4 ms·
"[During failure,] only a small amount of water transfer (about ten garden hoses worth) is necessary [...] to keep the reactor stable." How much is "ten garden
by delinka 11y ago
"[During failure,] only a small amount of water transfer (about ten garden hoses worth) is necessary [...] to keep the reactor stable."
How much is "ten garden hoses worth"? Are we talking running ten garden hoses at full capacity for a few seconds? A couple hours? I very much dislike comparisons like these in reporting, especially when discussing safety.
- astrodust 11y agoThey didn't say how long the hose is. Maybe it's a mile? To the moon and back?
- jessaustin 11y agoI've never seen a garden hose a mile long. Typically they are shorter than 25m.
- shawn-furyan 11y agoI would think the author means flow rate, not static volume. That would be a hopelessly unintuitive, and as you demonstrate, incomplete analogy if it were meant to represent the static volume needed.
- kevinchen 11y agoI think it's meant as a rate measurement, not volume. They want readers to visualize spraying the reactor with 10 garden hoses as the rate of cooling needed in an emergency.
- afarrell 11y agoWater transfer seems like it calls for a measurement of flowrate. So the real question is: how fast is the water flowing through a pipe of that diameter?
- TeMPOraL 11y agoDoesn't that depend on pressure as well?
- JorgeGT 11y agoYes. As a first approximation we can use Bernoulli equation, which states that 0.5V^2+gz+p/rho [1] remains constant through a streamline. If we assume incompressible flow, no friction, perfect discharge, no altitude change etc., and that speed in the pressurized reservoir is null, we can obtain that p_reservoir = 0.5 V^2+p_ambient and thus V = sqrt(2(p_reservoir - p_ambient)) if I'm not mistaken. --- [1] Where V=velocity, g=gravity accel., z=height, p = pressure, rho=density