Precipitation is the descent from the sky of several kinds of moisture, particularly rain and snow, and its deposit on buildings and terrain. If the moisture falls on a building, its upward surfaces shed the water or snowmelt by gravity, usually to nearby terrain; if the moisture falls on terrain, its vegetation absorbs the moisture and its topography drains it to lower levels. Whether shed by a building or topography, the water often flows through an open channel known as a gutter or a storm drain. Even if the channel is covered, as are storm drains, the drainage is still open; because the waterflow, even if it completely fills the channel, moves by gravity and not pressure as it does in plumbing supply systems.

Every open channel that carries water has a wetted perimeter (the portion of the channel's perimeter that is submerged by the waterflow) and a hydraulic radius (the section area of the channel's waterflow divided by its wetted perimeter). Thus the formula for a channel's hydraulic radius is

Formulas for obtaining the hydraulic radius of certain circular pipes, box channels, and half-hexagonal channels are given below.

2C21 Pipe flowing half full:

0.5 n r2/0.5 x 2 n r = 0.5 r Pipe f owing full: n r2/2 n r = 0.5 r Box channel flowing full:

Half-hex channel flowing full:

2C21- WETTED PERIMETER

The half-hex channel section above is also known as a semi-hexagonal or regular trapezoidal open channel. It is generally the most efficient noncircular channel section.

2C25

2C22.

2C23.

2C24.

2C24.

Fig. 2-6. Open channel geometry 1.

Solution: Finding A in the above formula is easy (A = 0.33 nr2), but finding Pw is often tedious. It is done as follows:

Fig. 2-6. Open channel geometry 1.

• Much of the information regarding the open channel formulas in this section j|

Referring to the sketch at right, the shaded area S is a spherical segment whose area is

Thus A for any fractional or decimal portion of a circle when the portion is a spherical segment may be written as below, wherein F equals the fractional or decimal portion of the circle:

A = Fn r2 = n {0.5 [h (r- h)]0 5 + h2} a F = n {0.5 [h (r- h)]0 5 + h2}/r2

Solve for h. Then, knowing h and r, find r- h as sketched to the right.

From cos 0 = (r- h)/r, find 0 Then from r0 = 57.3 Pw, find Pw

Due to the above complicated mathematics, Rh is generally solved only for the simplest circular relationships or where an open channel's section dimensions are known. Thus one usually begins by knowing a channel flow's section area A (or its flow in cfm) and its wetted perimeter Pw, then uses these values to find the section's hydraulic radius Rh, whose value is then used to size almost any channel of gravity-induced waterflow, from a custom gutters to an irrigation canals. However, these formulas are accurate only if the waterflow is smooth-flowing, which usually occurs if its slope is below about 11° (0.19 or 2.33:12 pitch). For steeper slopes and turbulent waterflows, the formulas may still be useful for general estimating.

The formulas that follow for sizing gutters, leaders, footing drains, and storm drains are simplified variations of the above mathematics. Also, it is a good idea to size channels at 0.8 capacity, to leave room for extreme situations. A round pipe's hydraulic radius at 0.8 capacity = 2.26 r.

Ageneral formula forfinding the maximum waterflow in any open channel of any cross-section is

■ section area of waterflow in channel (in.)167i

wetted perimeter

kr = roughness coefficient of channel's inner surface, from Table 2-3. Q = volume of waterflow, cfm

A = average slope of channel from inlet to outlet, ft/ft. This equals

(upper elevation - lower elevation) * horizontal distance between upper and lower elevations.

This formula may also be used to estimate the highest flood stage level of a stream or river valley.

was obtained from Introduction to Fluid Mechanics, William Haberman & James

A gutter collects rainwater draining from a roof, then leads it away from the building it falls on. These long narrow channels may be made of aluminum, copper, galvanized steel, plastic, and wood; and their sections may be the common ogee profile, semicircular, rectangular, any other shape, or simply a trough of flashing installed near the bottom of a roof, as were the gutters of Frank Lloyd Wright's Robie House (see Fig 2-7A). Such elegance is testimony to this man's genius, as it shows how he often created beauty out of simple utility —an outstanding ability of his that has gone largely unnoticed down through the decades.

Today the most common gutter section nationwide for residences and small commercial buildings of almost any size, economic situation, and climatic exposure is one with a rectangular back and an ogee front known as K-style. This section is shown in Fig. 2-86.

