ChemicalThermodynamicsmediumFE ExamCoursework

Enthalpy of Wet Steam

Steam at $1\ \text{MPa}$ has a quality of $x = 0.80$ .

At this pressure: $h_f = 762.8\ \text{kJ/kg}$ and $h_{fg} = 2015.3\ \text{kJ/kg}$ .

Find the specific enthalpy of the wet steam.

Given

x	0.8 -	Steam quality
hf	762.8 kJ/kg	Saturated liquid enthalpy
hfg	2015.3 kJ/kg	Enthalpy of vaporisation

Hint 1

Quality is the fraction of the mass that has been vaporised.

Hint 2

h = h_f + x·h_fg.

Hint 3

Sanity-check: the answer must sit between h_f and h_f + h_fg.

Worked solution — try the problem first

Quality $x$ is the mass fraction that is vapour. Any specific property of a saturated mixture is the liquid value plus the quality times the vaporisation increment:

$h = h_f + x\,h_{fg}$

$= 762.8 + 0.80(2015.3) = 762.8 + 1612.2 = 2375.0\ \text{kJ/kg}$

$\boxed{h = 2375\ \text{kJ/kg}}$

Bounds check. The answer must lie between $h_f = 762.8$ (saturated liquid, $x=0$ ) and $h_g = h_f + h_{fg} = 2778.1$ kJ/kg (saturated vapour, $x=1$ ). At $x = 0.8$ we expect a value 80% of the way up that range: $762.8 + 0.8(2015.3)$ — exactly what we computed.

The same relation works for $u$ , $v$ , and $s$ — replace $h$ with the property of interest and $h_{fg}$ with the corresponding $u_{fg}$ , $v_{fg}$ , or $s_{fg}$ .

Concepts:Steam qualitySaturated mixturesEnthalpy

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