Fluid machinery problem double overhung impulse turbine...

FLUID MACHINERY PROBLEM: A double-overhung impulse-turbine installation is to develop 15,000 bkW at 257 rpm under a net head of 340 m. Determine: a. Specific Speed, Ns b. Wheel pitch diameter, Dp c. The wheel diameter, Do d. The approximate Jet Diameter

Transcript

00:01 Hello student let's solve this question given that given that p is equal to 15 ,000 15 ,000 kilowatt now n is equal to n is equal to 257 rpm now h net h net is equal to 340 meter 340 meter now a part now a specific speed specific speed n s is equal to n root p divided by n power 5 divided by 4 now put the value to 57 root 257 root 15 000 15 000 divided by 340 340 power 5 divided by 4 now solve this net speed ns is equal to 21 .55 we know the power of turbine when to be developed p is equal to rho g r g qh rho g qh now put the value 10 to power 3 multiply 15 ,000 15 ,000 is equal to 1000 multiply 9 .81 multiply 9multiply 9 multiplied 340 now solve this we will get q is equal to 1 ,000 multiply 9 .81 multiply 340 now solve this we will get q is equal to q is equal to q is equal to q is equal to now solve this we will add q is equal to 4 .4 .49 cube per second and absolute absolute velocity of z v1 is equal to root 2gh root 2gh now v1 is equal to now v1 is equal to root 2 multiply 9 .81 multiply 340 this is equal to 881 .81 now q is equal to now q is equal to av1 av1 now this is equal to pi divided by 4 multiply 5 squared multiply 81 .67 now solve this 4 .497 is equal to pi divided by 4 multiply 5 square multiply 81 .67 now 5 is equal to 5.

02:25 Is equal to 0 .264 meter now be part now be part wheel pitch diameter taking 5 taking 5 taking 5 0 .452 0 .4d now u is equal to 5 root root 2g 2g h net 2g h net now u is equal to now u is equal to 0 .95 0.

02:53 95 0 .95 0 .95 0 .95 root root 2 2 multiply 9 .81 multiply 340 taking 5 is equal to 45 5 is equal to 45 now solve this now solve this we will get u is equal to u is equal to 36 .75 meter per second and also u is equal to pi d n 5 dn divided by 60 d is d is wheel pitch diameter we'll pitch diameter wheel pitch diameter wheel pitch diameter now put the value 36 .75 is equal to pi multiply d multiply 257 divided by 60 now solve this d is equal to 2 .7 8 meter d is equal 2 .7 8 meter now c part now now now see part now see part taking jet ratio of turbine is equal to 12 which is suitable for this case m is equal to d divided by d divided by d divided by d is equal to d is equal to wheel diameter and small d is small d is equal to jet diameter jet diameter jet diameter jet diameter now 12 is equal to d divided by zero 0 .264, 0 .264...

Question

FLUID MACHINERY PROBLEM: A double-overhung impulse-turbine installation is to develop 15,000 bkW at 257 rpm under a net head of 340 m. Determine: a. Specific Speed, Ns b. Wheel pitch diameter, Dp c. The wheel diameter, Do d. The approximate Jet Diameter

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