Question

A turbine designer is trying to select the appropriate blade shape to generate power efficiently. Based on the inlet and outlet structure of the device, they know that the absolute fluid velocities, V1 and V2, are oriented as sketched below: At the inlet radial position, R1, of 2.4 m, the velocity, V1, is 18 m/s at an angle, 01, of 55° relative to the circumferential direction. At the outlet radial position, R2, of 1.3 m, the velocity, V2, is 8.4 m/s at an angle, 02, of 779 relative to the circumferential direction. The turbine is expected to rotate at a speed, w, of 120 rpm. The blades depicted above are not to scale. a. Which direction should the turbine spin – clockwise or counterclockwise? b. Relative to the circumferential direction, what blade angle, B1 (in degrees), should be selected at the inlet? C. Relative to the circumferential direction, what blade angle, B2 (in degrees), should be selected at the outlet?

Answer 1

a) The turbine should spin in clockwise direction.

b) Blade angle at inlet =

143.47%

c) Blade angle at outlet =

150.46

a b) The Purline should holate in clockwise direction. To calculate the blade angle By , we will be ruse at inlet, velocity buangle Ck B U - In the velocity triangle down aboue, Vers is sclatie velocity of fluid with respect to blade. Uz is velocity of blade at inlet. • Calculation of Uf: W = 120 spm Use unit conversion 14pm = 2o had W = 120 x 21 W = 12.566 rad U = WX RI. Subskelute Rs =- 2.4m and w = 12.566 rad U1 = 12.566 X2-4 Uı = 30.159 M • Calculation of BS In briangle ABC) AB = - Vou Bing — BC = Veicosø ③ In sight angled triangle ABD BD = V2000, -> AB = Vi simo—

Equate the 2 expression of Rength AB in Equation 0 and Veel sin 6 = Visina, Veri = Vi sine, 6. sino The erfression for side BD com also be written as, BD = CD-BC. Subshilade BD- Vicoso, , BC= Vay cos co and CD= U, V, 06U, = -Ven caso tu. Subblitult Ver = . Vi samo from equation , simo V. COBO, = _Vi sinon coBQ +U, . smo VUBO = Visino, col + U. Subblitula Vi=18 M, &- 550, 8 U-30-159 1808855º = - 18 sin 55° cołø +30-159 18 81M 55°cot $ = 306159 – 1860555° cot 6 =, 19.835 1881m55 Ook o = 1.345. 0 = 3663° Subslilule 0= 36-63° im eshirassion 180 -ķ= 0. 180- Be= 36.63° B = 143: 370 | C). To calculate the blade angle Re, we will use velocity Biangle at oullet. By - ro K8=180-$, &

In the velocity briangle drawn above, V42 is relatuje velocity of fluid with seshoect to blade. Uz is velocity of blade at oullet • Calculation of U2; U2 = WR Substitute w= 12.566 sad and Re = 4.3 m U2 = 12.566 X 1.3 U2 =. 16-336 • Calculation of By: an right angled triangle par, PR = Verz sind 6 QR = V2 Los 2 → In sight angled Briangle PRS, . PR = Ve sinds 0 RS = V2c802 -> Equate the 2 erfression of length PR in equation 6 and 8, Vez sinds -- Vz sim os Ver2 = ksino2 0 sind The Expression for side as com le willen as, QS = QR + RS. Substitule. AS = U2 , QR-Vasacos de and RS= Vocoron U2 = Ver2 cos d2 + V₂ COSO2 Subollule Varz = k simos simo, U2 = V2 sim 02 00602 + Ve 2880, sind, V2 Sumo, col O2 = U2 - V2.00802

Substitute Vy = 8.4 and 09 = 77º and. Ug = 16-336 m 8.4 sim 770 cold, = 16-336 - (8.4 X 60377°) 14.946 8.4 8in77. 03 -- 29.53° Subsbilule D = 29.53° in equalion 180 --Ry = 29.53° 180 – 8 - 23:53° B = 150.46°.K Blade angle al oilet