EQUAL FRICTION METHOD ROUND DUCT Example 1 Given

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EQUAL FRICTION METHOD (ROUND DUCT) Example 1 : Given : Duct system layout executive

EQUAL FRICTION METHOD (ROUND DUCT) Example 1 : Given : Duct system layout executive office is shown in Fig. 2. Initial velocity is to be equal to or less than 1800 fpm. Total air quantity is equal to 4500 cfm. Elbows are 45° smooth, R/D = 1. 5. Find : Round duct sizes. Use Fig 2 for duct system conditions.

SECTION Fan to A A-B B - 13 13 - 14 14 - 15

SECTION Fan to A A-B B - 13 13 - 14 14 - 15 15 - 16 16 - 17 17 - 18 AIR QUANTITY (cfm) EQUIVALENT LENGTH (ft) ROUND DUCT SIZE (in) ACTUAL VELOCITY (fpm)

Friction rate = …………………in. WG/100 ft. Total equivalent length, ft. =…………ft. Friction loss =…………….

Friction rate = …………………in. WG/100 ft. Total equivalent length, ft. =…………ft. Friction loss =……………. . in. WG. Regain =……………. . . =………………. . in. WG. Total static pressure required by duct system =…………………. in WG.

EQUAL FRICTION METHOD (ROUND DUCT) Example 1 : Given : Duct system layout for

EQUAL FRICTION METHOD (ROUND DUCT) Example 1 : Given : Duct system layout for conference room is shown in Fig. 3. Initial velocity is to be equal to or less than 1450 fpm. Total air quantity is equal to 4500 cfm. Elbows are 90° smooth, R/D = 1. 5. Find : Round duct sizes. Use Fig 3 for duct system conditions.

SECTION Fan to A A-B B - 13 13 - 14 14 - 15

SECTION Fan to A A-B B - 13 13 - 14 14 - 15 15 - 16 16 - 17 17 - 18 AIR QUANTITY (cfm) EQUIVALENT LENGTH (ft) ROUND DUCT SIZE (in) ACTUAL VELOCITY (fpm)

Friction rate = …………………in. WG/100 ft. Total equivalent length, ft. =…………ft. Friction loss =…………….

Friction rate = …………………in. WG/100 ft. Total equivalent length, ft. =…………ft. Friction loss =……………. . in. WG. Regain =……………. . . =………………. . in. WG. Total static pressure required by duct system =…………………. in WG.