and power engineering career are crucial in finding insights into electrical load requirements and consumption patterns. Accurate demand calculations enable efficient power system planning, design, and operation.

According to National Renewable Energy Laboratory (NREL) findings, up to 70% of total electricity costs are based on demand charges, making it essential to estimate and manage demand accurately. Power engineers can optimize capacity, ensure reliable supply, and implement energy efficiency and cost savings strategies by analyzing the market demand. It helps in

• Load forecasting and capacity planning,
• Proper equipment sizing and selection,
• Efficient resource allocation,
• Optimization of power system operation,
• Voltage stability analysis,
• Load flow and fault analysis,
• Billing and tariff determination,
• Demand response programs,
• Energy management,
• Power conservation,
• Renewable energy integration,
• Grid resilience and reliability assessment.

## Power System Components

Discussing the different components and their behavior is vital to understanding Demand Calculations in the PE Power exam. This section will explore the critical components that help you master demand calculations in the PE Power exam.

### Power transformers

The power transformer is crucial in electrical engineering by facilitating alternating current (AC) conversion to direct current (DC). This transformative process, also known as Rectification, has earned the power transformer alternate names such as AC-DC converter, aptly capturing its ability to convert between the two current types.

The underlying principle that empowers the power transformer is electromagnetic induction, employed to modify the voltage of AC electric power using magnetic fields. Transformers find extensive employment as single-phase and three-phase power supply units within power generation plants, substations, distribution systems, and industrial equipment.

Moreover, they serve as indispensable components known as step-down regulators, which effectively lower the voltage levels of electric power in industrial machinery.

### Types of Transformers

Let’s explore the 3 common types of transformers: Step-Up Transformers, Step-Down Transformers, Center-Tapped or Brdige Transformers. We will examine the functioning of diodes in converting AC to DC, the usage and purpose of diodes, and the primary and secondary coils of each transformer to get a deeper understanding of this crucial power system component.

#### 1. Step-Up Transformer

A Step-Up Transformer is designed to increase the voltage from the primary to the secondary coil. It is commonly used in power transmission to raise the voltage for long-distance transmissions, reducing power losses.

Let’s explore the functioning of diodes and their role in the conversion process.

Functionality

When an AC signal is applied to the primary coil, it induces an alternating magnetic field. This field then passes through the secondary coil, inducing a voltage. To convert the AC voltage into DC, diodes are utilized.

Diodes act as one-way valves, allowing current to flow in only one direction. In the case of a Step-Up Transformer, diodes are used to rectify the AC voltage into pulsating DC. During the positive half-cycle of the AC input, the diode is forward-biased and conducts current, allowing it to pass through to the load. During the negative half-cycle, the diode becomes reverse-biased and blocks the current flow.

Let’s assume a Step-Up Transformer has a primary coil with 500 turns and a secondary coil with 1000 turns. The input voltage is 120V AC. After rectification, we’ll calculate the final DC voltage, considering a voltage barrier of 0.7V for the silicon diode.

The voltage step-up ratio can be calculated by dividing the number of turns in the secondary coil by the number of turns in the primary coil:

Voltage Step-Up Ratio = Secondary Turns / Primary Turns

= 1000 / 500

= 2

The voltage output can be calculated by multiplying the input voltage by the voltage step-up ratio:

Voltage Output = Input Voltage × Voltage Step-Up Ratio

= 120V × 2

= 240V

However, it is important to account for the voltage barrier of the diode. Subtracting the 0.7V barrier, the final DC output voltage would be:

Final DC Voltage = Voltage Output – Diode Barrier Voltage

= 240V – 0.7V

= 239.3V

Therefore, the final DC voltage would be approximately 239.3V after the step-up transformation and rectification process.

#### 2. Step-Down Transformer

A Step-Down Transformer operates in the opposite manner of a Step-Up Transformer. It is designed to decrease the voltage from the primary to the secondary coil. Step-Down Transformers are commonly used to reduce high voltage for domestic and industrial applications.

