• https://me.yahoo.com

# Impulse and Reaction Turbines

Velocity triangles for both Impulse and Reaction Turbines and the Force on the Blades

## Introduction

The vast majority of Turbines consist of a number of curved blades or cups which are attached to a wheel and move. This is actually not always true. The author has seen in Southern Sweden a rudimentary wooden turbine which consisted of a wheel with a vertical axis and which was fitted with a number of flat wooden vanes set radially around the circumference. Water entered through three or four wooden "nozzles" of square cross section and after hitting the vanes fell down out of the machine. The efficiency of this arrangement is not known but it is unlikely to be high! It should be added that the head available to the mill was not high.

In Britain water power traditionally relied on Waterwheels.Two sorts were common. The undershot and the overshot. The efficiency of both were poor compared to a modern turbine.

This design worked on a modest head but suffered from the leakage from the wheel. The efficiency was further reduced by the need for the wheel to "push" water along the tail race. It would have been possible to site the wheel clear of the tail race but this would have exacerbate the leakage problem.

This was more efficient. Clearly as the wheel rotated water spilled from the cups and as a result the wheel did not make full use of the available head

It should be stated that the heads available in Britain, particularly in the Southern part, are in general not more than a few feet and quite unsuitable for many designs of modern turbine.

## Velocity Triangles

To analyse the flow through moving curved vanes it is necessary to draw Velocity triangles The following symbols are used in their construction.

• $V_r&space;=&space;\text{The&space;Relative&space;Velocity&space;and&space;&space;is&space;&space;tangential&space;&space;to&space;&space;the&space;&space;blades.}$
• $v&space;=&space;\text{Blade&space;Velocity&space;and&space;is&space;Added&space;to&space;}\;V_r)$
• $V&space;=&space;\text{The&space;Absolute&space;&space;Velocity&space;and&space;is&space;the&space;vector&space;sum&space;of}&space;v&space;\text{and&space;}V_r$
(Note The arrows of v and $\inline&space;v_r$ must follow each other around the triangle)

• $V_w&space;=&space;\text{The&space;velocity&space;of&space;whirl&space;i.e.The&space;Component&space;of&space;V&space;in&space;the&space;direction&space;of&space;v}$
• $V_f&space;=&space;\text{The&space;velocity&space;of&space;flow}&space;\;(\text&space;{Component&space;of&space;V&space;normal&space;to&space;direction&space;of&space;v})$
• The suffix 1 refers to he outlet triangle.
• $\inline&space;\alpha$ and $\inline&space;\beta$ are the inlet and outlet angles of absolute velocity
• $\inline&space;\theta$ and $\inline&space;\phi$ are the inlet and outlet angles relative to the blade velocity.

### Axial Flow Turbines

• At Low Speed the velocity triangles are as follows.

• At High Speed The outlet triangle remains the same but the inlet triangle is now:-

Note
$v&space;=&space;v_1\;\;\;\;and\;\;\;\;V_r&space;=&space;V_{r1}&space;\;\;\text{if&space;there&space;is&space;no&space;friction}$

### Pelton Wheel ( Circumferential )}

The two velocity triangles are for low and high flow. The inlet triangle is a straight line.

## The Force On The Blades

For both types of flow (a and b)

$\text{Jet&space;velocity}&space;=&space;C_v\;\sqrt{2gH}$

Where H is the head behind the nozzle and $\inline&space;C_v$ is the Velocity coefficient.

