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#### spirally

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##### rather simple question on steady flow
« on: February 24, 2007, 08:50:41 AM »
hi there, i am new to this forum,and i have a question.assuming the efficiency of a turbine to be 0.85,
and a fluid flows into a turbine with intial enthalpy as h1, and leaves with and enthalpy of h2.also the mass flow rate is m.taking intial and final velocity as negligible,and the height to be same,what is the power developed.i know how to find the work rate which will be the power,but the problem in class is will i multiply to the efficiency or divide.i did divide,i hope its right.pls i expect an answer soon.

#### Dzoni

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##### Re: rather simple question on steady flow
« Reply #1 on: February 28, 2007, 07:32:06 AM »
hi there, i am new to this forum,and i have a question.assuming the efficiency of a turbine to be 0.85,
and a fluid flows into a turbine with intial enthalpy as h1, and leaves with and enthalpy of h2.also the mass flow rate is m.taking intial and final velocity as negligible,and the height to be same,what is the power developed.i know how to find the work rate which will be the power,but the problem in class is will i multiply to the efficiency or divide.i did divide,i hope its right.pls i expect an answer soon.

Remembering whether to multiply or divide is a rote way of learning and is not what engineers do and will get you nowhere. You need to understand why something is, and then deduce it when you need it, without the risk of forgetting.

So, to hopefully further your understanding, here's an explanation:

Firstly, it doesn't make sense to say "taking initial and final velocity as negligible" - that means that the fluid isn't flowing (or it's a pipe with an impractically big diameter)! You can, however, say "taking the difference between final and initial velocity to be negligible"...

85% efficiency means that of the all the possible work done on the turbine, only 85% of it is used "usefully" by the turbine. That tells you that the power developed is 85% of the work that the system is doing on the turbine, not so?

Using an energy balance, you can calculate how much work the system is doing, therefore the turbine is producing... (I'll let you finish that sentence)

If you still have trouble understanding, let me know.

#### Donaldson Tan

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##### Re: rather simple question on steady flow
« Reply #2 on: February 28, 2007, 03:55:18 PM »
turbine power (kW) = efficiency * m * (h1 - h2) where
1. hi: kJ/kg
2. m: mass flow, kg/s
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### mbeychok

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##### Re: rather simple question on steady flow
« Reply #3 on: March 02, 2007, 01:05:01 PM »
Geodome:

Your equation is correct except that you forgot to mention that it is a constant entropy expansion. That is, the ?H is at constant entropy.
Milton Beychok
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#### Donaldson Tan

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##### Re: rather simple question on steady flow
« Reply #4 on: March 02, 2007, 01:15:48 PM »
That is, the ?H is at constant entropy.

Constant entropy in a real turbine?

"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### mbeychok

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##### Re: rather simple question on steady flow
« Reply #5 on: March 02, 2007, 02:56:37 PM »
Geodome:

To calculate the power developed by a turbine, one first calculates the theoretical power (i.e., 100% efficient) from the ?H at constant entropy. One then multiplies that by the efficiency to develop the delivered horsepower.

Since your equation equation included the efficiency, then the ?H should have been denoted as being at constant entropy.

Now, if one had a real turbine and actually measured the inlet and outlet enthalpies, the equation would not need to include the efficiency.
« Last Edit: March 02, 2007, 03:40:28 PM by mbeychok »
Milton Beychok
(Visit me at www.air-dispersion.com)