Question

1st stage: (1$\rightarrow$2) 2 $\alpha_1$ $V_{U1}$ $U_1$ $V_1$ $\beta_1$ $W_1$ $V_{m1}$ $\alpha_2$ $V_{U2}$ $U_2$ $V_2$ $W_2$ $\beta_2 = 90^\circ$ $V_{m2}$ $V_{U3}$ $U_3$ $\alpha_3$ $V_3$ $W_3$ $\beta_3 = 90^\circ$ $V_{m3}$ $U_4$ $\beta_4$ $\alpha_4 = 90^\circ$ $V_{m4}$ $V_4$ $W_4$ 2nd stage: (3$\rightarrow$4) FIGURE 3: Theoretical velocity triangles of a cross-flow turbine runner.

          1st stage: (1$\rightarrow$2)
2
$\alpha_1$
$V_{U1}$
$U_1$
$V_1$
$\beta_1$
$W_1$
$V_{m1}$
$\alpha_2$
$V_{U2}$
$U_2$
$V_2$
$W_2$
$\beta_2 = 90^\circ$
$V_{m2}$
$V_{U3}$
$U_3$
$\alpha_3$
$V_3$
$W_3$
$\beta_3 = 90^\circ$
$V_{m3}$
$U_4$
$\beta_4$
$\alpha_4 = 90^\circ$
$V_{m4}$
$V_4$
$W_4$
2nd stage: (3$\rightarrow$4)
FIGURE 3: Theoretical velocity triangles of a cross-flow turbine runner.

1st stage: (1→2)
2
α1
VU1
U1
V1
β1
W1
Vm1
α2
VU2
U2
V2
W2
β2 = 90^∘
Vm2
VU3
U3
α3
V3
W3
β3 = 90^∘
Vm3
U4
β4
α4 = 90^∘
Vm4
V4
W4
2nd stage: (3→4)
FIGURE 3: Theoretical velocity triangles of a cross-flow turbine runner.

Added by Raul B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adam Moon

Step 1

First, we need to understand the variables in the schematic diagram: - Alpha (α): The angle between the absolute velocity (Va) and the axial direction of the turbine runner. - Beta (β): The angle between the relative velocity (Vr) and the axial direction of the Show more…

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Thanks for your feedback!

From the schematic diagram of Crossflow turbine below, what is the formula to get the angle of Beta? and make the example to get the Beta from certain Alpha! (source image: Numerical Investigation of the Internal Flow in a Banki Turbine, Jesus De Andrade, 2011). VU1 U1 Tx1 T B1 Vi > 1st stage: (1 - 2) Wi Vm1 VU2 K U2 a2 W2 V2 32= 90 Vm2 U3 X3 33=90 Vm3 U4 B4 2nd stage: (3 - 4) a4=90 V m4 V4 W4 FiGuRE 3: Theoretical velocity triangles of a cross-flow turbine runner.