$\text{weight&space;of&space;water&space;per&space;second}\;,W&space;=&space;w\,a.\;V$

$\text{The&space;blade&space;speed}\&space;,v&space;=&space;\frac{\pi&space;\,d\,N}{60}$

$\text{Force&space;on&space;the&space;vanes}&space;=\text{&space;mass&space;of&space;water/second}\times&space;\text{&space;Change&space;in&space;velocity}$

$=&space;\frac{W}{g}\left(V_w&space;-&space;V_{w1}&space;\right)$

$\text&space;&space;{Work&space;done&space;on&space;the&space;vanes}&space;=\text{&space;Force}\times&space;\text{&space;Velocity}$

$=&space;\frac{W}{g}\left(V_w&space;-&space;V_{w1}&space;\right)$

$\text{The&space;Kinetic&space;energy&space;supplied}\;&space;=&space;\frac{W\,V^2}{2g}$

$\text{The&space;efficiency}\;,\eta&space;&space;=&space;\frac{\text{Work&space;done}}{\text{K.E.\;supplied}}$

$\therefore\;\;\;\;\;\eta&space;&space;=&space;\frac{2\,v(V_1&space;-&space;V_{w1})}{V^2}$

##### For A Pelton Wheel Only
$V_w&space;=&space;V$

$V_{r1}&space;=&space;V_r\;\;\;\;\text{If&space;friction&space;is&space;ignored.}$

$\therefore&space;\;\;\;\;\;V_{r1}&space;=&space;V&space;-&space;v$

$But\;\;\;\;\;V_{w1}&space;+&space;v_1&space;=&space;V_{r1}\,cos\phi$

$=&space;(V&space;-&space;v)\;cos\phi$

$\therefore\;\;\;\;\;V_{w1}&space;=&space;(V&space;-&space;v)\;cos\phi&space;&space;-&space;v$

$\therefore\;\;\;\;\;\eta&space;&space;=&space;\frac{2v}{V^2}\left(V_w&space;-&space;V_{w1}\right)}$

$=&space;\frac{2v}{V^2}\left[(V&space;-&space;v)&space;+&space;(V&space;-&space;v)cos\phi&space;\right]$

For the maximum $\inline&space;\eta$ at a given head and blade angle

$\frac{d\eta&space;}{dv}&space;=&space;0$

Which occurs when v = V/2 i.e. The bucket speed is half the jet speed. This is a theoretical figure and in practice, due to frictional losses, the maximum efficiency is when $\inline&space;v\;\approx&space;\;0.47\;V$

### Example 1

The diagram shows a section of a Pelton wheel which has a 2 in. Diam. Jet which produces 2 cubic ft. of water per second. The blade speed is 40 ft/sec and due to friction $\inline&space;\;V_{r1}&space;=&space;0.9\;V_r$. ( 1 cubic foot of water weighs 62.4 lbs. and g = 32.2 ft/s. Find the Kinetic Energy supplied by the jet per second and the efficiency of the turbine.

To find the force in the X direction.

$Jet\;V&space;=&space;\frac{Q}{area}&space;=&space;\frac{2}{\frac{\pi&space;}{4}\times&space;\frac{1}{36}}&space;=&space;91.7&space;ft/sec.$

$and\;\;\;\;\;\therefore\;\;\;V_w\;=91.7\,ft/sec$

$v_r&space;=&space;V&space;-&space;v&space;=&space;91.7&space;-&space;40&space;=&space;51.7ft/sec$

$And\;\;\;\;\;V_{r1}&space;=&space;0.9V_r&space;=&space;46.53ft/sec$

From the outlet triangle

$(v_1&space;-&space;V_{w1})&space;=&space;V_{r1}\times&space;cos45^0&space;=&space;\frac{46.53}{\sqrt{3}}&space;=&space;32.9ft/sec.$

$And\;\;\;\;\;V_{r1}&space;=&space;40&space;-&space;32.9&space;=&space;7.1ft/sec.$

$Also\;\;\;\;\;V_{f1}&space;=&space;32.9ft/sec$

In the X direction the Force on the vanes = mass of water X change of velocity

$=&space;\frac{2\times&space;62.4}{32.2}\times&space;\left(91.7&space;-&space;7.1&space;\right)&space;=&space;328&space;lbs.$

$\text{Force&space;on&space;vanes&space;in&space;the&space;Y&space;direction}&space;=&space;\frac{2\times&space;62.4}{32.2}\left(0&space;-&space;32.9&space;\right)&space;=&space;127.5lbs.$

