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It is a tank-type 6 megawatt reactor [2] that is moderated and cooled by light water and uses heavy water as a reflector. It is the second largest university-based research reactor in the U.S. (after the University of Missouri Research Reactor Center) and has been in operation since 1958. [7] It is the fourth-oldest operating reactor in the ...
The cost of raw uranium contributes about $0.0015/kWh to the cost of nuclear electricity, while in breeder reactors the uranium cost falls to $0.000015/kWh. [54] Nuclear plants require fissile fuel. Generally, the fuel used is uranium, although other materials may be used (See MOX fuel). In 2005, prices on the world market for uranium averaged ...
Those blueprints contained designs for a power plant with a 100-megawatt pressurized water reactor. [3] The OPEN100 plans aim to standardize nuclear power plant construction to increase speed and cost-effectiveness, allowing plants to be built in under two years for a cost of $300 million. [ 4 ]
It's one of the world's biggest suppliers of high-grade uranium, selling $2.6 billion worth of nuclear fuel last year, turning $339 million of it into net income thanks to its low-cost operation ...
The reactors, known as Unit 3 and Unit 4 were originally expected to be completed in 2017 and cost $14 billion, but the second reactor only started commercial operations in April this year.
The most profound issue for microreactors is the cost per kWh, as microreactors lose the power-of-scale advantages for economic efficiency. Design, operation and maintenance costs can make these low-power nuclear reactors prohibitively expensive. [13]
For example, if a 3000 MW thermal (equivalent to 1000 MW electric at 33.333% efficiency, which is typical of US LWRs) plant uses 24 tonnes of enriched uranium (tU) and operates at full power for 1 year, the average burnup of the fuel is (3000 MW·365 d)/24 metric tonnes = 45.63 GWd/t, or 45,625 MWd/tHM (where HM stands for heavy metal, meaning ...
To understand how is used, consider a reactor operating at 20 MW and Q = 2. Q = 2 at 20 MW implies that P heat is 10 MW. Of that original 20 MW about 20% is alphas, so assuming complete capture, 4 MW of P heat is self-supplied. We need a total of 10 MW of heating and get 4 of that through alphas, so we need another 6 MW of power.