Introduction to Fluid Machines (Lectures 49 to 53)

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1 Introduction to Fluid Machines (Lectures 49 to 5) Q. Choose the crect answer (i) (ii) (iii) (iv) A hydraulic turbine rotates at N rpm operating under a net head H and having a discharge Q while developing an output power P. The specific speed is expressed as N Q (a) 4 ( gh ) (b) (c) N P ( gh ) 54 N P ( gh ) 4 N Q (d) ( gh ) 54 [Ans.(b)] A hydraulic pump rotates at N rpm operating under a net head H and having a discharge Q. The specific speed is expressed as N Q (a) 4 ( gh ) (b) (c) N P ( gh ) 54 N P ( gh ) 4 N Q (d) ( gh ) 54 [Ans.(a)] Two hydraulic turbines are similar and homologous when there are geometrically similar and have (a) the same specific speed (b) the same rotational speed (c) the same power output (d) the same Thoma s number [Ans.(a)] A turbomachine becomes me susceptible to cavitation if (a) velocity attains a high value (b) pressure falls below the vapour pressure (c) temperature rises above a critical value (d) Thomas cavitation parameter exceeds a certain limit

2 (v) Governing of turbines means (a) the discharge is kept constant under all conditions (b) allow the turbine to run at runaway speed (c) the speed is kept constant under all conditions(loads) (d) the power developed is kept constant under all conditions [Ans.(b)] [Ans.(c)] Q. A Pelton wheel operates with a jet of 50 mm diameter under the head of 500 m. Its mean runner diameter is.5 m and it rotates with a speed of 75 rpm. The angle of bucket tip at outlet is 5, coefficient of velocity is 0.98, mechanical losses equal to % of power supplied and the reduction in relative velocity of water while passing through bucket is 5%. Find (i) the fce of jet on the bucket, (ii) the power developed (iii) bucket efficiency and (iv) the overall efficiency. Inlet jet velocity is Cv gh m/s π Flow rate of water Q d 4 π ( ) m /s 4 Bucket speed is DN π π 44.8 m/s From inlet velocity diagram, elocity of jet relative to wheel at inlet r m/s Tangential component of inlet jet velocity w 86.8 m/s It is given that m/s r r Inlet w r w Outlet r 5 From outlet velocity triangle as shown, w rcos β cos m/s

3 Fce imparted on the bucket is F Q ( w w) 0.75( ) N 65.5 kn Power developed P F kw 7. 0 kw 7. MW Bucket efficiency 7. 0 η b % ( ) Overall efficiency η η ( 0.0) o b % Q. An inward flow reaction turbine wking under a head of H has inlet and outlet vane angles both equal to 90. Show that the peripheral velocity is given by gh + C tan α where C is the ratio of velocity of flow at outlet and inlet and α is the guide vane angle. Also show that the hydraulic efficiency can be expressed as η h + C tan α The inlet and outlet velocity triangles are shown in the figure below. α Inlet velocity triangle f r β r Outlet velocity triangle f From the inlet velocity triangle, we have w tan α f w w fcotα

4 When water flows through the vanes, we get f H w g g H w + g g f Cf H w g + g [ f C f ] C H fcotα fcotα + g g + C tan α H f g tan α tanα f gh + C tan α f gh tanα + C tan gh + C tan α α f Hydraulic efficiency is given by w η h gh f f tanα tanα + C tan α g f g tan α, η h + C tan α Q4. A Kaplan turbine operating under a net head of 0 m develops 6 MW with an overall efficiency of 80%. The diameter of the runner is 4. m, while the hub diameter is m and the dimensionless specific speed is rad. If the hydraulic efficiency is 90%, calculate the inlet and exit angles of the runner blades at the mean blade radius if the flow leaving the runner is purely axial. Power delivered Power available overall efficiency Therefe, Q

5 Q 0.94 m /s Dimensionless specific speed NP KST gh Therefe, ( ) 54 6 ( 6 0 ) N ( 000) ( 9.8 0) 54 which gives N 7.4 rad/s Blade speed at mean radius ( 4. + ) m m/s Again w m gh ηh ghηh w m m/s 6.98 m β r a m β r Inlet velocity triangle Outlet velocity triangle From inlet velocity diagram tan β which gives β 5 From outlet velocity diagram 9.5 tan β 6.98 which gives β ( ) Q5. A conical type draft tube attached to a Francis turbine has an inlet diameter of m and its area at outlet is 0 m. The velocity of water at inlet, which is 5 m above tail race level, is 5 m/s. Assuming the loss in draft tube equals to 50 % of velocity head at outlet, find (i) the pressure head at the top of the draft tube, (ii) the total head at the top of the draft tube taking tail race level as datum, and (iii) power lost in draft tube. 5

6 elocity of water at outlet from draft tube 5 π.77 m/s 4 0 (i) Let p be the pressure at inlet to the draft tube. Applying energy equation between the inlet and outlet of the draft tube, we have p 5 (.77) (.77) g p (.77) 5 g 9.8 p 6.0 m g (ii)the total head at inlet (.77) m 9.8 (ii) Head lost in draft tube (.77) m 9.8 Q6. Calculate the least diameter of impeller of a centrifugal pump to just start delivering water to a height of 0 m, if the inside diameter of impeller is half of the outside diameter and the manometric efficiency is 0.8. The pump runs at 000 rpm. Least diameter of impeller f a pump to just start would crespond to a situation when centrifugal head developed static lift manometric efficiency 0 Therefe, 7.5 m g 0.8 (subscripts and refer to the inlet and outlet of the impeller). D Again, D 4 Hence, 7.5 g m/s m/s Hence least diameter of impeller 6

7 60. D 0.6 m π 000 Q7. A centrifugal pump is required to wk against a head of 0 m while rotating at the speed of 700 rpm. If the blades are curved back to an angle of 0 to the tangent at outlet tip and velocity of flow through impeller is m/s, calculate the impeller diameter when (i) all the kinetic energy at impeller outlet is wasted and (ii) when 50% of this energy is converted into pressure energy in pump casing. From the outlet velocity triangle w cot β w f cot f Energy given to the fluid per unit weight r (.46 w ) g g (a) nder the situation when the entire kinetic energy at impeller outlet is wasted (.46) 0 g g From the outlet velocity triangle ( ) f + w Therefe, ( ) ( ) g g which gives 0. m/s Impeller diameter is D 0.55 m πn π 700 (b) When 50% of the kinetic energy at impeller outlet is converted into pressure energy in pump casing, we can write ( ) ( ) g g The feasible solution is 7.5 m/s Hence, D 0.48 m π 700 7

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