$\therefore\;\;\;\;\text{The&space;resultant&space;force}&space;=&space;\sqrt{328^2&space;+&space;127.5^2}&space;=&space;352lbs.$

The resultant is at
$tan^{-1}\;\frac{127.5}{328}\;\;\;\;i.e.\[\[\[at;21^0\,14'\&space;;\text{To&space;&space;the&space;&space;direction&space;&space;of&space;motion.}$

Work done per second on the vanes = The force in the X direction times blade velocity

$=&space;328\times&space;40&space;=&space;13,120\;ft\,lbs/sec$

$\text{The&space;Kinetic&space;energy&space;supplied&space;by&space;the&space;jet&space;per&space;second}&space;=&space;\frac{W\,V^2}{2g}&space;=&space;\frac{2\times&space;62.4\times&space;91.7^2}{2\times&space;32.2}&space;=&space;16,280\;ft\,lbs/sec.$

$Thus\;\text{the&space;efficiency&space;of&space;&space;the&space;turbine}\;&space;=&space;\frac{13220}{16280}&space;=&space;80.6\%$

### Example 2

Obtain an expression for the work done per lb. of flow by a Pelton Wheel in terms of the mean bucket velocity U, the Jet velocity $\inline&space;v_1$ and the outlet bucket angle $\inline&space;\theta$, neglecting all friction losses,

If the loss due to bucket friction and shock can be expressed by:-

$\frac{k_1}{2g}\left&space;(&space;v_1-U&space;\right&space;)^2$

And that due to bearing friction by :-

$\frac{k_2}{2g}\times&space;U^2$

Where $\inline&space;k_1$ and $\inline&space;k_2$ are constants , show that the maximum efficiency occurs when

$\frac{U}{v_1}=\frac{1-\cos\theta+k_1}{2(1-\cos\theta)+k_1+k_2}$

A Pelton wheel runner having a bucket angle of $\inline&space;165^0$ gave on test a maximum efficiency of 0.8 If $\inline&space;U/v_1$ is 0.47. Find$\inline&space;k_1$ and $\inline&space;k_2$ and hence express the losses as a percentage of the jet energy. (B.Sc. Part 2)

From the velocity diagram is can be seen that:-

$\text{The&space;change&space;in&space;relative&space;velocity}\;=v_{r1}+v_{r2}\times&space;\cos(180^0-\theta)$
$\text{But&space;with&space;no&space;losses}\;=v_{r2}=v_{r1}=v_1-U$
$\therefore&space;\;\;\;\;\text{The&space;change&space;in&space;velocity}\;=(v_1-U)(1-\cos\theta)$
$\text{And&space;the&space;work&space;done&space;per&space;lb.}\;=\frac{U}{g}(v_1-U)(1-\cos\theta)$

However there are losses so that:-

$\text&space;{Useful&space;work&space;done&space;per&space;lb.}\;=\frac{U}{g}(v_1-U)(1-\cos\theta)-\frac{k_1}{2g}(v_1-U)^2-\frac{k_2}{2g}U^2$

$\text{The&space;Jet&space;Kinetic&space;Energy&space;per&space;lb.}\;=\frac{{v_{1}}^{2}}{2g}$
$\therefore&space;\;\;\;\;\text{The&space;efficiency}\;\eta=2(n-n^2)(1-\cos\theta)-k_1(1-n)^2-k_2n^2$
$\text{Where}\;\;\;\;\;\;n+\frac{U}{v_1}$

Differentiating and equating to zero gives a maximum value for the efficiency when:-

$2(1-2n)(1-\cos\theta)+2k_1(1-n)-2k_2n=0$
$i.e.\;\;\;\;\text{When}\;\;\;\eta=\frac{U}{v_1}=\frac{1-\cos\theta+k_1}{2(1-\cos\theta)+k_2+k_2}$

Substituting values into the above equation:-

$0.47=\frac{1+\cos15^0+k_1}{2(1+\cos15^0)+k_1+k_2}$
$\therefore&space;\;\;\;\;\;k_2=0.248+1.13k_1$

