Need help looking up thermal properties

[bp123]

Homework Statement

I'm trying to look up the specific enthalpy (h) at 4.8Mpa and 40 degrees celsius for water vapor/steam. It should be a saturated liquid.

The Attempt at a Solution

I know the specific enthalpy should be ~171.78KJ/Kg but I can't for the life of me figure out how the textbook example comes up with that.
The value of specific enthalpy for a saturated liquid at 40ºC is 167.54KJ/Kg, obviously not quite right.. Any assistance would be great, probably a quick easy answer of something I'm missing / don't know.

 

[rock.freak667]

If it was stated that it is a saturated liquid then your enthalpy would just be the h_f value at 40C or 4.8 MPa.

If they did not state that it was a liquid, you might need to find the quality, x, and find the enthalpy using the equation h=h_f+xh_fg

 

[bp123]

If it was stated that it is a saturated liquid then your enthalpy would just be the h_f value at 40C or 4.8 MPa.

If they did not state that it was a liquid, you might need to find the quality, x, and find the enthalpy using the equation h=h_f+xh_fg

I guess a little more information may be needed.. This is part of a problem dealing with a simple rankine cycle, however the example also accounts for pump/turbine efficiencies and pressure/temperature drops along the lines. It gives the pressure directly out of the pump as 5Mpa and once the working fluid has reached the boiler (just before going in, this is point 3) it is now at the 4.8Mpa & 40ºC.

Because it just came out of the pump and has yet to go into the boiler I know for a fact that the fluid is still a saturated liquid. As well as that piece of information, if you enter the 4.8/40º into an online thermo table it will tell you the enthalpy is 171.78 and the quality is N/A. This is an example out of the book, the auther just failed to explain how they arrived at this number.

Online thermo calc for steam/water I found while trying to figure this out earlier.. 1bar=100kPa, so my pressure of interest is 48bar.
http://www.steamtablesonline.com/steam97web.aspx


Thanks for the quick reply!

 

[rock.freak667]

Oh you mean the point is outside of the curve if that is the case, then I think you need to use
dh=v dP

$$\int_3 ^4 dh = \int_3 ^4 vdP$$

$$h_4 - h_3 = v_3 (P_4-P_3)$$

where v₃ is the specific volume at 3 (v_f). Try that.

 

[bp123]

Oh you mean the point is outside of the curve if that is the case, then I think you need to use
dh=v dP

$$\int_3 ^4 dh = \int_3 ^4 vdP$$

$$h_4 - h_3 = v_3 (P_4-P_3)$$

where v₃ is the specific volume at 3 (v_f). Try that.

I'm assuming you mean where state three is the values at saturation and 4 is what I'm looking for?

Substituting with that:
$$h_4 - 167.54 = .001008(4800-7.384)$$
$$h_4=172.371KJ/Kg$$

Still not quite the 171.77 they have, but its a lot closer then what I found. I'm curious how the online thermo calculator goes about doing it..

 

[rock.freak667]

I am not exactly sure how the calculators do it. That is how I learned to do the basic Rakine cycle. Point 4 was usually calculated using the formula I gave you. I do not know if the calculators use a different method or formula.