# Thermodynamics:based on first law

• anisha
In summary: The isentropic enthalpy, h2s, is the theoretical enthalpy at the nozzle exit if the expansion is isentropic (i.e. no heat transfer, reversible process). In our problem, the actual expansion is not isentropic due to the presence of friction and other irreversibilities in the nozzle. So, to find the actual enthalpy at the nozzle exit, we use the isentropic

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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!

Last edited:
anisha said:

## 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.

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

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

You found h2s=2741.333KJ/Kg

anisha said:
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

SteamKing said:
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.

SteamKing said:
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.

SteamKing said:
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.

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

anisha said:
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.

kindly explain in more detail!

## 1. What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, but can only be converted from one form to another.

## 2. How does the first law of thermodynamics apply to systems?

The first law of thermodynamics applies to all closed systems, which are isolated from their surroundings and do not exchange matter or energy with the outside world. This law states that the total energy within a closed system remains constant.

## 3. Can the first law of thermodynamics be violated?

No, the first law of thermodynamics is a fundamental law of physics and cannot be violated. It has been tested and proven to hold true in all situations.

## 4. How does the first law of thermodynamics relate to heat and work?

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. This means that any heat added to a system will either increase the internal energy or do work, but the total energy will remain constant.

## 5. What is an example of the first law of thermodynamics in action?

An example of the first law of thermodynamics can be seen in a car engine. The chemical energy in the fuel is converted into heat energy, which is then converted into mechanical energy to move the car. The total energy remains constant, but it is converted into different forms throughout the process.

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