From equations (49) and (52)

$0.8=2(0.25)(1.966)-k_1(0.53)^2-(0.248+1.13k_1)\times&space;0.22$
$\therefore&space;\;\;\;\;\;k_1=0.243\;\;\;\;\;\text{And}\;\;\;\;\;k_2=0.522$

Using Equation (49)
$\text{The&space;Shock&space;Loss}=100k_1(1-n^2)=24.3(0.28)=6.8\%$
And
$\text{The&space;Windage&space;Loss}=100k_2n^2=52.2\times&space;0.22=11.5\%$

Note The reminder of the total loss $\inline&space;20-11.5-6.8=1.7\%)$ is the loss due to the water leaving the turbine with Kinetic Energy

### Example 3

A Pelton wheel buckets are square to jet when they are $\inline&space;10^0$ ahead of the normal position A. The bucket speed is 0.47 of the jet speed and the water leaves the bucket along BC with a relative velocity of $\inline&space;85\%$ of the ongoing relative velocity. The supply lake is 1000 ft. above the nozzle and of this head $\inline&space;5\%$ is lost in friction (f=0.0075) in the pipe which is 2000 ft long. The nozzle diameter is 1.25.in. and $\inline&space;C_v$ is 0.97

Draw the velocity vector figures for the bucket in the position B and find :-

• The horse power developed.
• The necessary diameter of the pipeline.

(B.Sc.Part 2)

The vector diagram shows the inlet and outlet velocities for the "B" position in which:-

$v_{r2}=0.85v_{r1}\;\;\;\;And\;\;\;\;U_1=\text{The&space;component&space;of&space;the&space;bucket&space;speed&space;U&space;parallel&space;to&space;the&space;jet}$
$\text{And}\;\;\;\;\;\frac{U}{v_1}=0.47$
$\text{The&space;Efficiency},\;\eta\;=\frac{\frac{U_1}{g}(v_{r1}+v_{r2}\cos15^0)}{{v_{1}}^{2}/2g}$
$=\frac{2U_1}{{v_{1}}^{2}}(v-U_1)(1+0.85\cos15^0)$

From the diagram it can be seen that $\inline&space;U_1=U\cos10^)$ and combining this with the above two equations:-

$\text{The&space;efficiency},\;\eta=0.905$

If

• v is the pipeline velocity.
• d is the supply pipe diameter.
• a is the pipe cross sectional area.
• $\inline&space;a_1$ is the cross sectional area of the jet.

Applying the Bernoulli equation for the pipeline and jet

$1000=h_f&space;(For\;pipe)+nozzle\;loss+\frac{{v_{1}}^{2}}{2g}$
$\text{Thus}\;\;\;\;1000=\frac{5}{100}\times&space;1000+\left&space;(&space;\frac{1}{0.97^2}-1&space;\right&space;)\frac{{v_{1}}^{2}}{2g}+\frac{{V_{1}}^{2}}{2g}$
$\therefore&space;\;\;\;\;\;\frac{{v_{1}}^{2}}{2g}=892ft\;\;\;\;\;\text{And&space;thus}\;\;\;\;\;v_1=240\;ft./sec.$

From Equations (63) and (66) The useful head from the wheel, H is given by:-

$H=\eta\times&space;\frac{{v_{1}}^{2}}{2g}=0.905\times&space;892=808\;ft.$

$W=wa_1v_1$
$\therefore&space;\;\;\;\;\text{From&space;equation&space;(67)},\;W=62.4\times\frac{\pi}{4}\times&space;\left&space;(&space;\frac{1.25}{12}&space;\right&space;)^2\times&space;240=128\;lb./sec.$
$\therefore&space;\;\;\;\;\text{The&space;horse&space;power&space;developed}=\frac{WH}{550}=\frac{128\times&space;808}{550}=188&space;h.p.$

For the pipeline 5% of the total head is lost to friction and hence using the Darcy equation:-

