using probability to determine medical gas flow

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    Using Probability to Determine Medical Gas Flow, Part I

    By Edward E. Hale, BSME, CPD, CMGIPosted: March 2, 2005

    When designing medical gas systems, the maximumanticipated flow must first be calculated in order todetermine accurate pipe sizing.

    Issue: 3/05

    Plumbing designers are often called upon to design medical gas piping systems, includingoxygen, nitrous oxide, nitrogen, carbon dioxide, medical vacuum, WAGD (Waste AnestheticGas Disposal) and medical compressed air. As with any gas piping system, in order to

    determine pipe sizes accurately, we must first determine the maximum anticipated flow. Theonce the flow has been determined, pipe sizes may be selected using standard tables or bydirect calculation using various gas-flow formulae.

    However, calculating medical gas flows and pipe sizes, especially for large facilities such ashospitals, can be a rather complicated process. To determine flow, designers havetraditionally counted the number of outlets/inlets served by a section of piping without regarto the location of those outlets/inlets, applied a use factor from a standard table, thenmultiplied that product by the average flow for that particular gas. Pipe diameters are oftensimply selected from tables based on the number of outlets/inlets and maximum pipe length

    These methods would be acceptable if the medical gas demand in all areas of the facility wasequal. However, thats not the case. Certain areas in medical facilities have a greater demanfor medical gases than others. For example, critical care areas have a higher usage of oxygethan do patient rooms, etc. Moreover, the number of outlets/inlets doesnt always directlycorrelate with the demand. If these facts are not taken into consideration, the flowscalculatedas well as the associated gas pipingcould be considerably oversized orundersized.

    Indeed, there seems to be a great deal of bewilderment among designers regarding the bestway to determine medical gas flows. This is not surprising, as there appears to be scantinformation available for such calculationsother than that mentioned previously. There hav

    been, over the years, various tables developed to assist in determining medical gas demandfor an entire facility. These tables typically assign use factors for each gas in the variousareas of the facility to be multiplied by the total number of operating rooms, beds, etc.1,2However, these factors may not be used to calculate flows for only small portions of thepiping. But, as well see, they do provide valuable information in determining medical gasflows.

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    Jacob Bernoulli(1654-1705)

    discovered many of

    the fundamental

    principles ofprobability, including

    the Theory of Large

    Numbers. These are

    discussed in the workArs Conjectandi,

    published in Basel,

    Switzerland, in 1713,

    eight years after hisdeath.

    Medical Gas Demand and Probability

    Whenever scientists, engineers and other researchers are called on to calculate certainquantitative information, they often turn to the principles of probability. Probability is a brancof mathematics that is used to determine what is likely to occur (not necessarily what willoccur) in a given event or sequence of events. Swiss mathematician Jacob Bernoulli(1654-1705) is credited with the discovery of many of the fundamental principles ofprobability, including the binomial equation used to generate probability distribution curves,as well as the Theory of Large Numbers. These theorems are discussed in the work ArsConjectandi, published in Basel, Switzerland, in 1713, eight years after his death.

    Determining medical gas demand is one exercise that conforms to the classic laws ofprobability. Therefore, the key to accurately determining gas flow is the understanding and

    application of these laws.

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    Equation 1

    First, lets look at a binomial probability distribution. A binomial probability distribution is useto determine the probability of a certain number of favorable results for a given number oftrials with dichotomous outcomes (i.e., a trial with only two possible mutually exclusiveoutcomesalso referred to as a Bernoulli trial). For example, we may use it to determine thechances of getting eight heads from 15 tosses of a coin, or the chances of getting four twos 10 tosses of a die, etc.

    The binomial formula (Equation 1) is shown at right.

    Equation 1

    Where:

    P(x) = Probability of getting x positive outcomes

    n = number of trials

    p = probability of a single trial being successful (e.g., p = 0.5 for Probability of getting a heaon a single toss of a coin and p = 1/6 for the probability of getting a two on a single toss of die, etc.)

    x = number of positive outcomes

    Equation 1proven by Jacob Bernoulli in 1685is used to generate the well-known binomiaprobability distribution curve, the familiar bell-shaped curve.

    A few things to keep in mind regarding the binomial distribution curve are:

    1. The total number (n) of trials is indicated along the x-axis.

    2. The probability of getting x number of positive results in n trials is indicated along the

    y-axis.

    3. The area under the curve between two values of x represents the probability of obtainina positive result in that range (e.g., the probability of getting six or more heads in 50 tossesof a coin, etc.)

    4. The value of x at the peak of the curve is called the mean (). Its value is equal to theprobability (p) of obtaining a positive result times the number of trials (n) or = p(n) (see

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    Equation 2

    Figure 1

    Figure 1).

    5. One standard deviation is equal to (np(1-p))1/2.

    6. Plus or minus one standard deviation covers approximately 68% of the area under the

    curve (mean 34%), two standard deviations about 95%, and three standard deviationsabout 99.7%.

    7. We may use the binomial formula to determine the probability of obtaining x or lesspositive results for a given number n trials.

    To predict a result with a particular confidence level, we multiply the standard deviationmentioned above by a z factor (found in standard statistical tables) and add this result tothe mean, as indicated in Equation 2. For example, a z factor of 1.96 will result in aconfidence level of 97.5%.

    Equation 2 is shown at right.