How Do You Calculate Enthalpy Using Steam Tables and Dryness Fraction?

[sperrya]

Homework Statement

Find the enthalpy of steam at 0.4 MPa and 0.7 dryness

Homework Equations

The Attempt at a Solution

At a pressure of 0.4MPa dry saturated steam has a total enthalpy of hg = 2,739 kj/kg.

∴ 0.7 kg of steam has an enthalpy of (0.7 x 2,739) = 1917.3 kj

The total heat of 0.3kg of water is 0.3hf (at 0.4MPa), so 0.3 x 604.8 = 181.44 kj

∴ Total heat for wet steam = 181.44 + 1917.3 kj/kg
= 2,098.74 kj/kg


This is my first time using steam tables so am struggling to know if I am gripping it or not. Any help you can offer in order for me to answer this question is VERY much appreciated.

Regards

 

[rock.freak667]

Yes, your answer is correct.

Your method is correct but it might take a little longer to do if you are in an exam. (the same formula with less lines would be)

h = h_f+ xh_fg= h_f+x(h_g-h_f)

(Your steam table might have values for h_fg sometimes)


Formula is the same for specific entropy and specific volume in that

s=s_f+xs_fg

v = v_f+xv_fg

 

[sperrya]

Thank you rock.freak667

 

[sperrya]

Hi rock.freak667,

Sorry, can you just clarify how my calculations would fit into that equation?

I can follow it through but I am unsure what your X stands for? Can you clarify.

hf + x (hg - hf)

∴ 604.8 + x (2739-604.8)

∴ 604.8 + x (2134.2)

Can you clarify for me?

thanks a lot

 

[rude man]

Hi rock.freak667,

Sorry, can you just clarify how my calculations would fit into that equation?

I can follow it through but I am unsure what your X stands for? Can you clarify.

hf + x (hg - hf)

∴ 604.8 + x (2739-604.8)

∴ 604.8 + x (2134.2)

Can you clarify for me?

thanks a lot

x = dryness = 70% = 0.7.