Appendix A Sizing and Capacities of Gas Piping

There are other systems, however, where the required inlet pressure to the different appliances may be quite varied. In such cases, the greatest inlet pressure required must be satisfied, as well as the farthest appliance, which is almost always the critical appliance in small systems.

There is an additional requirement to be observed besides the capacity of the system at 100-percent flow. That requirement is that at minimum flow, the pressure at the inlet to any appliance does not exceed the pressure rating of the appliance regulator. This would seldom be of concern in small systems if the source pressure is ¹/₂ psi (14-inch w.c.) (3.5 kPa) or less but it should be verified for systems with greater gas pressure at the point of supply.

To determine the size of piping used in a gas piping system, the following factors must be considered:

For any gas piping system, or special appliance, or for conditions other than those covered by the tables provided in this code, such as longer runs, greater gas demands or greater pressure drops, the size of each gas piping system should be determined by standard engineering practices acceptable to the code official.

To obtain the cubic feet per hour of gas required, divide the total Btu/h input of all appliances by the average Btu heating value per cubic feet of the gas. The average Btu per cubic feet of the gas in the area of the installation can be obtained from the serving gas supplier.

1. Flanged fittings have three-fourths the resistance of screwed elbows and tees.
2. Tabular figures give the extra resistance due to curvature alone to which should be added the full length of travel.
3. Small size socket-welding fittings are equivalent to miter elbows and miter tees.
4. Equivalent resistance in number of diameters of straight pipe computed for a value of (f - 0.0075) from the relation (n - k/4f).
5. For condition of minimum resistance where the centerline length of each miter is between d and 2¹/₂d.
6. For pipe having other inside diameters, the equivalent resistance may be computed from the above n values.

Application of the gravity factor converts the figures given in the tables provided in this code to capacities for another gas of different specific gravity. Such application is accomplished by multiplying the capacities given in the tables by the multipliers shown in Table A.2.4. In case the exact specific gravity does not appear in the table, choose the next higher value specific gravity shown.

TABLE A.2.4
MULTIPLIERS TO BE USED WITH TABLES 402.4(1)
THROUGH 402.4(22) WHERE THE SPECIFIC GRAVITY
OF THE GAS IS OTHER THAN 0.60

To determine the size of each section of gas piping in a system within the range of the capacity tables, proceed as follows (also see sample calculations included in this Appendix):

When a large number of piping components (such as elbows, tees and valves) are installed in a pipe run, additional pressure loss can be accounted for by the use of equivalent lengths. Pressure loss across any piping component can be equated to the pressure drop through a length of pipe. The equivalent length of a combination of only four elbows/tees can result in a jump to the next larger length row, resulting in a significant reduction in capacity. The equivalent lengths in feet shown in Table A.2.2 have been computed on a basis that the inside diameter corresponds to that of Schedule 40 (standard-weight) steel pipe, which is close enough for most purposes involving other schedules of pipe. Where a more specific solution for equivalent length is desired, this may be made by multiplying the actual inside diameter of the pipe in inches by n/12, or the actual inside diameter in feet by n (n can be read from the table heading). The equivalent length values can be used with reasonable accuracy for copper or brass fittings and bends although the resistance per foot of copper or brass pipe is less than that of steel. For copper or brass valves, however, the equivalent length of pipe should be taken as 45 percent longer than the values in the table, which are for steel pipe.

The sizing of the 2 psi (13.8 kPa) section (from the meter to the line regulator) is as follows:

The low pressure section (all piping downstream of the line regulator) is sized as follows:

Follow the procedures described in the Longest Length Method for Steps (1) through (4) and (9).

For each piping segment, calculate the pressure drop based on pipe size, length as a percentage of 100 feet (30 480 mm) and gas flow. Table A.3.4 shows pressure drop per 100 feet (30 480 mm) for pipe sizes from ¹/₂ inch (12.7 mm) through 2 inches (51 mm). The sum of pressure drops to the critical appliance is subtracted from the supply pressure to verify that sufficient pressure will be available. If not, the layout can be examined to find the high drop section(s) and sizing selections modified.