$1000\times\frac{5}{100}=\frac{4\times&space;0,0075\times&space;2000&space;\times&space;v^2}{2gd}$
$\therefore&space;\;\;\;\;\;v^2=53.6\times&space;d$

But since $\inline&space;W=wav$

$128=62.4\times&space;\frac{\pi}{4}d^2\times&space;v$
$\therefore&space;\;\;\;\;\;v=\frac{2.62}{d^2}\;\;\;\;\text{or}\;\;\;\;v^2=\frac{6.8}{d^4}$
$\text{Hence}\;\;\;\;d^5=\frac{1}{7.88}$
$\text{Thus&space;&space;the&space;required&space;diameter&space;of&space;the&space;supply&space;pipe&space;}\;\;=\frac{1}{1.51}\;ft.\;=7.95\;in.$

## Turbine With Curved Vanes And An Inward Radial Flow ( Francis Or Gerard

Turbine )

The following diagram shows the velocity triangles for both low and high speed

Let:-

• The weight of water/second striking the vanes be W lb/sec.
• Tangential momentum/second at entry $\inline&space;=&space;\frac{W}{g}V_w$
• Moment of momentum at entry =$\inline&space;\frac{W}{g}V_w\times&space;v$
• Moment of momentum at outlet =$\inline&space;\frac{W}{g}(V_{w_1}\times&space;v_1)$

Then the Torque on the vanes equals the change of moment of momentum per second

$=\frac{W}{g}\left(V_w\,v&space;-&space;V_{w1}\,v_1&space;\right)$

The work done per second on the vanes equals the Torque times the angular velocity

$=&space;\frac{W}{g}\left(V_w\,v&space;-&space;V_{w1}\,v_1&space;\right)\Omega$

$\text{But}\Omega&space;r&space;=&space;v\;\;\;\;\text{and}\;\;\;\;\Omega&space;r_1&space;=&space;v_1$

$\therefore\;\;\;\;\text{work&space;done/second}&space;=&space;\frac{W}{g}\left(V_w\,v&space;-V_{w1}\,v_1\right)$

This is the EULER equation which can be applied to any type of turbine or centrifugal pumps

### Example 4

Derive an expression for the hydraulic efficiency of a turbine in terms of the tangential velocities of the runner, the velocities of whirl at inlet and outlet and H the supply head. Take all velocities in the direction of the runner as positive.

An inward flow reaction turbine discharges radially and the velocity of flow is constant and equal to the velocity of discharge from the suction tube. Show that the hydraulic can be expressed by:-

$\eta&space;=1/1+\frac{\frac{1}{2}\tan^2\alpha}{1-(\tan\alpha/\tan\theta)}$

Where $\inline&space;\alpha$ and $\inline&space;\theta$ are the guide vane angles at inlet (B.Sc. Part 2)

Please refer to the velocity triangles in the diagram.

$\text{The&space;Force}=\text{The&space;rate&space;of&space;change&space;of&space;Momentum}$
$\therefore&space;\;\;\;\;\text{The&space;available&space;tangential&space;force&space;at&space;the&space;wheel&space;entry}=\frac{W}{g}\times&space;v_{w1}\;lb.$
$\text{And}\;\;\text{The&space;available&space;useful&space;power&space;at&space;the&space;wheel&space;entry}=\frac{W}{g}\times&space;v_{w1}U_1\;ft.lb/sec.$
$\text{Similarly&space;the&space;useful&space;power&space;expelled&space;at&space;exit}=\frac{W}{g}\times&space;v_{W2}\times&space;U_2$

$\text{But&space;the&space;power&space;available}=WH\;ft.lb./sec.$

$\therefore&space;\;\;\;\;\text{The&space;power&space;given&space;to&space;the&space;wheel}=\frac{W}{g}\left&space;(&space;v_{W1}U1-v_{W2}U_2&space;\right&space;)\;ft.lb./sec.$
$\therefore&space;\;\;\;\;\eta=\frac{&space;v_{W1}U1-v_{W2}U_2&space;}{gH}$