Functionality

Like the Step-Up Transformer, a Step-Down Transformer also utilizes diodes to convert AC to DC. Let’s examine the functioning of diodes in this context:

In a Step-Down Transformer, diodes are employed to rectify the AC voltage into pulsating DC. During the positive half-cycle of the AC input, the diode becomes forward-biased and blocks the current flow. During the negative half-cycle, the diode is reverse-biased and conducts current, allowing it to pass through to the load.

Consider a Step-Down Transformer with a primary coil of 1000 turns and a second wave of 500. The input voltage is 240V AC. We will calculate the final DC voltage after rectification, considering a voltage barrier 0.7V for the silicon diode.

The voltage step-down ratio can be calculated by dividing the number of turns in the primary coil by the number of turns in the secondary coil:

Voltage Step-Down Ratio = Primary Turns / Secondary Turns

= 1000 / 500

= 2

The voltage output can be calculated by dividing the input voltage by the voltage step-down ratio:

Voltage Output = Input Voltage / Voltage Step-Down Ratio

= 240V / 2

= 120V

Accounting for the diode barrier voltage of 0.7V, the final DC output voltage would be:

Final DC Voltage = Voltage Output – Diode Barrier Voltage

= 120V – 0.7V

= 119.3V

Hence, after the step-down transformation and rectification process, the final DC voltage would be approximately 119.3V.

*Both the Step-up and Step-down transformers we discussed above operate in half-wave rectification. It means they are missing the DC cycle when the diode turns off.

#### 3. Bridge or Center-Tapped Transformer

A Bridge or Center-Tapped Transformer is a type of transformer with a center tap on the secondary coil. This center tap is a connection point for various applications, allowing different voltage outputs.

Functionality

Diodes convert the AC voltage into pulsating DC in a Center-Tapped Transformer (Bridge Rectifier). The diodes function similarly to those in Step-Up and Step-Down Transformers, rectifying the AC signal.

In a Center-Tapped Transformer, two diodes are used in a full-wave bridge rectifier configuration. A bridge rectifier, or a bridge transformer, is a configuration commonly used to convert alternating current (AC) into direct current (DC) in power systems. It combines diodes arranged in a bridge circuit to achieve full-wave rectification.

This section will discuss the functioning of diodes in a bridge transformer, the purpose of the configuration, and the resulting DC output. These diodes are oriented in a way that ensures at least one diode is forward-biased and conducting current during each AC cycle. Let’s discuss the usage and purpose of these diodes:

A bridge transformer consists of four diodes connected in a bridge circuit. The diodes are arranged to create alternate paths for the current flow during each half-cycle of the AC input. Let’s examine how the function of the diode in the bridge transformer:

• Diode 1 – During the positive half-cycle of the AC input, Diode 1 is forward-biased and conducts current, allowing it to pass through to the load.
• Diode 2 – During the positive half-cycle of the AC input, Diode 2 is reverse-biased and blocks the current flow.
• Diode 3 – During the positive half-cycle of the AC input, Diode 3 is reverse-biased and blocks the current flow.
• Diode 4 – During the positive half-cycle of the AC input, Diode 4 is forward-biased and conducts current, allowing it to pass through to the load.

In the next cycle, the diode operations reversed alternatively. It means you will be getting a DC cycle at each AC cycle, signifying the impact of Full Wave Rectification.

Purpose of the Configuration

The bridge transformer configuration serves two primary goals:

• Full-Wave Rectification – The bridge configuration enables full-wave rectification, which means that the AC input’s positive and negative half-cycles are converted into a continuous DC output. This results in a smoother and more consistent DC waveform than half-wave rectification.
• Improved Efficiency – By utilizing all four diodes, the bridge transformer configuration minimizes power loss and increases the efficiency of the rectification process. The diodes ensure the current flows in the desired direction, allowing maximum input power utilization.

DC Output of Bridge Transformer

The output of a bridge transformer is a pulsating DC waveform. The positive and negative half-waves of the AC input are rectified, resulting in continuous production. However, the DC output still contains some ripple, which is the residual alternating component of the waveform. Additional filtering components like capacitors are often used to further smoothen the DC output.

Let’s consider a bridge transformer with a primary voltage of 220V and a secondary voltage of 12V. We will calculate the final DC output voltage after rectification and subtracting the voltage barrier of the diodes.

