Applying cooling at the Condenser....

[pranj5]

We have an STG based thermal power plant in our hand and the steam flow rate is around 10 kg/sec or 36000 kg/hr. The Condenser is kept at 10 kPa pressure and the temperature there is 45°C. Now, a cooling system of capacity 142 ton of ice has been applied to the Condenser. I don't want to give details of the cooling system here, but consider that 142 ton of ice at 0°C per day i.e. 5.923 ton ice at 0°C per hour has been continuously added to the Condenser.
I want to know what can be the possible temperature and pressure at the Condenser if such a cooling has been added. With the previously mentioned set up, the output was 10 MW. I also want to know how much increase in output can be achieved with that kind of cooling.

 

[oz93666]

You say the low pressure end was 10kPa (tenth of an atms) ...and that is indeed the VP of water at 45C ... you are reducing this to at least , perhaps 8C where the VP is 1kPa ...so can expect a pressure change of around 9kPa to your advantage ... (reducing to 0C from 8C will only get another 0.4KPa)

But you haven't told us the high pressure figure , this is needed to calculate the power increase.

 

[pranj5]

It's 50,000kg/hr 10 barG (11 barA) at 345C. I hope that will be enough. Kindly keep in mind that the output is 10 MW by keeping the pressure at the Condenser 10 kPa and temperature to be 45°C.

 

[oz93666]

It's 50,000kg/hr 10 barG (11 barA) at 345C. I hope that will be enough. Kindly keep in mind that the output is 10 MW by keeping the pressure at the Condenser 10 kPa and temperature to be 45°C.

This is not my area of expertise, hopefully another member can offer some input , failing that this page may help .. https://en.wikipedia.org/wiki/Rankine_cycle .

It occurred to me that efficiency could also be increased by keeping electrical generators and transformers as cold as possible.

 

[russ_watters]

There is no simple answer to this question because the components of the cycle all work together and changing what happens to one changes what happens to the others until a new equilibrium is found (or a control system objects and changes it back!). In addition, depending on how the ice is added to the condenser, it may interfere with the normal operation of the condenser and could even reduce the heat rejection capability of the condenser (by interrupting airflow).

However, based on the amount of steam you are condensing and amount of ice you are adding, the cooling capacity of the ice is a tiny fraction (much less than 1%) of the condenser capacity, so at best the impact will be negligible.

 

[Mech_Engineer]

You'll need to calculate the process efficiency for your two cases, and then compare. Based on your lengthy conversations with several members here about the finer points of thermodynamic analysis (Chester for example), I'm guessing you have everything you need.

If you need some extra information, I'd recommend taking a look at these resources:

• MIT Class Notes: Rankine Power Cycles
• University of Virginia Excel Calculation Sheets: http://www.faculty.virginia.edu/ribando/modules/xls/Thermodynamics/
• Steam Fluid Properties: NIST Thermophysical Properties of Fluid Systems

 

[pranj5]

There is no simple answer to this question because the components of the cycle all work together and changing what happens to one changes what happens to the others until a new equilibrium is found (or a control system objects and changes it back!). In addition, depending on how the ice is added to the condenser, it may interfere with the normal operation of the condenser and could even reduce the heat rejection capability of the condenser (by interrupting airflow).

I have said that the cooling is equivalent to that amount of ice. That doesn't mean I want to insert ice directly at the Condenser. So, no question of Condenser performance disruption by introduction of ice.

However, based on the amount of steam you are condensing and amount of ice you are adding, the cooling capacity of the ice is a tiny fraction (much less than 1%) of the condenser capacity, so at best the impact will be negligible.

At least I can say that it will reduce the temperature and pressure at the Condenser and that means increased efficiency and output. The question is how much.

 

[russ_watters]

I want to know what can be the possible temperature and pressure at the Condenser if such a cooling has been added.

You seem to be ignoring all of the caveats and this sounds like a real application, so I'm hesitant to provide theoretical answer that is meaningless in the real world. But if that's what you want*, here's what I would do:

Take the current enthalpy of the water leaving the condenser and subtract the enthalpy being removed by the ice. Then use a steam table to find the saturated water temperature and pressure at that enthalpy.

With the previously mentioned set up, the output was 10 MW. I also want to know how much increase in output can be achieved with that kind of cooling.

Are you also increasing the boiler and feedwater pump capacity in order to keep the supply steam pressure and temperature constant?

