# Constant Temperature Process (Steam)

• Jameseyboy
In summary, the law of P1V1 = P2V2 does not apply in an isothermal process where the dryness fraction of steam is given. The pressure and volume values may change depending on the specific conditions, such as the wet/dry mix and temperature. To find the final values of U or H, further properties of water and steam must be used. The pressure may not necessarily remain constant throughout the process, and the specific volume of the steam must be determined in order to accurately calculate the final values.
Jameseyboy

## Homework Equations

In an isothermal process, where the dryness fraction of steam is given - Does the law of P1V1 = P2V2 still stand? Mass is also given here.

I know that I should not be using PV = mRT for steam.

Thanks

Jameseyboy said:

## Homework Statement

In an isothermal process, where the dryness fraction of steam is given - Does the law of P1V1 = P2V2 still stand?

Hi Jameseyboy. Welcome to Physics Forums!

No. The law does not still apply. If you want some more help, you need to tell us more about your specific problem, and what your thinking is about how to approach it. Please try to "dope out" your ideas for us.

Chet

Thanks for getting back to me with my question!

Basically, I am away from the actual question at the moment:

- Constant temperature process
- The process begins with a wet/dry mix
- Dryness fraction given
- P1 is given
- V2 is given

I found out U1 by using steam tables u = uf + x(ug - uf)
I also found V1 by using S.T with the given P1
I found T1 by S.T at P1

I multiplied both of these by the mass to find the actual values, not the per KG value.

However I am not sure how to go about finding the final value of U or H.

The value of V2 does not indicate it is superheated (with the same temp given by P1), so I am basically stumped.

Sorry this is vague but I can only remember a couple of actual values.

Jameseyboy said:
Thanks for getting back to me with my question!

Basically, I am away from the actual question at the moment:

- Constant temperature process
- The process begins with a wet/dry mix
- Dryness fraction given
- P1 is given
- V2 is given

I found out U1 by using steam tables u = uf + x(ug - uf)
I also found V1 by using S.T with the given P1
I found T1 by S.T at P1

I multiplied both of these by the mass to find the actual values, not the per KG value.

However I am not sure how to go about finding the final value of U or H.

The value of V2 does not indicate it is superheated (with the same temp given by P1), so I am basically stumped.

Sorry this is vague but I can only remember a couple of actual values.
You should have found V1 by using the same kind of equation as you used for u.

Now to get the new set of conditions, you need to determine whether V2/m lies between the specific volumes of the liquid and the saturated vapor at T1. If it is, then the only thing that changes is x. The pressure is still P1. Does it lie between these values?

Chet

So to gain V1 I have used "Futher properties of water and steam" using 5 bars of pressure i.e. V1 = vf + x(vg - vf) (obviously multiplying this by the mass).

How am I to assume that pressure is still 5 Bar for the second part of the process?

Edit: The V2/m does not lie at the same T1 and is actually smaller than Vg at T1. I assumed this means its superheated, however after checking the tables the new V1 does not correspond with the value. Really not sure what I am supposed to make of this.

Thanks again for the help

Last edited:

## 1. What is a constant temperature process in steam?

A constant temperature process in steam refers to a thermodynamic process in which the temperature of the steam remains the same throughout the process. This means that the steam is neither gaining nor losing heat, resulting in a constant temperature.

## 2. How is a constant temperature process achieved in steam?

A constant temperature process in steam can be achieved by controlling the heat input and output of the system. This can be done by using a heat exchanger to maintain a constant temperature or by regulating the fuel input in a boiler.

## 3. What are the advantages of a constant temperature process in steam?

One of the main advantages of a constant temperature process in steam is that it allows for better control and stability in industrial processes. It also helps to prevent damage to equipment and reduces energy consumption.

## 4. Can a constant temperature process in steam be used for all types of processes?

No, a constant temperature process in steam is not suitable for all types of processes. It is most commonly used in industrial processes such as power generation, chemical reactions, and heating and cooling systems.

## 5. How does a constant temperature process in steam differ from a constant pressure process?

In a constant temperature process, the temperature remains constant while the pressure may vary. In a constant pressure process, the pressure remains constant while the temperature may vary. Additionally, a constant temperature process is often more energy efficient compared to a constant pressure process.

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