We have:

Primary Voltage (VP) = 220V

Secondary Voltage (VS) = 12V

Voltage Barrier of Diodes (Vd) = 0.7V

To calculate the final DC output voltage, we need to account for the voltage drop across the diodes. Each diode has a voltage barrier of 0.7V when forward-biased.

Positive Half-Cycle

During the positive half-cycle of the AC input, Diode 1 and Diode 2 are forward-biased and conduct current, while Diode 3 and Diode 4 are reverse-biased and block current flow.

The DC output voltage during the positive half-cycle will be equal to the secondary voltage (VS) minus the voltage drop across the diodes (2 * Vd) since two diodes are in series:

VDC_positive = VS – (2 * Vd)

VDC_positive = 12V – (2 * 0.7V)

VDC_positive = 12V – 1.4V

VDC_positive = 10.6V

Negative Half-Cycle

During the negative half-cycle of the AC input, Diode 3 and Diode 4 are forward-biased and conduct current, while Diode 1 and Diode 2 are reverse-biased and block current flow.

The DC output voltage during the negative half-cycle will be equal to the secondary voltage (VS) minus the voltage drop across the diodes (2 * Vd) since two diodes are in series:

VDC_negative = VS – (2 * Vd)

VDC_negative = 12V – (2 * 0.7V)

VDC_negative = 12V – 1.4V

VDC_negative = 10.6V

Since the bridge transformer provides full-wave rectification, the magnitude of the positive and negative DC output voltages is the same.

Final DC Output Voltage

To calculate the final DC output voltage, we can take the average of the positive and negative DC voltages:

VDC_average = (VDC_positive + VDC_negative) / 2

VDC_average = (10.6V + 10.6V) / 2

VDC_average = 21.2V / 2

VDC_average = 10.6V

Therefore, the final DC output voltage after rectification in the bridge transformer is approximately 10.6V, considering the voltage barrier of the diodes.

### Transmission and Distribution Lines

Transmission and distribution play a crucial role in the efficient and reliable delivery of electrical energy from power generation sources to end consumers. This section explores the significance of transmission and distribution lines in power systems. An optimized power distribution ensures:

• Efficient Power Transfer – Transmission lines carry high-voltage electrical power over long distances from power plants to distribution substations. These lines are designed to minimize power losses, enabling efficient power transfer over vast geographical areas. By transmitting power at high voltages, transmission lines reduce resistive losses and allow for the transportation of large amounts of electricity.
• Voltage Regulation and Control – Transmission lines facilitate voltage regulation and control within the power system. Through the use of transformers and voltage regulators, transmission lines can adjust voltage levels to ensure a stable and reliable power supply to consumers. Voltage control is essential to maintain system stability and optimize power transfer efficiency.
• Grid Interconnection – Transmission lines enable the interconnection of multiple power sources and regional grids. They facilitate power transfer between generating stations, helping balance supply and demand across the grid. Interconnecting grids also improves system reliability and allows for integrating renewable energy sources in other locations.
• Distribution of Power to Consumers – Distribution lines carry power from distribution substations to consumers’ homes, businesses, and industries. These lines operate at lower voltages than transmission lines, ensuring safe and reliable power delivery to end users. Distribution lines are designed with a minimal voltage drop to maintain adequate voltage levels for consumers at various locations.

### Circuit Breakers and Protective Relays

Circuit breakers and protective relays are essential devices used in power systems to ensure the safety and protection of equipment, personnel, and the overall electrical infrastructure. This section highlights the significance of circuit breakers and protective relays in power systems. These crucial power system components ensure:

• Fault Protection – Circuit breakers and protective relays are critical in detecting and isolating electrical faults within the power system. These faults can include short circuits, overloads, and ground faults. Interrupting current flow during a fault condition, circuit breakers safeguard equipment from damage and prevent potential hazards, such as electrical fires and explosions.
• System Reliability – Circuit breakers and protective relays contribute to the reliability of the power system by isolating faulty sections and preventing cascading failures. When a fault occurs, protective relays sense abnormal conditions and send signals to circuit breakers, initiating their operation. This rapid response helps minimize downtime, improve system reliability, and reduce the impact of faults on the overall power grid.
• Selective Coordination – Protective relays and circuit breakers are coordinated to provide selective tripping during faults. This coordination ensures that only the specific circuit breaker nearest the fault is activated, isolating the faulted section while leaving the rest of the system operational. Selective coordination enhances system reliability, reduces disruption, and facilitates efficient troubleshooting and maintenance.

