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Reliability with redundancy



There has been some discussion here lately about electrical network
reliability. With increasing deregulation, some electrical utilities have
made some economic decisions to save money at the expense of decreased
network reliability. For instance, some utilities deliberately undersize
new transformers in power-handling capability, with secondary fuses rated
for, say 10 times the sum of the primary loads. So, if all the customers
attached to a transformer, from 1-5 houses, where I live, were to turn on
all of their lights, electric stoves, etc, at once, the secondary voltage
would go down and the transformer might actually burn up before the
safety fuse on the pole blew. The fuses are there for fault safety,
not sustained load.

Utilities recognize that sustained maximum loads like this rarely happen,
perhaps only once in 10 years, and it is less expensive to suffer an outage
in this rare event (especially if people don't want to pay much for
reliability, as reflected in PUC policies) than to spend money providing
a more conservatively-designed, "worst-case" system, like an aircraft system.
So there is a small but non-zero failure rate built into this utility policy.
Of course additional failures result from acts of God.

Since this is SCI.energy, I'd like to talk for a moment about how to
calculate the reliability of redundant systems, eg 2 vs. 1, if either
can do the job, and we know when one breaks, and we fix it. Having
2 subsystems (I would say "2 redundant subsystems," but that would be
redundant :-) can greatly increase the reliability of an overall system,
roughly squaring the MTTF. MTTF means "Mean Time To Fail," and in most
systems, it is very close to the MTBF, the Mean Time Between Failures, which
is a little bit longer, because it includes the repair time of a system.
If a system works for a long time, then it fails, then it is fixed in a
relatively short time, and then it works for a long time again, the MTTF
is the average amount of time that it stays working, and the MTBF is the
average time between failures, including the average repair time, MTTR. 

If a single system with no redundancy has a 10,000 hour MTBF, it fails about
once every 10,000 hours, ie the average failure rate is 1/10,000 hours, or
about once a year. If fixing it takes an average of 8 hours (counting the
time for the repairperson to get some parts and drive to the system, etc.)
then we'd expect it to be down for about 8 hours a year. This corresponds
to an Availability, or probability that the system is working, of

A = MTBF/(MTTF+MTTR) = 10,000/(10,000+8) = 0.992 or an Unavailability of

U = 1-A = 0.0008,

which means that the system is not working about 0.08 % of the time. In
a year, we would expect the overall system to be down, ie unavailable, for
0.0008 x 366 days x 24 hours/day = 7 hours (which is a bit less than
8 hours because there are only 8766 hours in an average year.) Most
people are used to better electrical service than this, and much better
telephone service... US Telephone companies calculate and budget and try
to deliver user unavailabilities measured in seconds per year. They try to
design telephone networks so that an average telephone is out of order
less than 10 seconds a year. That may increase too, under deregulation.

What happens if we have 2 electrical systems? A power line and a solar
electric system, or a power line and a standby generator, or a power line
and a UPS, or a standby generator and a wind generator with some batteries
and an inverter. Or 2 telephone systems, or 2 cars, or 2 of any sort of
systems, (eg 2 heads, better than one?) If the systems are relatively
independent, and either can do the job by itself, the the reliability and
availability of the combined system goes up, the cost probably goes up, and
the unavailability can go down, which is good. For this to work well,
we have to know when one system is broken (an "alarm"), and we have to
fix it fairly quickly when it breaks. For this discussion, we might
say that a solar electric system is temporarily "broken," ie unavailable
when the batteries are discharged, or more permanently broken, in the
usual sense, when something goes wrong with the inverter. 

Suppose we have 2 electrical power systems, each with an MTTF of 10,000
hours and an MTTR of 8 hours, and they are independent, and we can easily
and quickly switch from one to the other, and we know when one is broken
and start fixing it right away, the overall system can be in one of 3
conditions: 2 systems are working, only 1 is working, and 0 are working,
ie both broken. This leads to the following simple Markov model and state
diagram for the overall system:

 ---------    2L    ---------    L     ---------
|     2   |------->|    1    |------->|    0    |
| (both   |        | (one    |        | (none   |
| working)|<-------| working)|<-------| working)|
 ---------    M    ---------     2M    ---------

At a particular time, the system can be only in 1 of 3 states, numbered 2
1 and 0. Each state has an associated probability, P(2), P(1) and P(0).
In states 2 and 1, we have electrical power. In state 0, we don't.

The unavailability of the system is P(0). We want to minimize that.
The model has state transition rates based on the failure rate L=1/MTTF
and repair rate M=1/MTTR of a subsystem. Since both subsystems are working
in state 2, and either can fail, the transition rate from state 2 to
state 1 is twice the rate of the transition rate from state 1 to 0. The
repair rate from state 0 to 1 is twice the repair rate from state 1 to 2,
assuming 2 repairpersons are at work, when both subsystems are broken. 

The unavailability is easy to calculate: since the overall system can only
be in 1 state at a time, and there are only 3 states, the probabilities
have to add up to 1:

P(2) + P(1) + P(0) = 1                  (1), and from the transition rates, 

P(1) = 2L/M P(2)                        (2), and

P(0) = L/(2M) P(1)                      (3), so

M/(2L) P(1) + P(1) + P(0) = 1           (4), from (1) and (2), and

M/(2L) 2M/L P(0) + 2M/L P(0) + P(0) = 1 (5), from (4) and (3), so

P(0) = 1/(M^2/L^2-2M/L+1).

We could do this calculation exactly, but the first term in the denominator,
M^2/L^2, is over a million, and the second term is 2,500 and the third is 1,
which is how these calculations usually turn out, so we can approximate
P(0) = 1/(M^2/L^2) = (L/M)^2 = (MTTR/MTTF)^2. In our example, this is
0.00000064. So the expected number of seconds a year of electrical outage,
when  neither  system is working, is

0.00000064 x 8766 hr/yr x 60 min/hr x 60 sec/min = 20 seconds.

Much better than 8 hours. Almost as good as the telephone company :-)

Nick