LNCT GROUP OF COLLEGES Name of Faculty: Amrendra kumar Amar Designation: Asst.Prof Department: Mechanical Engineering Subject: Turbomachinary Unit: Fourth Topic: NUMERICAL OF BLOWERS AND FANS Fan and Blowers (1) LNCT GROUP OF COLLEGES Q.No-1 A Centrifugal blower takes air at 100kPa and 309K. It develops a pressure head of 750 mm W.G, while consuming a power of 33 kW. If the blower efficiency is 79% and mechanical efficiency is 83 %, determine the mass rate and volume rate and exit properties of air. Solution:- 5 2 Given, P1 = 100 kPa = 1 x 10 N/m , T1= 309 K, ΔH = 750 mm W.G., Pin = 33 kW, η in = 33 kW, ηb = 79% = 0.79, ηm = 83% =0.83. From ideal gas equation, at inlet p1v1 = mRT1 5 3 Or m/v1 = 훒₁ = p1/ RT1 = (1 x 10 )/(287 x 309) = 1.128 kg/m Rise in pressure, Δp = 훒gΔH = 1000 X 9.81 X 0.75 (⸫ 훒 for water = 1000 kg/m3) 2 = 7357.5N/m Ideal work done per kg of air W ideal = Δp/훒₁ = 7357.5/1.128 = 6522.61 J/kg Actual work done per kg of air W Actual = W Ideal / η B = 6522.61/0.79 = 8256.47 J/kg Actual power input, (Pin)Actual = Pin X η m = 33 X 0.83 = 27.39 kW Therefore, the mass flow rate, (푃 푖푛)퐴푐푡푢푎푙 3 m = = (27.39 X 10 )/8256.47 = 3.317 kg/sec Ans. 푊퐴푐푡푢푎푙 Volume rate, 3 Q = (m / 훒 1 ) = ( 3.317/1.128) = 2.94 m /s Ans. Fan and Blowers (2) LNCT GROUP OF COLLEGES Pressure rise, Δp = p2 – p1 7357.5 = p2 – 100000 2 ⸫ p2 = 107357.5 N/m = 107.36 kPa Ans. Actual work done per kg of air W Actual = cp (T2 – T1) 8256.47 = 1005 (T2 – 309) (⸪ For air cp= 1005 J/kg K) Or T2 – 309 = 8.22 = 317.22 K Ans. Q.NO 2:- Air flows through a blower where its total pressure is increased by 150 mm of water. At the inlet, p0 = 1.05 bar and t0 = 15°C. The total efficiency is 70%. Find the exit total pressure, the exit total temperature, and the isentropic and actual changes in total enthalpy. Sol. Given, h = 150 mm of H2O, p01 = 1.05 bar, t01 = 15°C. Or T01 = 15+273 = 288 K, η tt = 70% = 0.7. Pressure difference, h = 150 mm of H2O 150 푋 9.81 푋 1000 Or 훥푝 = 1000 = 0.01474 푏푎푟 1 푋 105 The exit total pressure, P02 = Δp + p01 = 0.0147 + 1.05 = 1.0647 bar Ans. Total efficiency of turbine is given by, Fan and Blowers (3) LNCT GROUP OF COLLEGES Or T02 – 288 = 1.1463/0.7 = 1.638 ⸫ T02 = 288 + 1.638 = 289.64 K Or t02 = 289.64 – 273 = 16.64°C Actual change in total enthalpy, Δh0 = cp (T02 – T01) = 1.005(289.64 – 288) = 1.6482 kJ/kg Ans. And isentropic change in total enthalpy, ’ Δh0 = Δh0 X η tt = 1.6482 X 0.7 = 1.1537 kJ/kg Ans. Q.No-3. A backward-swept centrifugal fan develops a pressure of 75mm of water. It has an impeller diameter of 89 cm and runs at 720 r.p.m. The blade air angle at tip is 39° and the width of impeller is 10 cm. Assuming a constant radial velocity of 9.15 m/s and density of 1.2 kg/m3, determine the – (i) Fan efficiency (ii) Discharge (iii) Power required (iv) Stage reaction (v) Pressure coefficient. Solution- Given, ΔH = 75 mm of water, D2 = 89 cm = 0.89 m, 3 N = 720 r.p.m, β2 = 39°, b2 = 10 cm = 0.1 m, Vf = 9.15 m/s, 훒 = 1.2 kg/m . Fan and Blowers (4) LNCT GROUP OF COLLEGES (i) Fan efficiency 3 Ideal work done/kg = (Δp/훒) = (훒 w g.ΔH)/훒 = (10 X 9.81 X 0.075)/1.2 = 613.13 J/kg Actual work done/kg = ( W/m =u2 Vw2) Fig. Velocity Triangle for a Radial Impeller Tan β 2 = Vf / (u2 – Vw2) or Vw2 = u2 – ( Vf / tan β 2) ...(i) ⸪ u 2 = ( П D2N/60) = ( П X 0.89 X 720)/ 60 = 33.55 m/s Putting the value of u2 in equation (i), we get Vw2 = 33.55 – (9.15/tan 39°) = 22.25 m/s ⸪ W/m = 33.55 X 22.25 = 746.5 J/kg 퐼푑푒푎푙 푤표푟푘 푑표푛푒/푘푔 613.13 Now, Fan efficiency, ηf = = = 0.821 퐴푐푡푢푎푙 푤표푟푘 푑표푛푒/푘푔 746.5 = 82.1 % 3 (ii) Disharge Q = П D2b2Vf = П X 0.89 X 0.1X 9.15 = 2.558m /s (iii) Power required mass of air flow, m = 훒Q = 1.2 (2.558) = 3.07 kg/s ⸫ Power required W = m(W/m) = 3.07 X 746.5 = 2291.7 W = 2.2917 kW Ans. (iv) Stage reaction R = 1 – ½(Vw2/u2) = 1- ½(22.25/33.55) = 66.84 % 푉푤2.푢2 푉푤2 22.25 (v) Pressure coefficient Ψ p = 2 = 2 = 2( ) 푢2 푢2 33.55 2 = 1.326 Ans. Fan and Blowers (5) LNCT GROUP OF COLLEGES Fan and Blowers (6) .