TURBO MACHINERY AXIAL FLOW PUMPS Dr Umi Fazara

TURBO MACHINERY AXIAL- FLOW PUMPS Dr Umi Fazara Md Ali umifazara@unimap. edu. my

1. The axial flow pump acts like a propeller enclosed in housing as shown in Figure below. 2. The rotating element, the impeller, causes a pressure change between the upstream and downstream section of the pump 3. Application- to deliver relatively low heads and high flow rates

Head and Discharge Coefficient for Axial Flow Pump • Analyses of pumps – CH (Head coefficient) = ∆ H / (D 2 n 2 /g) – CP/Ẇ (Power coefficient) = P / ( ρ D 5 n 3) – CQ (Discharge coefficient) = Q / (n D 3) Definitions ( H: Head (m); D: Diameter (m); n: rotation per second (s-1); g: 9. 81 m s-2 ; P: Power (J s-1); ρ: Density (kg m-3 ); Q: Discharge (m 3 s-1) ) Please REMEMBER all equations

Net Positive Suction Head (NPSH) • NPSH is the difference in pressure between the suction side of the pump and the vapor pressure of the liquid being pumped. • In practice, engineers express this difference in term of pressure head , called NPSH. CNPSH = n. Q 1/2 / g 3/4 (NPSH)3/4 NPSH is in meter • Please REMEMBER all equations

Example 1 • For the pump represented by Fig 1 and Fig 2, what discharge of water in cubic meters per second will occur when the pump is operating against a 2 m head and at a speed of 600 rpm? What power in kilowatts is required for these conditions?

Fig 1

Fig 2

Solution • Problem Definition - Situation: Axial flow pump with water - Find: • Discharge (in m 3 s-1) • Power ( in k. W) - Sketch: N = 600 rm Axial- flow pump ∆H=2 m

Assumptions: Assume ρ = 1000 kg/ m 3 - Plan 1. 2. 3. 4. Calculate CH From Fig 1 and Fig 2 find CQ and CP Use CQ to calculate discharge Use CP to calculate power

Solution 1. Rotational rate is (600 rev/min) / (60 s/min) = 10 rps (rotation per second). D = 35. 6 cm (from Fig 2) CH = (2 m)/ [ (0. 356 m)2 (102 s-2) / (9. 81 m/s 2) ] = 1. 55 2. From Fig 1, CQ = 0. 40 and CP = 0. 72 3. Discharge is Q = C Q n D 3 Q = 0. 40 (10 s-1) (0. 356 m)3 = 0. 180 m 3 /s

Cont solution 4. Power is P = 0. 72 ρ D 5 n 3 = 0. 72 ( 103 kg / m 3) (0. 356 m)5 (10 s-1)3 = 4. 12 km. N/s = 4. 12 k. J/s = 4. 12 k. W

Example 2 If a 30 cm axial- flow pump having the characteristics shown in Fig 1 is operated at a speed of 800 rpm, what head ∆H will be developed when the water-pumping rate is 0. 127 m 3 /s? What power is required for this operation?

Problem Definition Situation: 30 cm axial flow pump with water 1. Head (in meters) developed 2. Power (in k. W) required Sketch: ∆H=? P=? N = 800 rpm Axial flow pump Q = 0. 127 m 3/ s

Assumptions: Water with ρ = 1000 kg/ m 3 - Plan 1. Calculate the discharge coefficient, CQ 2. From Fig 1, read CH and CP 3. Use this equation to calculate head produces • CH (Head coefficient) = ∆ H / (D 2 n 2 /g) 4. Use this equation to calculate power required • CP (Power coefficient) = P / ( ρ D 5 n 3)

Solution 1. Discharge coefficient is Q = 0. 127 m 3 /s n = 800/ 60 = 13. 3 rps D = 30 cm CQ = (0. 127 m 3 /s )/ [ (13. 3 s-1 )(0. 30 m) 3) ] = 0. 354 2. From Fig 1, CH = 1. 70 and CP = 0. 80 3. Head produced is ∆ H = (CH D 2 n 2 /g) = [1. 70 (0. 30 m)2 (13. 3 s-1)2] / (9. 81 m/s 2) = 2. 76 m

Cont solution 4. Power required is P = C P ρ D 5 n 3 = 0. 80 ( 103 kg / m 3) (0. 30 m)5 (13. 3 s-1)3 = 4. 57 k. W