$\text{Since}\;v_2\;\text{is&space;radial}\;\;\;\;v_{W2}=0$

$\text{The&space;work&space;done&space;per&space;lb.&space;of&space;water}=v_{W1}\times&space;\frac{U_1}{g}$

From the vector triangles it can be seen that:-

$\frac{v_f}{v_{W1}}=\tan\alpha\;\;\;\;\text{i.e.}\;\;\;\;&space;v_{W1}=\frac{v_f}{\tan\alpha}$
$\text{Also}\;\;\;\;\frac{v_f}{U_1-v_{W1}}=\tan(180^0-\theta)=-\tan\theta$

From the above equations

$U_1=v_f\times&space;\frac{\tan\theta-\tan\alpha}{\tan\theta\tan\alpha}$
$\text{The&space;work&space;done&space;per&space;lb.}=\frac{{v_{f}}^{2}}{g}\times&space;\frac{\tan\theta-\tan\alpha}{\tan\theta\tan^2\alpha}$

If there are no guide vane losses

$H=\text{Work&space;done&space;per&space;lb.}+\frac{{v_{f}}^{2}}{2g}$
$\eta=\frac{\text{Work&space;done&space;per&space;lb.}}{\text{Work&space;done&space;per&space;lb.}+\frac{{v_{f}}^{2}}{2g}}$
$\therefore&space;\;\;\;\;\eta=\frac{1}{1+\frac{{v_{f}}^{2}}{2g}/\text{work&space;done&space;per&space;lb.}}$

From Equations (94) and (97)

$\eta=\frac{1}{1+\frac{\frac{1}{2}\tan^2\alpha\tan\theta}{\tan\theta-\tan\alpha}}=1/1+\frac{\frac{1}{2}\tan^2\alpha}{1-(\tan\alpha/\tan\theta)}$

### Example 5

State briefly the reasons for fitting a draft tube to a reaction turbine and sketch three common types.

In a vertical Francis Turbine, the runner speed is 380 r.p.m. and the available head across the turbine is 150 ft. The inlet runner is 6 ft.above the tail race level and the area and diameter of the runner at inlet are 2.6 sq.ft. and 3 ft respectively. The guide and runner vane angles at inlet are shown in the diagram. The water enters a draft tube without whirl 5.25 ft. above the tail-race and the draft tube diameter here is 1 ft. 10 in. At outlet the draft tube diameter is 2 ft. 4 in. If the frictional losses in the runner amount to 9 ft.lb/lb and in the draft tube 5 ft.lb./lb. and the overall efficiency is 0.9 X the hydraulic efficiency find

• a) The B.H.P. of the Turbine
• b) The hydraulic efficiency.
• The pressure head at the inlet to the runner in ft.lb./lb.
• The pressure head at the entry to the draft tube in ft.lb./lb.

(B.Sc.Part 2)

The reasons for fitting a draft tube are that it allows the turbine to be mounted above the tail race without there being a significant loss of head a loss of head. In addition because of the gradual increase in cross section the discharge velocity from the turbine is not all wasted as some is converted into useful a pressure head.

Three types of draft tubes are shown in the following sketch. From the left these are a simple cone; Spreading and elbow types;

For the calculations please refer to the following two diagrams

$U_1=R_1\times&space;\frac{2\piN}{60}=\frac{1.5&space;\pi\times380}{30}=59.7\;ft./sec.$
$\frac{v_1}{\sin110^0}=\frac{U_1}{\sin55^0}\;\;\;\;\;\text{i.e.}\;\;\;\;v_1=\frac{59.7\times&space;0.94}{0.82}=68.5\;ft./sec.$
$v_{W1}=v_1\cos15^0=68.5\times&space;0.996=66.2\;ft./sec.$
$\text{The&space;work&space;done&space;per&space;lb.}=\frac{U_1v_{W1}}{g}-\frac{59.7\times&space;66.2}{g}=123\,ft.lb./lb,$
$\text{Thus&space;the&space;Hydraulic&space;Efficiency}=\eta_h=\frac{\text{Work&space;done&space;per&space;lb.}}{H}=\frac{123}{150}=0.82$
$\therefore&space;\;\;\;\;\text{&space;The&space;overall&space;Efficiency}=\eta_o=0.9\times0.82=0.738$
$W=wA_1v_{f1}=wA_1v_1\sin15^0$
$=62.4\times&space;2.6\times&space;68.5\times&space;0.259=2880\;lb./sec.$

