Outlets should be designed to receive discharges at such a rate as to enable the outlet to discharge as a circular weir.

Rectangular weir

Circular weir

Flat roof application (no gutters)

Rainwater pipes with tapered inlets D

Q Discharge rate in m3/sec. V Velocity in metres/sec. A Cross-sectional area in m2 Cd Co-effi cient of discharge (0.64) g Gravitational acceleration 9.8 m/s2 h Depth of water at outlet in m ho Depth of water at outlet in mm B Mean breadth of weir in metres D Outlet diameter in metres D

RA Roof area in square metres

Taking theory of rectangular weir fl ow− fl ow rate=mean velocity of fl ow× effective crosssectional area of weir i.e. Q=V×A from which

(i) Allowing for 3 in. of rain per hour and seconds to hours

(ii) Flat roof application (no gutters)

(iii) Outlets from fl at roofs (no gutters) Outlets from large gutters or small ‘cesspools’, the head over the outlet can be allowed

to increase to (not as for fl at roofs)

Taking expression (ii)

(iv) For large gutter application It can be shown by hydraulic principles that for any given roof area, if the outlet takes

the form of a properly shaped transition piece, then some reduction of the diameter of the rainwater pipe compared with the diameter of the outlet is permissible. The following expression can be used-

(v) Note for tapered inlets only N.B. Owing to the problem of assessing the correct amount of rain to allow for exact

mathematical accuracy is not an important factor

Q Discharge rate in cu. ft./sec. V Velocity in feet/second A Cross-sectional area in sq. ft. Cd Co-effi cient of discharge (0.64) g Acceleration (gravity) 32 ft/s2 h Depth of water at outlet in ft. h

B Mean breadth of weir in ft. D Outlet diameter in feet D

RA Roof area in square feet