### Switchgear and Bus Systems

Switchgear and bus systems form an integral part of power distribution networks. This section highlights the significance of switchgear and bus systems in power systems. These critical components are instrumental in:

• Control and Protection – Switchgear is a central control point for electrical power distribution. It consists of various devices, including switches, circuit breakers, and protective relays, which enable the control and protection of electrical circuits. Switchgear allows for safe switching operations, isolation of faulty equipment, and restoring power supply.
• Reliable Power Distribution – Switchgear and bus systems ensure reliable power distribution by providing multiple paths for current flow. Bus systems connect various switchgear units and distribute power to different feeders or loads. They enable the flexibility to redirect energy in case of equipment failure or maintenance, ensuring uninterrupted power

### Resistive, Inductive, and Capacitive loads

Understanding the characteristics of resistive, inductive, and capacitive loads is essential for Power and Electrical engineers as it enables them to analyze and design circuits efficiently, considering the effects of these loads on power consumption, power factor, phase relationships, and overall system performance. Let’s discuss each in detail.

Resistive loads are electrical loads that primarily exhibit resistance as their main characteristic. Resistance is the property of a material or component to oppose the flow of electric current. In engineering terms, resistive loads are typically represented by purely resistive elements such as heating elements, incandescent light bulbs, and resistors.

• Power Dissipation – Resistive loads convert electrical energy into heat energy. The power dissipated in a resistive load can be calculated using Ohm’s Law and the formula P = V^2 / R or P = I^2 x R, where P is the power, V is the voltage, and R is the resistance.
• Linear Response – Resistive loads exhibit a linear relationship between voltage and current. This means that the current flowing through a resistive load is directly proportional to the applied voltage, following Ohm’s Law (I = V / R).
• Power Factor – Resistive loads have a power factor of 1, which means the current is in phase with the voltage waveform. This results in a unity power factor and does not introduce reactive power or distort the power supply waveform.

Inductive loads are electrical loads with inductance as their primary characteristic. Inductance is the property of a component or circuit element to oppose changes in current flow. Examples of inductive loads include electric motors, transformers, solenoids, and electromagnets.

• Energy Storage – Inductive loads store energy in their magnetic field when current flows through them. This energy is released back into the circuit when the current changes or is interrupted.
• Inductive Reactance – Inductive loads exhibit inductive reactance (XL), the opposition the inductor offers to the flow of alternating current. Inductive reactance increases with frequency and can be calculated using the formula XL = 2πfL, where f is the frequency, and L is the inductance.
• Phase Shift – Inductive loads introduce a phase shift between the voltage and current waveforms. The current lags behind the voltage waveform by 90 degrees in an ideal inductive load.

Capacitive loads are electrical loads that possess capacitance as their primary characteristic. Capacitance is the ability of a component or circuit element to store electric charge. Common examples of capacitive loads include capacitors, fluorescent lights with electronic ballasts, and electronic devices with power supplies incorporating capacitors.

• Energy Storage – Capacitive loads store energy in their electric field when a voltage is applied across them. This energy is released back into the circuit when the voltage changes or is interrupted.
• Capacitive Reactance – Capacitive loads exhibit capacitive reactance (XC), which is the opposition the capacitor offers to the flow of alternating current. Capacitive reactance decreases with frequency and can be calculated using the formula XC = 1 / (2πfC), where f is the frequency, and C is the capacitance.
• Phase Shift – Capacitive loads introduce a phase shift between the voltage and current waveforms. The current leads the voltage waveform by 90 degrees in an ideal capacitive load.

### Peak, Average, and RMS Demand

Peak, average, and RMS demand are essential concepts in demand calculations in the PE power exam that describe different power consumption aspects. Understanding these terms is essential for designing and managing power systems effectively. Let’s delve into these concepts

#### Peak Demand

Peak demand refers to the maximum amount of power consumed by an electrical load within a specific time period. It represents the highest point of power demand in a system. Peak demand is crucial for determining the capacity requirements of power generation, transmission, and distribution systems.