Therefore the B.H.P.generated is given by:-

$\frac{\eta_0\times&space;WH}{550}=\frac{0.738\times&space;2880\times&space;150}{550}=580\;&space;b.h.p.$

For the Draft tube

$Q=\frac{W}{w}=Av=\text{&space;A&space;Constant&space;and&space;therefore&space;if&space;}v_d\;\text{is&space;the&space;exit&space;velocity&space;from&space;the&space;draft&space;tube}$
$\frac{2880}{62.4}=\frac{\pi}{4}\times&space;\left&space;(&space;\frac{22}{12}&space;\right&space;)^2\times&space;v_{f2}=\frac{\pi}{4}\times&space;\left&space;(&space;\frac{28}{12}&space;\right&space;)^2\times&space;v_d$
$\therefore&space;\;\;\;\;v_{f2}=17.45\;ft./sec.\;\;\;\;\text{And}\;\;\;\;v_d=10.75\;ft./sec.$

$\text{The&space;total&space;head,&space;H}=\text{Guide&space;loss&space;+&space;Rotor&space;loss&space;+Work&space;done&space;+Draft-tube&space;loss&space;+}\;\frac{{v_{d}}^{2}}{2g}$
$\text{i.e.}\;150=\text{Guide&space;loss}+9+123+5+\frac{10.75^2}{2g}$
$\therefore&space;\;\;\;\;\text{Guide&space;loss}\;=11.14\,ft.$

If suffix 1 refers to the runner inlet and suffix 2 refers to the draft-tube inlet then:-

$H_1=\frac{p_1}{w}+\frac{{v_{1}}^{2}}{2g}+z_1$

Also

$h_2=H_1-\text{Work&space;done/lb.}-\text{Rotor&space;loss}=\frac{p_2}{w}+\frac{{v_{f2}}^{2}}{2g}+z_2$
$\text{i.e.}\;\;\;\;150-11.14-123-9=\frac{p_2}{w}+\frac{17.5^2}{2g}+5.25$
$\therefore&space;\;\;\;\;\frac{p_2}{w}=-3.14\,ft.\;\;\;\;\text{i.e.&space;3.14ft.below&space;atmospheric&space;pressure}$

$\therefore&space;\;\;\;\;150-11.14=\frac{p_1}{w}+\frac{{68.5^2}}{2g}+6$
$\text{Thus}\;\;\;\;\;\frac{p_1}{w}=59.86\;ft.$

### Example 6

In a Francis Turbine the guide angle is $\inline&space;8^0$, the inlet angle of the moving blades is $\inline&space;110^0$ and the outlet angle is $\inline&space;20^0$

Both fixed and moving vanes reduce the flow area by 15%. The runner is 24 in. outside diameter and 16 in. inside diameter and the widths at entrance and exit are 2 in. and 3 in. respectively. The pressure at entry to the guides is + 87 ft. head and the kinetic energy can be neglected. The pressure at discharge is - 6 ft. head

If the losses in the guides and moving vanes are taken as $\inline&space;8f^2/2g$ where f = the radial component of flow calculate:-

1) The speed of the runner in rev/min. for tangential flow on to the runner vanes. 2) The Horse power given to the runner by the water

(B.Sc. Part 2)

Let

• The wheel inlet area be $\inline&space;A_1\;ft.^2$
• The wheel outlet area be $\inline&space;A_2\;ft.^2$

