Heat Rate

A2) when the boiler water level starts to drop.

Q3.1) Define the acronym "MW".

Dear Mr.PWSlack,

The acronym "MW" refers to MEGAWATT.

DHAYANANDHAN.S

Dear Mr.RAMConsult,

I have one observation on your posting and you have mentioned "Turbine heat rate is for a specific piece of equipment (hint: Turbine) while the Plant heat rate is for the overall system including the Turbine."

The Plant Heat Rate (also Turbine Heat Rate) refers to the THERMO DYNAMIC ASPECT and THERMO DYNAMIC EFFICIENCY.

The Overall System Load - i.e., is the POWER CONSUMED to operate the Boiler and to generate steam and use it at the turbine, the Power for Boiler I.D, F.D, SA Fans, Feed Water Pumps, Fuel and Ash Handling System, Cooling Tower etc.etc. is TERMED AS "AUXILIARY LOAD" and there are standards for this Auxiliary Power consumption. For 64 Kg/cm^2 and 87 Kg/cm^2 the Norm is 8% or below 8% is a GOOD PERFORMANCE.

These definitions I have taken from the Text Book and I will try to copy and post here.

Pl. clarify.

DHAYANANDHAN.S

Dear Mr.mohnish-powermech,

1. TURBINE HEAT RATE: is the RATIO of ELECTRICAL OUTPUT CONVERTED TO HEAT EQUIVALENT/ NETT HEAT INPUT TO TURBINE.

Example: If Turbine Out Put is 10 MW = 10,000 KW.Hr = 10,000 X 860 = 86,00,00 K.Cal....(A)

NETT HEAT INPUT can be calculated from the Steam Flow/Hr. and the difference of heat content at inlet and exit of the turbine is the NETT HEAT DROP/Kg. of Steam...(B)

Now TURBINE HEAT RATE = (A)/(B). 2500 to 2600 K.Cal (or) 10,400 to 10,850 K.Joules for Plant Heat Rate or Turbine Heat Rate is a GOOD PERFORMANCE.

If it is in K.JOULES, it will be 10 MW = 10,000 KW x 3600 = 3,60,00,000 KJOULES.

To my knowledge TURBINE HEAT RATE and PLANT HEAT RATE is one and the same. On the other hand there is a factor called PLANT LOAD FACTOR. This is the Ratio of Power output in a given time to the ACTUAL POWER that could have been developed.

EXAMPLE: Let us say the Turbine Capacity is 10 MW, in a day of 24 Hours, the Turbine should have theoriticaly produced 240 MWHrs or 2,40,000 KW.Hrs. Suppose it has developed only 2,00,000 KW.Hrs. in 24 Hours, then PLANT LOAD FACTOR = 2,00,000/2,40,000 = 0.833. A good plant should give a P.L.F of 0.98.

2. When to start the Feed Pump:

Obviously based on the Low Permissible Water Level. Instead of ON/OFF for the Regular Feed Water Pump, a Start-Up Pump of 30% capacity of Feed Water Pump is to be installed to take care of ON/OFF of the Regular Feed Water Pump, and during start-up time, so that it can continuously work. This a better handling principle.

3. For LOW MW, capacity Super Critical Pressure Boiler will NOT BE ECONOMICAL or VIABLE. The Rate of Return on Investment and Operational Cost will decide the VIABILITY, and Minimum Capacity in MW for Super Critical Boiler.

DHAYANANDHAN.S

Mr. Dhayanandhan,

The more postings I read the more I become aware of the differences between the commercial world and the utility world. Although we talk about the same basic ideas, there are many areas where there is more than one way to describe the situation, which in turn leads to similar but not identical solution sets. Here's how a utility engineer defines Heat Rate, nothing is left to chance in this legal document, it also coincidentally answers OP's question more precisely than my original response.

This excerpt taken from the TXU 10-Q filed Nov 9, 2006:

Net Heat Rate Calculation

Net Heat Rate (NHR) is not directly measurable and will be determined analytically based on the average of tests using the NHR calculations described below:

Where

Where

I hope this helps explain where I was coming from with my Turbine Heat Rate answer.

Regarding these questions, I suspect that they are (good) interview questions, #1 tests specific engineering knowledge, #2 is a trick question since the answer has to be qualified by the type of boiler, once through boilers don't have a drum so the BFP has to be started before the burners are, and #3 tests specific industry knowledge, 500 MW is about the lower limit for the economics of supercritical once through boilers.