#### Average Demand

Average demand, also known as the mean demand, represents the average power consumed by an electrical load over a given time period. It provides a measure of the overall power consumption pattern. Average demand is useful for evaluating long-term power usage trends, load forecasting, and sizing power infrastructure.

#### RMS (Root Mean Square) Demand

RMS demand measures the effective or equivalent value of the demand waveform. It is calculated using the root mean square method, considering the time-varying nature of alternating current (AC) waveforms. RMS demand considers both the magnitude and the frequency of the waveform, providing a more accurate representation of the power consumed.

Let’s consider a residential area with the following power demand measurements recorded over 24 hours:

1. Peak Demand

To determine the peak demand, we identify the highest power demand recorded over the 24 hours. In this case, the peak demand is 5.7 kW, occurring at hour 9.

2. Average Demand

To calculate the average demand, we sum up all the power demand values and divide by the number of measurements:

Average Demand = (1.5 + 2.0 + 1.7 + … + 2.3 + 1.9) / 24

Average Demand = 81.4 kW / 24

Average Demand ≈ 3.4 kW

RMS Demand

To calculate the RMS demand, we square each power demand value, calculate the mean of the squared values, and then take the square root:

RMS Demand = √[(1.5^2 + 2.0^2 + 1.7^2 + … + 2.3^2 + 1.9^2) / 24]

RMS Demand = √[92.48 / 24]

RMS Demand ≈ √3.853

RMS Demand ≈ 1.96 kW

The peak demand in the given 24-hour period is 5.7 kW, the average demand is approximately 3.4 kW, and the RMS demand is approximately 1.96 kW. These values provide insights into the maximum, average, and adequate power consumption, aiding in power system planning, load management, and infrastructure sizing.

Diversity and load factor are essential concepts in Demand Calculations in the PE Power exam that help assess the efficiency and utilization of electrical loads. Understanding these terms is crucial for power system planning, capacity management, and optimizing energy usage. Let’s discuss each concept in detail.

#### Diversity

Diversity refers to the phenomenon where the total power demand of a group of electrical loads is lower than the sum of their peak demands. It arises due to the statistical probability that all loads within a group will not operate at peak demand simultaneously.

By considering the diversity factor, power system designers can optimize the sizing and capacity of power generation, transmission, and distribution infrastructure. The diversity factor allows for more efficient utilization of resources and helps avoid over-dimensioning of equipment.

Load factor measures the average energy utilization of an electrical load over a given period, typically expressed as a percentage. It quantifies the ratio of the average power consumed to the peak power demand during that period.

A higher load factor indicates a more efficient use of electrical energy. The load factor helps assess the overall system efficiency, determine billing rates, and optimize the capacity of power infrastructure. By improving the load factor, utilities and consumers can reduce energy wastage, increase system reliability, and maximize the cost-effectiveness of the power supply.

A load duration curve, also known as a load duration curve or an energy duration curve, is a graphical representation that shows the cumulative time a specific load or demand level is exceeded over a given period.

A load duration curve plots the load on the vertical axis and the cumulative time on the horizontal axis. The load is typically represented in percentage or power units (kW or MW). The curve starts from the highest load level and gradually decreases as the cumulative time increases.

#### Load Duration Curve – Key Insights & Takeaways

• Load Distribution – The load duration curve visually represents the load distribution over a specified period, usually in hours, days, or months. It illustrates the duration or percentage of time the load remains at various levels.
• Peak Load – The highest point on the load duration curve represents the peak load, the maximum load demand experienced during the given period.
• Base Load – The lower portion of the load duration curve, corresponding to extended periods with low load levels, represents the base load. The base load typically consists of essential loads that are continuously required to be supplied.
• Load Profile – By examining the load duration curve, it is possible to analyze the load profile and identify periods of high demand, peak hours, and variations in load patterns. This information is valuable for load forecasting, system planning, and resource allocation.

