# Coal Fired Power Station Unit

Under construction.

## Performance

The following performance calculations are available on station instances:

 Calculation Reference Output Tags Sent Out Heat Rate ASME PTC 4 & 6 .efficiency.net.heatrate Generated Heat Rate .efficiency.gross.heatrate Sent Out Efficiency .efficiency.net.use . .efficiency.net.percentage Generated Efficiency .efficiency.gross.use . .efficiency.gross.percentage

## Calculations

Heat Rate

Heat rate is a method of expressing efficiency. Heat rate defines the ratio of heat input in the form of coal at the Boiler, to the electrical power output of the generator. It is usually expressed in kJ/kWhr and varies with load.

Heat rate can be expressed in two ways: Units Generated (UG: uses generateor terminal voltage), and Units Sent Out (USO: includes auxiliary energy). To convert Heat Rate to efficiency you use the formula:

$$HR = \frac{3600}{\eta}$$

Where:

• $\eta$ = efficiency (%)

In CAS terms:

$$efficiency.gross.heatrate = \frac{3600}{efficiency.gross.use}$$

Generator Efficiency

To determine the overall unit efficiency you require the total energy extracted as electricity divided by the heat added to the boiler. The gross efficiency as calculated below:

$$\eta _{GEN} = \frac{Q_{GEN}}{Q_{rF}}$$

Where:

• $Q_{GEN}$ = Generated Output; and
• $Q_{rF}$ = Fuel Input.

But as stated in PTC 46 Section 3.1.2.1, another method for determining boiler efficiency is via the “energy balance” method. This is considered to be more accurate as it doesn’t rely on mass flow measurement. changing the equation to:

$$\eta _{GEN} = Q_{GEN} \times \frac{\eta blr}{Qblrout}$$

Where:

• $\eta blr$ = blr efficiency by losses (%); and
• $Qblrout$ = net boiler heat output (kW).

In CAS terms:

$$efficiency.gross.use = turb.gen.c2out.energyFlow.use \times \frac{blr.efficiency\_losses.use}{blr.netheatoutput.use}$$

Sent Out Power

The sent out power as caculated is:

$$Q_{SO} = Q_{GEN} - Q_{AUX}$$

In CAS terms:

$$turb.gentrf.c2out.energyFlow.use = turb.gen.c2out.energyFlow.use - turb.auxPower.use$$

Turbine Auxillary Power

The turbo auxiliary power as calculated is:

$$Q_{AUX} = Q_{UNITAUX} + Q_{TRANLOSS}$$

In CAS terms:

$$turb.auxPower.use = unit \ auxiliaries + transformer \ losses$$

Transformer losses are considered to remain constant and are therfore not calculated.

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