Question 8.16: Assume that the reactor in the preceding problem suffers a c......
Assume that the reactor in the preceding problem suffers a control failure and undergoes a power transient P(t) = P(0)[1 + 0.25 t] where t is in seconds. The shut down system trips the reactor if the outlet temperature rises by more than 40 °C :
a. Determine the coolant outlet temperature transient and plot your result until the temperature rises by 40 °C .
b. At what time does the reactor shutdown system terminate the transient?
a)
Using equations [8.49] and [8.57], initial conditions y, and the stiff differential equation solver Radau (Mathcad), a vector Y was obtained. The first data column of Y after the time step column is the fuel temperature and the second column is average coolant temperature.
To find the core outlet temperature a third column needs to be added algebraically using the second column (average coolant temperature) and the inlet temperature (T_i)
R_{\bf f}:= \frac{\tau}{M_{\bf {fCf}}} = 2.963 \times 10^{-7}
T_{c0} := \frac{P}{2 \cdot W \cdot c_p} + T_i = 428.571
\mathbf{y}:={\left(\begin{array}{l}{{T}_{\mathrm{f0}}}\\ {{T}_{\mathrm{c0}}}\end{array}\right)}
Y:=\mathrm{Radau}({\bf y},0,300,10000,\mathrm{D}_{1})
\mathrm{YY}:=Y_{0,3}
\mathrm{YYY}:=Y_{0,3} + 40
P(t) := P \cdot (1 + 0.25 \cdot t)
T_{\bf f0}:= \bigg(R_{\bf f} +\frac{1}{2 \cdot W \cdot c_p}\bigg) \cdot P + T_i = 1.14 \times 10^3
D_1(t , y) := \begin {bmatrix}\frac{P(t)}{\mathrm{M}_{\mathrm{fCf}}} – \frac{1}{\tau}\bigl({\bf y}_0-\mathrm{T}_{{i}}\bigr) \\ \Bigg[\frac{1}{\mathrm{R}_{\mathrm{f}}}.\mathrm{(y_0-y_1)}-2\cdot\mathrm{W.c}_{\mathrm{p}}.\mathrm{(y_1-T_{i})}\Bigg].\mathrm{\frac{1}{M_{\mathrm{ccp}}}}\end {bmatrix}
{Y_{\mathrm{{n,3}}}:}=\left(2\cdot {Y_{\mathrm{{n,2}}}}\right)-{T_{\mathrm{{i}}}}
n := 0 , 1 .. 10000
b)
The reactor terminates the transient temperature when the temperature has risen by 40 degrees. This occurs at approximately 3.7 seconds as is seen on the above graph.