Gutters must be designed precisely if they are to maximize the longevity of the parent architecture; yet, unfortunately, designing gutters will never be an exact science. This is primarily due to the vagaries of rainfall, whose intensity in any area can only be estimated. For this and numerous other reasons, an ^ designer must be especially alert when selecting and sizing this component. Thus he or she should know the subject well, never hesitate to oversize this relatively low-costing component, and always insist that the installer meticulously adhere to construction specifications. Beyond this, long-lasting gutters are designed as follows:

fl A gutter can be any width, but this usually varies from 4-8 in. A gutter is usually at least 4 in. wide so one's hand can clean it. Narrower gutters are often installed below very small roofs. fl A gutter's depth should be 0.5-0.75 its width and its top should be slightly wider than its bottom. Half-round gutters drain better than do other sections and less water generally remains in one's curved bottom. For this reason flat-bottomed gutters should pitch more steeply than half-round ones to keep stagnant water from remaining in them which can breed mosquitoes. fl Every gutter should pitch at least 1/16 in/— to facilitate drainage. Do not for aesthetic reasons run a gutter level for appearance; because the only reason for a gutter is functional —and if this is violated, much damage can occur to the parent building. fl A gutter exceeding 50 ft in length requires an expansion joint, which must be at a pitch peak because water cannot flow over an expansion joint; thus such a gutter requires a downspout on each side. Downspouts are usually located at a gutter's ends or quarter-lengths; thus in the latter position a properly-pitched gutter would have a very slight "W" elevation profile. fl A gutter's top front edge should be (1) at least 1/2 in. below the

John (Prentice-Hall, Englewood Cliffs, NJ, 1971); pp. 224-31. • A primary ||

outbr. rim v lowé.r than inner, rim inner rim outer rim expansion joint only at pitcm pzkks

half-round réct &eveelêd ogee gutter profiles inner rim outer rim expansion joint only at pitcm pzkks

Fig. 2-8. Gutter details.

gutter at wriqht's a ro&ie house

plane of the roof's pitch above to prevent sliding branches, clumps of leaves, and snow from tearing the gutter from the building; (2) no more than £ in. below the roof's pitch plane to keep descending roof rainwater from shooting over its top; and (3) at least 1 in. lower than the gutter's back edge so if it fills it will overflow away from the building. Remember that the steeper the roof, the greater is the rainwaterflow's velocity at its base.

► The sheeting for aluminum gutters (the most popular kind) is usually .028 or .032 in. thick. The thicker sheeting is only slightly more expensive but is considerably more durable. Copper gutters, especially custom-made ones, should be thicker.

► Where a roof valley drains into a gutter, during heavy rains the intersecting roof planes may concentrate the collected waterflow so much that it can overshoot the gutter at its corner or tear the gutter from the building. This is especially a problem if the roof is large and steep. Then an L-shaped sheet metal baffle may be installed in the valley above the gutter.

► Copper gutters develop an attractive verdigris patina after a few years, but water dripping from them can form green stains on surfaces below, and cedar-shingle roof runoff corrodes copper.

► A gutter is usually attached to the building by one of several kinds of hangers. The best hangers do not extend over the top of the gutter but cradle it from underneath, as then the gutter is easier to clean from below.

► Large flat roofs may have interior rain-leader traps located at slight depressions toward which the surrounding roof slopes at least % in/—. Such roofs typically have at least 4 in. high cant strips around their perimeters.

source for the information in this and the following two sections was Andy Engel's

Fig. 2-9. Rainfall intensity map.

Fig. 2-9. Rainfall intensity map.

Unfortunately, there is probably not a single standardized gutter sold in the nation's building supply stores today that does a perfect job of satisfying all the criteria above. For example, a semicircular gutter's front is rarely 1 in. lower than its back, the most widely-sold gutter has a mosquito-breeding flat bottom, the vast majority of gutter hangers extend over the gutter's top, and valley baffles are rarely found in any lumberyard. Even costly custom-made gutters are typically formed by machines with standard dies, which do not satisfy all the above criteria. Thus if you want to design the perfect gutter for a client, today you almost have to teach someone how to do it. This must be one reason why Frank Lloyd Wright turned his office into a school or 'Fellowship', where his employees were often 'students' who went to the site to construct the Plans.

The primary determinants of a gutter's size are the maximum local rainfall intensity and the roof's watershed area. The rainfall intensity is generally the most intense 5-minute rate of rainfall to be expected in the region, which is taken to be 1.5 its maximum hourly rainfall intensity. A gutter's watershed area is not the surface area of the roof that drains into the gutter, or even its horizontally projected area, but its horizontally projected area plus half its vertically projected area; because rain rarely falls vertically. Thus a 4-in-12 pitch rectangular roof that is 12 ft wide and 4 ft high has a watershed area of (12 + V2 x 4) x the roof's length. Such algebra also leads to proper gutter sizing if a large facade rises above the roof that drains into the gutter; as then any rainfall intercepted by the facade is accounted for. If the seam of two pitched roofs at right angles to each other forms a valley, the waterflow from both roof planes drains into a continuous L-shaped gutter; then the roof's watershed length = 0.5 (roof eave length + roof peak length).