From the above:-

$A_1=\pi\times&space;2\times&space;\frac{2}{12}\times&space;0.85=\frac{0.85\pi}{3}$
$\text{And}\;\;\;\;A_2=\pi\times&space;\frac{16}{12}\times&space;\frac{3}{12}\times&space;0.85=\frac{0.85\pi}{3}$

$\therefore&space;\;\;\;\;\;A_1=A_2$

Thus it can be seen that $\inline&space;v_{f1}=v_{f2}=&space;f_v\;\;\;\text{say}$ From the velocity triangle:-

$\frac{v_{f1}}{U_1-v_{W1}}=\tan70^0=2.75$

$\text{And}\;\;\;\;\frac{v_f}{v_{W1}}=\tan8^0=0.1405$
$i.e\;\;\;\;\;v_{W1}=7.12v_f$

From equations (123) and (125)
$U_1=7.484v_f$
$\therefore&space;\;\;\;\;U_2=\frac{R_2}{R_1}U_1=\frac{2}{3}\times&space;7.484v_f=4.98v_f$
$\frac{v_f}{U_2-v_{W2}}=\tan20^0=0.364$
$i.e.\;\;\text{From&space;equation&space;(127)}\;\;\;\;2.74v_f=4.98v_f-v_{W2}$
$\therefore&space;\;\;\;\;\;v_{W2}=2.24v_f$

$\text{Work&space;done&space;per&space;lb.}=\frac{U_1v_{W1}-U_2v_{W2}}{g}$

Substituting in value from above.

$\text{Work&space;done&space;per&space;lb.}=\frac{{v_{1}}^{2}}{g}\left&space;(&space;7.484\times&space;7.12-4.98\times&space;2.24&space;\right&space;)=42.1\frac{{v_{f}}^{2}}{g}$
$\text{Also}\;\;\;\;{v_{2}}^{2}={v_{f}}^{2}+{v_{W2}}^{2}={v_{f}}^{2}+\left&space;(&space;2.24v_f&space;\right&space;)^2$
$\therefore&space;\;\;\;\;\;\frac{{v_{2}}^{2}}{2g}=6\times&space;\frac{{v_{f}}^{2}}{2g}$

The energy at the runner exit is given by:-

$\frac{p_2}{w}+\frac{{v_{2}}^{2}}{2g}+z_2$
$\text{Hence&space;}H_2=-6+\frac{{v_{f}}^{2}}{2g}+z_2$

where $\inline&space;z_2$ is the position head measured above the tail race.

Thus measured from the Tail race the Total Head equals the Work done per lb. + the guide vane losses + the losses in the rotor + $\inline&space;4H_2$

$\text{Thus}\;\;\;\;87+z_2=42.1\times&space;\frac{{v_{f}}^{2}}{g}+8\times&space;\frac{{v_{f}}^{2}}{2g}-6+6\frac{{v_{f}}^{2}}{2g}+z_2$
$\therefore&space;\;\;\;\;\;\frac{{v_{f}}^{2}}{g}=1.89\;\;\;\;\text{Or}\;\;\;\;v_f=7.8\;ft./sec.$

By substitution using equation (126)

$U_1=R_1\frac{2N\pi}{60}\;\;\;\text{Where&space;N&space;is&space;the&space;speed&space;of&space;the&space;rotor&space;in&space;r.p.m.}$

$\therefore&space;\;\;\;\;N=\frac{30\times&space;58.5}{1\times&space;\pi}=558\;r.p.m.$

From Equation (132)

$\text{The&space;work&space;done&space;per&space;lb.}=42.1\times&space;1.89=79.7\;ft.lb./lb.$
$W=wA_1v_f=62.4\times&space;7.8\times&space;\frac{0.85\pi}{3}=433\;lb/sec.$
$\therefore&space;\;\;\;\;\text{Horse-power&space;at&space;runner}=\frac{W\times&space;\text{work&space;done&space;per&space;lb.}}{550}$
$=\frac{433\times&space;79.7}{550}=63\;&space;h.p.$