The 2011 Fukushima nuclear disaster was a situation in which backup pumps had fa
ID: 1858909 • Letter: T
Question
The 2011 Fukushima nuclear disaster was a situation in which backup pumps had failed and emergency cooling water had to be pumped between fuel-rod assemblies in the reactor core. Imagine a similar situation in which water in the core must flow at an average velocity of 0.5 m/s. The water has a density of 1026 kg/m^3 before entry to the reactor vessel but its temperature inside the core can be assumed to be approximately constant at 984 kg/m^3. There are 400 fuel-rod assemblies of square cross-section 0.1 m x 0.1 m. The reactor vessel has a 3 m internal diameter
A typical friction factor for flow between fuel-rod assemblies is 0.025 and a Darcy-Weisbach equation based on the gap D_2 between the assemblies is known to apply. The cooling-water entry to the bottom of the reactor vessel is rounded with negligible head loss. The exit is to the atmosphere and also has negligible head loss. The reactor is 100 m from the sea, the bottom of the fuel-rod assemblies where the cooling water enters is 2 m above sea level and the top where the water exits is 6 m above sea level. Steel piping 0.5 m in diameter is available on site. Calculate:
a) The head the pump must supply
b) The mechanical power the pump must deliver
State all assumptions and show working.
Explanation / Answer
a.) Flow rate is equal to cross sectional area times the velocity:
In this case the cros sectional area is equal to the areas of the 3m diameter reactor vessle minus the area taken up by all of the fuel rods.
A = pi*(1.5)^2 - 400 * (.1)(.1) = 7.07 - 4 = 3.07m^2
V_dot = 3.07m^2 * .5m/s = 1.535 m^3 / s
b.) Head, this will be a combination of the head loss due to the physical heigh difference between sea water and top of reactor vessle (10 m), the friction in the reactor vessle, and the friction in the supply piping.
Head loss due to reactor vessle friction loss:
H_f_reactor = f * ( L/D ) * (v^2 / 2*g)
f = friction factor : (.025)
L = lenght of pipe: (4 m)
D = Hydraulic dia: (4*A / P) : A is area, P is wetted perimiter
v = velocity: (.5 m/s)
g = acceleration due to gravity: (9.81 m/s^2)
D = (4 * 3.07) / (P)
P = (.1 + .1 + .1 + .1) * 400 + pi * 3
P = (169.424 m)
D = (4 * 3.07) / (169.424)
D = .0725 m
H_f_reactor = .025 * ( 4 / .0725) * (.5^2 / 19.6)
H_f_reactor = .0176 m
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Friciton due to 100 m of .5m dia supply pipe:
Since you are not given the friction factor of the steel pipe you will need to determine this:
The friciton factor can be determined from a Moody diagram. To use this diagram we need to first find the Reynolds number and surface roughness. A typical value for surface roughness of a steel pipe is 5*10^-5
Reynolds number = p*v*L / u
p = density (look this up in table for sea water) = 1026 kg/m^3
v = velocity = Flow rate / area of pipe = 1.535 / (pi * .25^2) = 7.821m/s
L = characteristic lenght (diameter) = .5
u = Dynamic viscosity (look this up as well) = .00108 Pa*s
Re = (1026 * 7.821 * .5 ) / .00108
Re = 3.715*10^6
Use this moody chart: http://en.wikipedia.org/wiki/File:Moody_diagram.jpg
first find the roughness on the right hand side (5*10^-5) then follow that curve to the left and find when it meets a Re # of ~3.715*10^6 follow this point to the left to get a friction factor of about .009.
Use the Darcy-Weisbach equation again to find the head loss:
H_f_pipe = .009 * (100m / .5m) * (7.821^2 / 19.6) = 5.612m
Head due to rise in heigh = 10m
Total head = 10 + 5.612 + .0176 = 15.63m
c.) Power = V_dot * p * g * H
v_dot = volumetric flow
p = density
g= gravity
H = head
Power = 1.535 * 1026 * 9.81 * 15.63 = 241.5 kW
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