"All About Rain Gutters", Fine Homebuilding magazine (Taunton Press, New- ||

Example 1. If a residence nearTampa, FL, has a 54 x 32 ft gable roof with a 6/12 pitch and semicircular gutters, how wide should the gutter along each eave be if it has a downspout at each end? | ||||||||||||||||||||

2 Half-round section: 2,500 q k, (wd)167(w + 2 d)-°-67 2,500 q K, (4V'67^)«0'67 Ogee section: WR = 2 Half-round section: 2,500 q k, (wd)167(w + 2 d)-°-67 2,500 q K, (4V'67^)«0'67 W = watershed area of roof drained by each gutter, sf. W = area of horizontal projection + 0.5 area of vertical projection if roof slopes + 0.5 area of any facade above roof that drains into gutter. Assume gabled roof drains to gutters on each side 4 2 gutters. At 6/12 pitch, W = 54 x 32/2 + 0.5 x 54 x 6/12 x 16 = 1,080 sf. R = maximum rainfall intensity, based on 50-year-frequency storm, in/hr. From Fig. 2-9, R for northern Florida = 5 in/hr. q = number of leaders each gutter drains into. Here assume at least 1 leader at each end 4 q = 2. If 6 computed below is undesirably large, consider adding a third leader near the gutter's center. k^ = gutter slope coefficient based on its pitch, from bar graph below. The more visible the gutter, usually the shallower its pitch. Here gutter is prominently visible 4 use 1/16 in/— pitch 4 k^ = 0.072. kp = roof pitch coefficient 0.072 0.102 0.125 0.144 0.161 0.176 0.191 0.204 0.250 0.289 1/16 1/8 3/16 1/4 15/16 3/8 7/16 1/2 3/4 Roof pitch, in/— Fig. 2-10. Roof pitch coefficient bar graph. w = min. top width of gutter if it is an ogee or rectangular section, in. 6 = minimum diameter of gutter if it is a half-round section, in. d = minimum depth of gutter if it is a rectangular section, in. A = section area of gutter if it is any section, in2. Pw = wetted perimeter of gutter if it is any section, in. 1,080 x 5 = 390 x 2 0.072 x d2 67 ... d = 5.5 4 6 in. dia.
W = watershed area of roof drained by each gutter, sf. As roof is flat or nearly so and drains on only 1 side 4 1 gutter. A W = 62 x 46 = 2,850 sf. R = maximum rainfall intensity, based on 50-year-frequency storm. From Fig. 2-9, R for central Colorado = 3 in/hr. q = number of leaders gutter drains into. Assume 1 leader at each end 4 q = 2. ka = gutter slope coefficient depending on its pitch, from bar graph on previous page. Here the gutter is at rear of building where attractiveness matters little 4 use maximum pitch of/in/— 4 ka = 0.204. w = minimum top width of gutter if it is an ogee or rectangular section, in. Here try w =4 in. and solve for d. d = min. depth of gutter if it is a rectangular section, ? in. d s 3/in.
W = watershed area of roof drained by each gutter, sf. W = area of horizontal projection + 0.5 area of vertical projection if roof slopes + 0.5 vertical area of any facade above roof that drains into gutter. Roof is flat and drains on 2 sides 4 2 gutters. Whor =16 x 24/2 = 192 sf. Wver of wall above each roof gutter = 24/2 x 135 = 1,620 ft. a W = 192 + 0.5 x 1,620 = 1,000 sf. R = maximum rainfall intensity, based on 50-year-frequency storm. From Fig. 2-9, R for eastern Iowa = 4 in/hr. ka = gutter slope coefficient depending on its pitch, from bar graph on previous page. Here gutters are at side of canopy where attractiveness is neither a plus or minus; but the steeper the pitch, the smaller the gutter 4 try maximum pitch of 1/2 in/— 4 ka = 0.204. q = number of leaders each gutter drains into. 1 leader at the gutter's back end 4 q = 2. 6 = minimum width of semicircular gutter, ? in. 1,000 x 4 = 390 x 2 x 0.204 x d2 67 ... d = 3.34 4 4 in. Note: Since an ogee section's width w is usually 4.5 in, formula is commonly used to determine this section's number of leaders. |

Was this article helpful?

## Post a comment