Note: Other values can be obtained by using the following equation:

For example, if it is desired to get flow through ³/₄-inch (19.1 mm) pipe at 2 inches/100 feet, multiply the capacity of ³/₄-inch pipe at 1 inch/100 feet by the square root of the pressure ratio:

TABLE A.3.4
THOUSANDS OF BTU/H (MBH) OF NATURAL GAS PER 100 FEET OF PIPE AT VARIOUS PRESSURE DROPS AND PIPE DIAMETERS

where:

Q	=	Rate, cubic feet per hour at 60°F and 30-inch mercury column
D	=	Inside diameter of pipe, in.
P1	=	Upstream pressure, psia
P2	=	Downstream pressure, psia
Y	=	Superexpansibility factor = 1/supercompressibility factor
Cr	=	Factor for viscosity, density and temperature*
	Note: See Table 402.4 for Y and Cr for natural gas and propane.
S	=	Specific gravity of gas at 60°F and 30-inch mercury column (0.60 for natural gas, 1.50 for propane), or = 1488µ
T	=	Absolute temperature, °F or = t + 460
t	=	Temperature, °F
Z	=	Viscosity of gas, centipoise (0.012 for natural gas, 0.008 for propane), or = 1488µ
fba	=	Base friction factor for air at 60°F (CF = 1)
L	=	Length of pipe, ft
ΔH	=	Pressure drop, in. w.c. (27.7 in. H2O = 1 psi)

(For SI, see Section 402.4)

Solution:

Maximum gas demain for Outlet B:

Maximum gas demain for Outlet C:

Maximum gas demain for Outlet D:

FIGURE A.6.1
PIPING PLAN SHOWING A STEEL PIPING SYSTEM

Solution:

Note: It is not unusual to oversize the supply line by 25 to 50 percent of the as-installed load. EHD size 18 has a capacity of 189 cfh (5.35 m³/hr).

FIGURE A.6.2
PIPING PLAN SHOWING A CSST SYSTEM

Solution:

FIGURE A.6.3
PIPING PLAN SHOWING A COPPER TUBING SYSTEM

FIGURE A.6.4
PIPING PLAN SHOWING A MODIFICATION
TO EXISTING PIPING SYSTEM

If the volume of the piping system is unchanged, then the formula based on Boyle's and Charles' law for determining the new pressure at a reduced temperature is as follows:

Therefore, the gauge could be expected to register 18 psig (124 kPa) when the ambient temperature is 40°F (4°C).

For ¹/₂-inch pipe, ΔH = ^{20 feet}/_{100 feet} × 0.3 inch w.c. = 0.06 in w.c.

For ³/₄-inch pipe, ΔH = ^{15 feet}/_{100 feet} × 0.3 inch w.c. = 0.045 in w.c.

For 1 inch pipe: ΔH = ^{10 feet}/_{100 feet} × 0.2 inch w.c. = 0.02 in w.c.

For ³/₄-inch pipe: ΔH = ^{10 feet}/_{100 feet} × [0.5 inch w.c. + ^{(110,000 Btu/hr-104,000 Btu/hr)}/_{(147,000 Btu/hr-104,000 Btu/hr)} × (1.0 inches w.c. - 0.5 inch w.c.)] = 0.1 × 0.57 inch w.c.≈ 0.06 inch w.c.

Note that the pressure drop between 104,000 Btu/hr and 147,000 Btu/hr has been interpolated as 110,000 Btu/hr.

For 1-inch pipe: ΔH = ^{20 feet}/_{100 feet} × [0.2 inch w.c. + ^{(14,000 Btu/hr)}/_{(27,000 Btu/hr)} × 0.1 inch w.c.] = 0.05 inch w.c.

For ³/₄-inch pipe: ΔH = ^{20 feet}/_{100 feet} × 1.0 inch w.c. = 0.2 inch w.c.

Note that the pressure drop between 121,000 Btu/hr and 148,000 Btu/hr has been interpolated as 135,000 Btu/hr, but interpolation for the ³/₄-inch pipe (trivial for 104,000 Btu/hr to 147,000 Btu/hr) was not used.

For 1-inch pipe: ΔH = ^{30 feet}/_{100 feet} × 1.0 inches w.c. = 0.3 inch w.c.

For 1¹/₄-inch pipe: ΔH = ^{30 feet}/_{100 feet} × 0.2 inch w.c. = 0.06 inch w.c.

Note that interpolation for these options is ignored since the table values are close to the 245,000 Btu/hr carried by that section.

Minimum pressure drop to farthest appliance:

ΔH = 0.06 inch w.c. + 0.02 inch w.c. + 0.06 inch w.c. = 0.14 inch w.c.

Larger pressure drop to the farthest appliance:

ΔH = 0.06 inch w.c. + 0.06 inch w.c. + 0.3 inch w.c. = 0.42 inch w.c.

Notice that Section 2 and the run to B do not enter into this calculation, provided that the appliances have similar input pressure requirements.

For SI units: 1 Btu/hr = 0.293 W, 1 cubic foot = 0.028 m³, 1 foot = 0.305 m, 1 inch w.c. = 249 Pa.