Consider a residential building with the following electrical loads connected:

Lighting: 20 kW

Air conditioning: 50 kW

Appliances: 30 kW

Demand Factor Calculation

Demand factor is a measure used in power engineering to quantify the ratio of the maximum demand of a system or equipment to the total connected load. It represents the relationship between the maximum expected demand and the total capacity of the system or equipment. Suppose the maximum demand recorded in the building is 75 kW. To calculate the demand factor:

Demand Factor = Maximum Demand / Total Connected Load

Demand Factor = 75 kW / 100 kW

Demand Factor = 0.75

The demand factor, in this case, is 0.75 or 75%.

Load factor measures the average energy utilization of a system or equipment over a given period. It quantifies the ratio of the average power consumed to the peak power demand during that period. Load factor indicates the efficiency and utilization of the system or equipment, helping in capacity planning, system optimization, and cost-effective energy management.

Suppose the average power consumption over a given period is 60 kW. To calculate the load factor:

Load Factor = Average Power / Maximum Demand

Load Factor = 60 kW / 75 kW

The load factor, in this case, is 0.8 or 80%.

Key Take Aways

The demand factor of 0.75 indicates that, on average, only 75% of the total connected load is expected to be in use simultaneously. This implies that the electrical loads have a diversity factor of 0.75, meaning not all loads operate at peak demand simultaneously.

The load factor of 0.8 indicates that, on average, the system utilizes 80% of its maximum capacity. This suggests that the system operates at a relatively high efficiency and adequate utilization level.

Therefore, the demand factor helps determine the appropriate sizing and capacity of electrical infrastructure, ensuring it can handle peak demand without being over-dimensioned. The load factor aids in optimizing energy consumption, identifying periods of high and low demand, and developing strategies for load management, demand response, and energy efficiency.

Demand Calculation in the PE Power exam and load factors calculation provide valuable information for load forecasting, system planning, and resource allocation in power engineering.

## Power Factor Correction

Power factor is the ratio of real power (kW) to apparent capacity (kVA) in an electrical system. It represents the efficiency of power utilization and is affected by the presence of reactive power.

Power factor correction is the process of improving the power factor of an electrical system by reducing or eliminating reactive power. It involves the addition of capacitors to offset reactive power, leading to improved system efficiency and reduced energy consumption.

Suppose the Power factor is not properly optimized and corrected for errors. In that case, it can cause troubles like

• Voltage Drop – Low power factor causes increased current flow, leading to voltage drop along the transmission and distribution lines. This can result in decreased equipment performance and operational issues.
• Overloading – Inefficient power factor increases the current drawn from the system, potentially leading to the overloading of transformers, cables, and other equipment. This can result in premature equipment failure and increased maintenance costs.
• Harmonics – Power factor correction capacitors can introduce harmonic currents into the system, affecting power quality. Proper capacitor selection and harmonic filters help mitigate this issue.
• Voltage Flicker – Unbalanced or low power factor loads can cause voltage flickering, leading to undesirable effects on sensitive equipment and lighting systems.

There are different techniques to optimize and correct the Power factor for smooth power and electric operations, such as

• Capacitor Sizing – Capacitors used for power factor correction are sized based on the reactive power (kVAR) needed to compensate for the reactive load in the system. Proper sizing ensures optimal compensation and avoids overcorrection or under-correction.
• Capacitor Placement – Capacitors are strategically placed at the point of use, such as near large inductive loads, to mitigate the reactive power generated by these loads. This placement minimizes the reactive power flow and improves the power factor locally.

### Conclusion

As you have completed this insightful journey into the horizontal and vertical depths of demand calculations in power engineering, we hope you have gained a deeper understanding of their significance in the PE Power Exam and your power engineering career. Remember, mastering demand calculations in the PE Power exam is crucial for exam success and excelling in real-world power engineering scenarios.

If you are preparing for the PE exam, we highly recommend consulting Study for FE – a reputable online resource offering comprehensive and dedicated PE Power exam preparation. With Study for FE, you can access expertly designed study materials and resources created by PE-licensed engineers, enabling you to confidently navigate the complexities of demand calculations and other key exam topics.

Invest in your professional growth and future success by choosing Study for FE as your trusted study companion. Start your journey towards becoming a licensed professional engineer and unlock a world of opportunities in the dynamic field of power engineering. Licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with consulting experience in design, commissioning and plant engineering for clients in Energy, Mining and Infrastructure.