This is our ‘crown jewel’. Heat energy from whatever source is the ‘fuel’ for this power plant. The disadvantages of waters’ freezing temperature and low boiling temperature is offset by its advantages of being universally available and of widely known characteristics. At certain locations, low grade heat is basically in unlimited supply and of essentially no cost – geothermal sites, or using thermal-solar collection. The fuel could be nuclear ‘hot’ waste, or traditional power plant waste heat. Heat can be stored for later use. This type of power plant COULD use traditional fuels but it’s advantage is its ability to extract usable energy from low, easily achievable, temperatures.

The law of conservation of energy states “The total amount of energy in a closed system is constant.” This means that “energy can never be created nor destroyed but can only change form.” [ignoring Einstein’s E=MC**2 for a moment]

   We know very well that electric power can be produced by a Rankin cycle using fossil and nuclear fuels. That’s the way it’s almost universally done now and has been since Thomas Edison lit up that first block of New York City with his electric plant. Typically, today the superheated steam is 1,000 PSI at 1,000 ◦F on the boiler side and ambient temperatures on the condenser side for a difference of 900 ◦F.

 [η_max =   maximum energy conversion efficiency = 1 – (low temp. / high temp.)]                                                                                             waste heat / input heat

They would use higher temperatures if they could afford the metals which could maintain strength at higher temperatures but the economics of using cheaper mild steel dominates the decision. Consequently we don’t efficiently use the 2,500 ◦F+ combustion temperatures of fossil fuels. Why do we throw away still ‘hot’ spent reactor fuel? Because it is no longer capable of achieving those 1,000 ◦F steam temperatures – nuclear power plants are engineering design clones of coal fired power plants.

 η_max = 1 – (150°F + 460) / (1000°F + 460) =                                                                        
710°R / 1460°R = .4863                                          
1 - .4863 = .5136    51% maximum possible Carnot efficiency at 1,000ºF
                 =               76 % maximum possible Carnot efficiency at 2,500ºF

=If one were able to produce electric power with lower temperatures we would be able to place almost ALL of the radioactive waste BACK into productive use and use it for another 20,000 years or until it couldn’t sustain a 200 ◦F or so boiler temperature. We would no longer need to mine coal or burn natural gas nor heavy crude oil to produce electricity. We’d achieve energy independence from the OPEC nations. I’d say that this is a desirable thing to do. Given the recent power blackouts I assert that adding millions of ‘grass-roots’ power producers would add diversity and robustness to our national grid.

The 'nay Sayers' may point out that with low temperature differentials the Carnot 'efficiency' is low. May I point out that once the capital outlay for the plant is made and the power plant is operational, the ongoing fuel source / cost is essentially free - ever after.

A booming vibrant economy is closely tied to the availability of abundant, inexpensive energy. Of course other factors are important also such as respect of property rights, individual freedom and levels of education, but abundant, inexpensive energy is a key factor. If one was in a position to ‘save the world’ from greenhouse gas pollution, and be able to undercut the production cost of every other energy producer on the planet don’t you think that one would have an ‘economic advantage’? I do. The underdeveloped countries of the world could no longer blame the USA for damaging the environment. That dubious distinction would fall to others.

And then there is the potential, in the future, to find a way to capture the waste /condenser heat and recycle it back to the input reducing the waste to approaching zero. Click on this.

 

The Difference between kWh and kVA

"Question: Please explain the difference between billing based on KWH and KVA. The benefits and conversions. "

 

kWh versus kVA

kilowatt-hour (kWh) is the measure of energy consumption (power over time). As an example, a 60 watt light bulb burns power at the rate of 60 watts (or 0.060kW). In one hour, the bulb will consume 60 watts times 1 hour, or 60 watt hours). In 12 hours, that same bulb will have consumed 60 watts times 12 hours, or 720 watt hours (0.72kWh). Therefore, a kWh is a unit of energy.

 

There are two kinds of power being provided to a given facility.

 

Kilovolt-Ampere (kVA) is a unit of electrical power, often referred to as Apparent Power. This is what the utility must supply to its customers on the primary side of the facility transformer and is the basis for sizing power plants.

 

Kilowatt (kW) is also a unit of electrical power (equal to 1,000 watts), referred to as Real Power. This measure of power is also called demand. For a 60 watt bulb, the demand would be 60 watts.

 

Different facilities may use the same amount of real power (kW), but different supply levels of apparent power (KVA) depending on the kind of equipment in the facilities. Equipment with lots of magnetic fields (like motors) will require more apparent power to feed the same real power demand. If you know the kilowatt demand (real power), then you can calculate the electrical energy consumed over time (kW x time = energy).  There is also a relationship between real power and apparent power that is based on the concept of power factor. If you know your real power (kW) and your power factor, you can calculate your level of apparent power (KVA).

 

Determining kVA and kWh

 

The following is an equation that can be used to determine kVA from kW:

 

Power Factor = kW/KVA

            or

KVA = KW/Power Factor 

 

If the useful power that the customer consumes is described as "real power" (KW), then the component of lost power is sometimes referred to as reactive power, or kilovolt amps reactive (KVAR). The total amount of power that the utility has to supply to the customer is given in KVA.  If all electric loads were resistive (as in lights, resistance heaters etc), we would not have to worry about power factor. The electrical system would be operating at its highest efficiency. But there are also motors, transformers and capacitors that are components of electrical systems, and these devices and others create induction and capacitance. The net effect of these inductive and capacitive devices results in inefficiency. When this happens, some of the energy is lost to generate the magnetic field of the motor, or the energy is lost in the creation of the stored energy for the capacitor.

 

For example to determine the kWh charge:

 

Let's say that your organization has 200 bulbs that are 60 watts each that are turned on for 12 hours each day. In 30 days, your electric bill would amount to 4,320kWh (200 X 60w X 12hrs X 30 days divided by 1,000). This is the quantity of energy consumed, and may cost you $302 if you are paying $0.07/kWh.

 

However, at any given moment, your organization is provided power at the level of 12kW (200 bulbs X 60 watts each divided by 1,000 to get kilowatts). If your power factor is 0.90, then the kVA provided would be 13 kVA (12 kW/0.90 power factor). If your demand rate around $13 per kVA then your monthly bill may also include a $169 (13 kVA x $13/kVA) additional demand charge to the bill.