# How to use steam table to solve the following:?

A compressor adiabatically & reversibly compresses a mixture of saturated water and steam from a pressure 1.01 bar and specific volume of 1.5 m^3/kg to 0.313 m^3/kg. USE STEAM TABLES to find the following:

a) fraction of steam at the compressor entrance.

b) the exit entropy in kJ/kgK

c) The exit pressure in bar.

I have the Thermodynamic and Transport Properties of Fluids steam table (by Rogers ad Mayhew). I have no idea how to use it to solve for the above. Please help me. Best answer for the first person to offer help :-) Thanks!

### 2 Answers

- az_lenderLv 77 years agoBest Answer
Go to some page entitled "Saturated Water Properties, SI Units" or like that.

Part(a)

At exactly 1 bar, the specific volume of the steam portion would be 1.694 m^3/kg.

At 1.013 bar, the specific volume of the steam portion would be 1.673 m^3/kg.

Therefore, at 1.01 bar, the specific volume of the steam portion will be around 1.678 m^3/kg.

The specific volume of the water portion will be very close to 0.001 m^3/kg,

so if the specific volume of the mixture is 1.50 m^3/kg, you have

1.678 F + 0.001(1-F) = 1.500

where "F" is the fraction of steam.

1.677 F = 1.499 => F = 89.4%

(b) I think the "adiabatically & reversibly" means that delta-S = 0.

Therefore, we just have to find the ENTRANCE entropy,

On the tables I'm looking at, the entropies of water and steam respectively are

1.307 kJ/(kg K) and 7.358 kJ/(kg K for 1.01 bar.

For a mixture that's 89.38% steam, the specific entropy would be

(0.1062)(1.306 kJ/kg K) + (0.8938)(7.358 kJ/kg K) = 6.715 kJ/(kg K)

This is the entrance entropy, but I believe it's the exit entropy as well.

(c) So we need to find a pressure where a specific volume of 0.313 m^3/kg

could correspond to a specific entropy of 6.715 kJ/(kg K).

At 543.1 kPa, the specific volume of the steam portion is 0.3468 m^3/kg

and at 617.8 kPa, the specific volume of the steam portion is 0.3071 m^3/kg.

For the moment, I'm going to fudge the volume of the water portion, which will

account for no more than 1/300 of the overall volume.

If the specific volume of steam portion is (let's say) 0.314 m^3/kg,

then the pressure should be about

617.8 kPa - (7/38.7)(617.8 kPa - 543.1 kPa) = 604 kPa

but let's see what steam fraction that gives in entropy-land...

Specific entropy of saturated steam at 604 kPa would be about 6.758 kJ/(kg K),

and the specific s of saturated water at 604 kPa would be about 1.934 kJ/(kg K).

For this to end up as a specific entropy of 6.715 kJ/(kg K),

we would need

6.758 F + 1.934 (1 - F) = 6.715

4.824 F = 4.781 and F = 99%, which makes no sense to me.

I would be thinking the steam fraction should decrease in the compression.

So you can't trust my answer to part (c) [604 kPa absolute],

but I'm pretty comfortable with parts(a) and (b).