Question

Derive an expression for the entropy of the steam.

Answer 1

Answer on Question #80424 - Molecular Physics | Thermodynamics

Total Entropy of Steam

Entropy of Water

The change of entropy can be expressed as:

$$\mathrm{dS} = \log_{\mathrm{e}}(\mathrm{T}_1 / \mathrm{T})$$

where

$\mathrm{T} =$ absolute temperature (K)

The entropy of water above freezing point can be expressed as:

$$\mathrm{dS} = \log_{\mathrm{e}}(\mathrm{T}_1 / 273)$$

Entropy of Evaporation

Change of Entropy during evaporation

$$\mathrm{dS} = \mathrm{dL} / \mathrm{T}$$

where

$\mathrm{L} =$ latent heat (J)

Entropy of wet steam

The entropy of wet steam can be expressed as:

$$\mathrm{dS} = \log_{\mathrm{e}}(\mathrm{T}_1 / 273) + \zeta(\mathrm{L}_1 / \mathrm{T}_1)$$

where

$\zeta =$ dryness fraction

Entropy of superheated steam

Change of entropy during super-heating can be expressed as

$$\mathrm{dS} = c_p \log_{\mathrm{e}}(\mathrm{T} / \mathrm{T}_1)$$

where

$c_p =$ specific heat capacity at constant pressure for steam (kJ/kgK)

The entropy of superheated steam can be expressed as:

$$\mathrm{dS} = \log_{e} \left( \frac{T_1}{273} \right) + \frac{L_1}{T_1} + c_p \log_{e} \left( \frac{T_s}{T_1} \right)$$

where

$T_s$ = absolute temperature of superheated steam

$T_1$ = absolute temperature of evaporation

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