Summary 

• General presentation of the project   
• Problematic  
• Assenssement and improvement of energy loss in the inductive receptors 

 General presentation

Under the project of Senior Year, we are going to present us our project focused on the collective catering

We have chosen this thematic because she respond to the expectations regarding the issue of saving energy but also sustainable development.

General presentation

The goal of our project is to improve, in the context of the issue of sustainable delevelopement, the consomations and losses in the collective catering. This in order to fully optimize the use of energy in order to reduce costs and losses.

For this we will study the technological and sustainable scientific solutions able to respond our problem.

       Assenssement and improvement    of energy loss in the inductive receptors

The purpose of my part is to make a complete assessment of energy losses in the inductive receptors in the restaurant.

And explore scientific and technological solutions possible in order to minimize and reduce costs and optimize the use energy.

-> For this I will mainly study the phenomenon of compensation to determine if this is a solution to improving the energy losses.

       Assenssement and improvement    of energy loss in the inductive receptors

                                             Introduction

Our collective restaurant possess different engines such as ovens, the trays conveyor, hoods, ventilation box ...

All these systems use electric AC and involve two forms of energy: active and reactive power. In processes using electric power alone active energy is transformed in the production tool into mechanical, thermal, light energy etc ... The reactive power is used in particular in the power supply of magnetic circuits of the motors.

       Assenssement and improvement    of energy loss in the inductive receptors

                                             Implications

The flow of active and reactive power causes active losses and voltage drops in the conductors. Active losses reduce the overall network performance
The transport of reactive power on the restaurant's indoor network has the following disadvantages:
 

• overload or sizing of electrical installation (transformer, cables, etc..)

• most important active losses in these works

• Increase the bill.

       Assenssement and improvement    of energy loss in the inductive receptors

We have seen that all engines and all devices running AC and comprising a magnetic circuit absorb two forms of energy:

• A so-called active power, which is manifested by a work on the shaft of a motor for example.

• A so-called reactive power, which only serves to magnetize the iron magnetic circuit.

At each of these energies is an actif power (Ia) in phase with the line voltage and reactif power (Ir), also called magnetizing current. The latter being shifted 90 ° back to the active power. Both active and reactive power up vectorially to form the apparent power, offset by an angle Phi to the active power. This apparent power is the one that runs through the various conductors of the circuit from source to receiver included, which among other causes overheating of these conductors, so the energy loss Joule effect.

Solutions

       Assenssement and improvement    of energy loss in the inductive receptors

So the most obvious solution will reduce the phase reducing the cos (phi) to the close as possible to the value 1 (0.94).

For that we must seek the scientific and technological solution to reduce reactive power, or rather provide energy needed for circuit directly from the restaurant

Solutions

                                                                  Active Power               Reactive Power

Ventilation motor (ovens):                        2x 6,1kW                       2x3,78 KVar

Trays conveyor motor:                              0,454 kW                         0,605 KVar

Motor hoods                                                   1,5 kW                           1,36 KVar

So                                           ΣP = 14,15 kW              ΣQ = 9,52 kVar

tan(φ) = Q/P = 952/14,15 = 0,67

Donc φ =33,8 et cos(φ) = 0,83

Calculations of total active and reactive power

The goal is to bring the cos ( φ ) to = 0.94 therefore φ=19.95

=> Q2 must be equal to D x tan ( φ ) is 14.15 x tan ( 19.95) Q2 = 5.136 kVar

The capacitors must provide reactive power of 9.52 to 5.136 = 4.384 kVar

We will now determine the capacity of capacitors by type of coupling.

    Calculations capacitors capacity

.Couplage Etoile pour la batterie de condensateur :

The power consumed on a phase is identical on the 3 phases The capacitors are connected in star so they are subject to a simple voltage ( phase and neutral ) The power supplied by the capacitors

Qc=-V².C.w   (avec Qc = Q/3)

So Qc = 1,46x10^3 = (-230)²xCx2.π.F

=> C =87,8µF

.Couplage Triangle pour la batterie de condensateur :

The power consumed on a phase is identical on the 3 phases The capacitors are connected in triangle so they are subject to a phase voltage ( between phases) The power supplied by the capacitors

 Qc=-U².C.w (avec Qc = Q/3)

So Qc = 1,46x10^3 = (-400)²xCx2.π.F

=> C = 29µF

Interesting triangle coupling capacitor capacitance is less important ...

Calculations capacitors capacity