# Waterflow of coolant in nuclear reactor [closed]

The generator in a nuclear reactor delivers 580 MW electric power.

a) How much TWh electrical energy is delivered yearly if the reactor runs only 70 % of the time?

b) What flow of coolant (water) is required if the efficiency of the steam turbines is 35 % and the temperature increase of the water is 8 °C?

a) was easily solved but I am still posting it for contextual purposes; fact is that I am struggling with b).

Since the reactor is only running $70\,\%$ of the time, the energy-output (per second) is $406 \cdot 10^6 \text{ J}$. The steam turbines are only able to harness $35\,\%$ of that energy i.e. $1.421\cdot 10^8 \text{ J}$. Now, using $mc\cdot\Delta \text{T}$, we arrive

$1.421 \cdot 10^8 = m \cdot 4180 \cdot 8 \iff m \approx 4249 \text{ kg}$

A mass of $4249\,\text{ kg}$ (of water) entails approximately $4000$ litres but the key in my book says $32 \text{ m$^3$/s}$. What went wrong?

## closed as off-topic by user10851, tpg2114♦, Emilio Pisanty, Brandon Enright, Abhimanyu Pallavi SudhirNov 21 '13 at 16:32

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## 2 Answers

Since the reactor is only running 70% of the time, the energy-output (per second) is 406⋅106 J.

This is curious logic, and it demonstrates that you're thinking of the entire problem in a cumulative sense, just like how you thought the answer should be in $kg$, and not $kg/s$. Your solution uses different units, but we'll get to that later.

For starters, the plant is either on or off. It doesn't do any good to look at a running plant and say that it is using $50 kg$ of water right now. The entire grammatical sense of right now implies a rate, which should be intuitive to you. The question asks what "flow of coolant" is required to keep the reactor cool. In my opinion, that should imply mass per unit time (on this site we have no obligation to defend your course's solutions). Doing grading, it would probably be reasonable to accept either volume flow rate or mass flow rate.

When asked what the mass flow rate is, I would assume we're talking about the plant when it's running. We know this plant is running at $70 \text{ %}$ of full output, and we know the output, which is already in units of $J/s$, or Watts. Use your intuition to establish that the reactor's thermal output should be higher than the electrical output, and you know the conversion factor is $0.35$ between those two. Obviously, dividing the output by that gives a larger number, and this is what you should do. That give a thermal power of the reactor of $1657 \text{ MWth}$. Nuclear engineers often use the "th" addition to Watts in order to establish that it's thermal, and avoid confusion. The electrical power may be written $MWe$.

Now if we do what you're trying to do, it would be the following, as per $\dot{m} C_p \Delta T = Power$. I'll abuse notation with the SI prefixes a bit for illustration. The dots represent a rate of the variable it modifies.

$$\dot{m} = \frac{ \left( \frac{580 MWe}{0.35} \right) }{ \left( 4180 \frac{J}{kg ^{\circ} C} \right) \left( 8^{\circ} C \right) } = 0.05 \frac{M kg}{s} = 50,000 \frac{kg}{s}$$

$$\dot{V} = \frac{ 50,000 \frac{kg}{s} }{ \left( 1000 \frac{kg}{m^3} \right) } \approx 50 \frac{m^3}{s}$$

But this is wrong. The reason is that I intentionally used "everyday" units. For us, water is often around $25 ^{\circ} C$, so it does us good to remember the density of water at that temperature. After all, it's easy. Let us return to the question at hand:

b) What flow of coolant (water) is required if the efficiency of the steam turbines is 35 % and the temperature increase of the water is 8 °C?

Surely you know that a turbine runs off steam. But here, they give a temperature delta. That's because the flow they refer to isn't at the turbine, it's talking about the primary loop. The red loop in this diagram:

A nuclear thermal cycle works on the principle of a phase change, and lots of optimization tweaks that make the system more complicated. Giving a temperature delta for something on the secondary side would just be wrong and incoherent. Because of that, I can fairly reliably interpret the question as applying for the specific kind of reactor, which is a PWR, and for its primary loop flow.

I would assume that your question had other context available. Even so, $8^{\circ} C$ is too low for commercial reactors by a factor of 2 or 3, but I'm just splitting hairs here.

To solve the problem correctly, you need to use the method here, and use the correct values for:

• water density
• water specific heat

both at the temperature of the reactor primary loop. I don't know what this is, and it's necessary to solve the problem. Once you have that temperature, try your query in Wolfram alpha or something like that. Doing so is likely to revise the 50 m^3/s answer down to 32.

What went wrong:

Can you postpone some of the cooling of the full throttle reactor to the $30 \%$ of the time that the reactor is not running at full throttle? You have...

If the generators put out electrical energy at 580 MW, and that electrical energy represents only 35% of the thermal energy input, how much thermal energy must the cooling system accept each second?

EDIT: by multiplying the electrical power output by $70\%$ you are trying to spreading the cooling load over the whole year. You can't. You need to cool based on the power output at the moment. Does the fact that you drive your car fewer hours a day that your neighbour's identical car mean that your radiator can be smaller?

Secondly, you are applying the $35\%$ factor in reverse!

The generator is putting out electrical energy at 580 MW. This is 35% of the energy from the reactor. What's the total supplied from the reactor? Where does the other 65% go?

• So what you are saying is that we have to take into consideration the cooling of the reactor during the time it is shut down (i.e. $30\, \%$ of the time)? Well I'm not entirely sure but $0.35\cdot 580\cdot 10^6$? – QuantumClassicalPhysx Sep 21 '13 at 22:34
• I am afraid I am not following you, could you please show? – QuantumClassicalPhysx Sep 22 '13 at 13:25