Thermodynamics:based on first law

[anisha]

Homework Statement

In an adiabatic steam nozzle,steam is expanded from 10 bar and 473k to an exit pressure of 5 bar.nozzle has an isentropic efficiency of 90%.neglect kinetic energy at the inlet.assuming equilibrium condition at exit,find the velocity of nozzle at exit? given data : i)at nozzle inlet: enthalpy=2828KJ/Kg ; entropy = 6.70KJ/KgK
ii)saturated liquid properties at nozzle exit: enthalpy=640 KJ/Kg;entropy=1.86KJ/KgK
iii)saturated vapour properties at nozzle exit: enthalpy=2750 KJ/Kg;entropy=6.82KJ/KgK

Homework Equations

energy equation for nozzle: (h1/2)+((c1)^2/2)=(h2/2)+((c2)^2/2)
isentropic efficiency of nozzle : (h1-h2)/(h1-h2s)
where h2s is isentropic enthalpy at nozzle exit

The Attempt at a Solution

i didn't understand what is the use of giving the efficiency here and how it's related in finding the exit velocity.the isentropic enthalpy i found is h2s=2741.333KJ/Kg,h1=2828KJ/Kg,h2=2750KJ/Kg (given).the correct answer is482 m/s
please do help me ,any hint will be of great use and appreciated!

 

[SteamKing]

Homework Statement

In an adiabatic steam nozzle,steam is expanded from 10 bar and 473k to an exit pressure of 5 bar.nozzle has an isentropic efficiency of 90%.neglect kinetic energy at the inlet.assuming equilibrium condition at exit,find the velocity of nozzle at exit? given data : i)at nozzle inlet: enthalpy=2828KJ/Kg ; entropy = 6.70KJ/KgK
ii)saturated liquid properties at nozzle exit: enthalpy=640 KJ/Kg;entropy=1.86KJ/KgK
iii)saturated vapour properties at nozzle exit: enthalpy=2750 KJ/Kg;entropy=6.82KJ/KgK

Homework Equations

energy equation for nozzle: (h1/2)+((c1)^2/2)=(h2/2)+((c2)^2/2)
isentropic efficiency of nozzle : (h1-h2)/(h1-h2s)
where h2s is isentropic enthalpy at nozzle exit

The Attempt at a Solution

i didn't understand what is the use of giving the efficiency here and how it's related in finding the exit velocity.the isentropic enthalpy i found is h2s=2741.333KJ/Kg,h1=2828KJ/Kg,h2=2750KJ/Kg (given).the correct answer is482 m/s
please do help me ,any hint will be of great use and appreciated!

With regard to the isentropic efficiency:

η = (h1-h2)/(h1-h2s)

Now, if η = 1, that would imply η = 1 = (h1-h2)/(h1-h2s), which would only be true if h2 = h2s.

If η = 0.9, then you are given a hint how to calculate h2 given h1 and h2s.

 

[anisha]

Thanks,but there is no h2s given,instead h2 value is there.i am totally confused now.

 

[rock.freak667]

Thanks,but there is no h2s given,instead h2 value is there.i am totally confused now.

You found h_2s=2741.333KJ/Kg

isentropic efficiency of nozzle : (h1-h2)/(h1-h2s)
where h2s is isentropic enthalpy at nozzle exit

Using this formula, and the fact that you were told that the isentropic efficiency = 90 %, can you put

 

[anisha]

With regard to the isentropic efficiency:

η = (h1-h2)/(h1-h2s)

Now, if η = 1, that would imply η = 1 = (h1-h2)/(h1-h2s), which would only be true if h2 = h2s.

If η = 0.9, then you are given a hint how to calculate h2 given h1 and h2s.

 

[anisha]

With regard to the isentropic efficiency:

η = (h1-h2)/(h1-h2s)

Now, if η = 1, that would imply η = 1 = (h1-h2)/(h1-h2s), which would only be true if h2 = h2s.

If η = 0.9, then you are given a hint how to calculate h2 given h1 and h2s.

 

[anisha]

With regard to the isentropic efficiency:

η = (h1-h2)/(h1-h2s)

Now, if η = 1, that would imply η = 1 = (h1-h2)/(h1-h2s), which would only be true if h2 = h2s.

If η = 0.9, then you are given a hint how to calculate h2 given h1 and h2s.

sir,if i take given exit enthalpy to be h2s then my answer is 374 m/s as exit velocity were as the actual one is 482 m/s.

 

[anisha]

Can anyone tell me whether given enthalpy at the exit is h2s or h2,cause if it's h2s it's not giving me the right answer

 

[SteamKing]

Can anyone tell me whether given enthalpy at the exit is h2s or h2,cause if it's h2s it's not giving me the right answer

h2s is the enthalpy at the nozzle exit assuming that the expansion is isentropic, which the problem statement tells you it is not. In order to find h2s, you use the value of entropy at the inlet condition, namely s1 = 6.70 kJ/kg-K, and the saturated vapor and liquid conditions at the exit pressure of 5 bar, as given in the problem statement. Due to expansion thru the nozzle, the steam exiting may be a mixture of vapor and saturated liquid.

Having found h2s and knowing the isentropic efficiency of the nozzle, you should be able to find the enthalpy at the exit of the nozzle, h2.

In order to keep going back and forth on this point, please post your calculations, not just the final velocity of the steam.

 

[anisha]

kindly explain in more detail!