*I actually supplemented the condenser capacity of my home air conditioning system and measured capacity and efficiency improvements. But that was a 5 kW system, I owned it, and I fully understood the potential risks of messing with it. If you are *actually* planning on messing with the operation of a 10MW power plant, I would advise you stop and we will have to close the thread for liability reasons. Please answer if this is a (potential) real application or a theoretical exercise.

 

[Mech_Engineer]

The numbers that have been provided for before and after the turbine also appear to approach saturated steam. If this is the case, cooling the steam further will reduce the quality, and potentially damage the turbine. From what I understand you don't want to have steam with a quality lower than about 90% in a steam turbine because the droplets can be abrasive and damage the blades.

 

[pranj5]

Take the current enthalpy of the water leaving the condenser and subtract the enthalpy being removed by the ice. Then use a steam table to find the saturated water temperature and pressure at that enthalpy.

Point is, how much enthalpy can be subtracted from the ice can only be deduced if we know the final temperature.

The numbers that have been provided for before and after the turbine also appear to approach saturated steam. If this is the case, cooling the steam further will reduce the quality, and potentially damage the turbine. From what I understand you don't want to have steam with a quality lower than about 90% in a steam turbine because the droplets can be abrasive and damage the blades.

No. My main motto is to reduce both pressure and temperature at the Condenser and want to know how much output gain can be achieved by that.

The numbers that have been provided for before and after the turbine also appear to approach saturated steam. If this is the case, cooling the steam further will reduce the quality, and potentially damage the turbine. From what I understand you don't want to have steam with a quality lower than about 90% in a steam turbine because the droplets can be abrasive and damage the blades.

What you have said can be true if the pressure at the Condenser will remain constant. But with the temperature, the pressure too at the Condenser too will be reduced.
In fact, this thread is both theoretical and practical. There is no question of altering a 10 MW power plant. But, I want to know if a 10 MW power plant with the said input parameters can be built, where I can set the temperature and pressure of the Condenser.[/QUOTE]

 

[russ_watters]

Point is, how much enthalpy can be subtracted from the ice can only be deduced if the know the final temperature.

You can get pretty close by guessing the final temperature of the water, but if you want to calculate it, then it is just a temperature of a mixture problem. Still not too difficult.

 

[pranj5]

Just calculated it and found that just the latent heat of vaporisation of the amount of ice will bring the temperature to 20°C level and the pressure will be 0.023 barA. Now, I want to know how much power output can be increased with such lowering of temperature and pressure.

 

[pranj5]

At 45°C, the saturated steam pressure is 0.096 barA and the gross enthalpy of steam is 2582.45 kj/kg. That means the amount of enthalpy embedded in that amount of steam is 129122500 kJ. While the amount ice added to it at the Condenser is 7000 kg/hr and the amount of cooling just as the latent heat of freezing that they can deliver is 2352000 kJ. Therefore the gross enthalpy after introducing the cooling is 12670500 kJ and dividing that with 50000 kg means 2535.41 kj/kg. That is the enthalpy level of steam at 19°C and 0.022 barA pressure.

 

[Mech_Engineer]

@pranj5 go ahead and calculate all necessary parameters and the plug into the linked calculator I provided above (http://www.faculty.virginia.edu/ribando/modules/xls/Thermodynamics/). This will help you determine the cycle efficiency, and you should be able to modify numbers to see the effect of changing certain parameters.

In simple terms, all you need to do is determine the fluid state at process steps 1-4 in the Rankine cycle, once you have those you have everything you need to easily calculate efficiency.

 

[oz93666]

I'm finding it hard to believe this is a real project ...OP says .." 142 ton of ice at 0°C per day i.e. 5.923 ton ice at 0°C per hour has been continuously added to the Condenser..." just how would you do that in practical terms ... even if surrounded by snow this has to be mechanically moved to the condenser ...how would you do that? ...the cost would never be covered by the slight increase in efficiency ... if piping in the 'cold' with antifreeze via a heat exchanger the pipes would need to be constantly moved as the snow melted, also the heat transfer would be too slow ...pumping costs ...

 

[russ_watters]

Just calculated it and found that just the latent heat of vaporisation of the amount of ice will bring the temperature to 20°C level and the pressure will be 0.023 barA.

Vaporization? Do you mean fusion or are you actually thinking of vaporizing the water after you melt the ice? I'm not sure how you would even do that and if you did it would certainly interfere with the functioning of the condenser instead of just augmenting it.

If you actually just calculated using the latent heat of fusion, I think there might be something wrong with the calc: I got a much different answer. Can you post what, exactly, you did?

 

[Mech_Engineer]

@pranj5 any updates? I'd like to see your calculations regarding the Rankine process you're hoping to modify...