vane. The axial flow area of water through the runner is $25m^2$. If the runner blade angle at inlet is radial determine :

i) Hydraulic efficiency of the turbine

ii) Discharge through the turbine

iii) Power developed by the runner

iv) Specific speed of turbine

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 08

Years: MAY 2015

Given :

Runner diameter $D_o= 4.5 m$

Speed $N=40 r.p.m$

Guide blade angle $\alpha =145^\circ $

Runner blade angle at outlet $\phi = 25 ^\circ $

Flow area $a=25m^2$

Runner blade angle at inlet is radial

$\theta =90^\circ, Vf_1=Vr_1 $ and $U_1= Vw_1$

For Kaplan turbine, the discharge is given by the product of area of flow and velocity of flow

As area of flow is constant and hence $Vf_1=Vf_2$

$(\because Q= \text {Area of flow} \times Vf_1= \text {Area of flow } \times Vf_2)$

The tangential speed of turbine at inlet

$U_1=\dfrac {\pi D_o N}{60} = \dfrac {\pi\times 4.5\times 40}{60} \\ =9.42 m/s $

Also $U_2=U_1 = 9.42 m/s $

From inlet velocity triangle, $\tan (180^\circ-\alpha) = \dfrac {Vf_1}{U_1} \\ \tan(180^\circ - 145^\circ)= \tan 35^\circ = \dfrac {Vf_1}{U_1}\\ Vf_1 = U_1\tan 35^\circ \\ =9.452\tan35^\circ \\ =6.59$

Also $Vw_1=U_1=9.42 m/s $

From outlet velocity triangle

$\tan \phi = \dfrac {Vf_2}{U_2+Vw_2}\hspace{1.5cm}$ (Where $Vf_2=Vf_1=6.59$ and $U_2=U_1=9.42$)

$\tan 25^\circ = \dfrac {6.59}{9.42 +Vw_2} \\ \therefore Vw_2 +9.42 = \dfrac {6.59}{\tan25^\circ} \\ Vw_2 = 14.13-9.42 = 4.71 m/s \\ V_2 =\sqrt{Vf_1^2+Vw_2^2}= \sqrt{6.59^2+4.71^2}= \sqrt{43.43+22.18} \\ =8.1m/s $

Using equation

$H-\dfrac {V_2^2}{2g} = \dfrac 1g[Vw_1U_1-Vw_2U_2]$

Here -ve sign is taken as the absolute velocity at inlet and outlet (i.e $V_1$ and $V_2$) are in the same direction and hence change of velocity will be with a -ve sign

$H- \dfrac {8.1^2}{2\times 9.81} = \dfrac 1{9.81}[9.42\times 9.42-4.71\times 9.42]\\ H- 3.344 = \dfrac 1{9.81} [88.736-44.368]=4.522 m \\ H= 4.522+3.344=7.866m $

i) Hydraulic efficiency is given by equation

$N_n = \dfrac {Vw_1U_1 - Vw_2U_2}{g\times H}\\ = \dfrac {9.42\times 9.42-4.71\times 9.42}{9.81\times 7.866} = 0.575=57.5\% $

ii) Discharge through turbine is given by

Q = Area of flow $\times $ velocity of flow

$Q=25 \times Vf_! = 25\times 6.59=164.75m^3/s $

iii) Power developed by turbine = $\dfrac {\text {Work done per second }}{1000} \\ =\dfrac 1g \dfrac {[Vw_1U_1-Vw_2U_2]}{1000}\times \text {weight of water} \\ = \dfrac {1}{9.81}[\dfrac {9.42\times 9.42-4.71\times 9.42}{1000}]\times p\times g\times Q \\ =\dfrac 1{9.81}[\dfrac {9.42\times9.42-4.71\times 9.42 }{1000}]\times 1000\times 9.81\times 164.75\\ =6867 Kw$

iv) Specific speed is given by the relation

$N_s = \dfrac {N\sqrt P}{H^{5/4}} = \dfrac {N\sqrt{6867}}{7.866^{5/4}} = \dfrac {40\times \sqrt{6867}}{7.866^{5/4}} \\ = \dfrac {40\times 82.867}{13.173} \\ = 251.62 rpm$