Question 17.7: Consider a DC power supply consisting of a generator with a ......
Consider a DC power supply consisting of a generator with a constant failure rate of \lambda_{1} = 0.0003 and a standby battery with a failure rate \lambda_{2}= 0.0005. Assume that both the generator and the stand-by battery have exponential time to failure distributions. Assume that the switching circuit has a known reliability of 0.98 for one switching operation. When the generator fails, then the switch turns on the standby battery. The reliability block diagram of the circuit is shown in Figure 17.7.
(a) Find the reliability of the above system for 15 hours of operation.
(b) Find the MTBF for the system.
(a) Using Equation 17.32, we have
R_S(t)=R_1(t)+p_{S W} \int_0^t f_1(x) R_2(t-x) d x (17.32)\\ =e^{-\lambda_1 t}+p_{S W} \int_0^t \lambda_1 e_1^{-\lambda_1 x} e^{-\lambda_2(t-x)} d x \\ =e^{-\lambda_1 t}+p_{S W} \lambda_1 e_1^{-\lambda_2 t} \int_0^t e^{-\left(\lambda_1-\lambda_2\right) x} d x \\ =e^{-\lambda_1 t}+p_{S W} \frac{\lambda_1}{\lambda_2-\lambda_1}\left[e^{-\lambda_1 t}-e^{-\lambda_2 t}\right].
Hence,
R_S(15)=e^{-0.0003 \times 15}+0.98 \frac{0.0003}{0.0005-0.0003}\left[e^{-0.0003 \times 15}-e^{-0.0005 \times 15}\right] \\ =0.99551+0.98 \times 1.5 \times(0.99551-0.992528) \\ =0.99989 \text {. }(b) \text { MTBF }=\int_0^{\infty} R_S(t) d t=\int_0^{\infty}\left[e^{-\lambda_1 t}+p_{S W} \frac{\lambda_1}{\lambda_2-\lambda_1}\left[e^{-\lambda_1 t}-e^{-\lambda_2 t}\right]\right] d t \\ =\frac{1}{\lambda_1}+p_{S W} \frac{\lambda_1}{\lambda_2-\lambda_1}\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right] \\ =\frac{1}{\lambda_1}+p_{S W} \frac{1}{\lambda_2}=\frac{1}{0.0003}+0.98 \frac{1}{0.0005}=5293.333 \text { hours. }