Solve problems using data extracted from thermodynamic property tables

In this section we are concerned with the interpretation and extraction of data on thermodynamic properties of working fluids listed in tables as arranged by Rogers and Mayhew.

These tables, commonly known as ‘steam tables’, give values for the properties of steam and refrigerants over an extensive range of pressures and temperatures.

The ability to understand and extract data from the tables extends into the solution of problems in this section and also later when the steady flow energy equation is dealt with.

Thermodynamic Property Tables

For this section you will need to make reference to the steam tables – these are available under the requirements tab

Layout and use of thermodynamic tables for water and steam

The figure below duplicates the information given at the top of page 4 in the Rogers & Mayhew tables. The ‘s’ figures in the right hand columns of the actual tables relate to entropy values and these are not required for this unit.

© A.Henderson, UHI

The various symbols in the eight columns are identified with their quantities and units as below.

With reference to pages 3, 4 and 5 of the Rogers & Mayhew tables:

The first column headed p is the absolute pressure measured in bar, where 1 bar = 1 x 10⁵ N m^-2 or 100 kN m^-2.

The second column headed T_s is the temperature in °C at which the water boils (saturation temperature). Note how T_s changes relative to pressure.

The third column v_g is the specific volume in m³ of 1 kg of completely dry saturated steam. That is at pressure of 2 bar, saturation temperature is 120.2°C and 1 kg of dry steam occupies a volume of 0.8856 m³.

The fourth column u_f is termed the specific internal energy of saturated liquid. That is the heat energy required to raise the temperature of 1 kg of water from 0°C to saturation temperature in (kJkg^-1). This is a constant volume operation.

The fifth column u_g is the specific internal energy of 1 kg of completely dry saturated steam. That is the heat energy required to raise the temperature of 1 kg of water at 0°C to saturation temperature plus the heat energy required to completely evaporate it (specific enthalpy of evaporation) as if the operation were carried out at constant volume.

The sixth column h_f is the specific enthalpy of saturated liquid – i.e. the enthalpy of

1 kg of water from 0°C to saturation temperature at constant pressure. Note that u_f and h_f are identical down to 4.5 bar and then they gradually drift apart since h_f increases slightly faster than u_f.

The seventh column h_fg is the specific enthalpy of evaporation – i.e. the heat energy required to completely evaporate 1 kg of water at saturation temperature to 1 kg of dry saturated steam at constant pressure and at same temperature.

The eighth column h_g is the specific enthalpy of saturated vapour and is the enthalpy of 1 kg of dry saturated steam at constant pressure measured from water at 0°C.

On page 2 of the Rogers & Mayhew tables the same properties are given but are set out against the reference of saturated water temperature (T°C) in the first column.

The graph below illustrates three phases of steam formation and incorporates enthalpy values from tables.

© A.Henderson, UHI

From A to B, a heat transfer of 417 kJ kg-1 raises the temperature from 0°C to saturation temperature (boiling point) of 99.6°C at the pressure of 1 bar

From point B, if more heat is added, the boiling water will evaporate to form steam at the same temperature and pressure. At point C, the enthalpy of evaporation process is completed by a further heat energy transfer of 2258 kJ kg-1. At point C the steam is in a completely dry saturated state.

In the region C to D, further heat addition produces superheated steam.

Superheated Steam

So far we have considered pages 2 to 5 of the thermodynamic tables. These pages set out the properties and heat energy requirements for steam to be raised from water at 0°C to dry saturated steam at different pressures.

When steam has a temperature higher than its saturation temperature for a stated pressure, then the steam is in a superheat state in which case we use pages 6 to 9 of the steam tables.

The following explanation of the columns on these pages will enable their use:

The remaining columns list these same properties corresponding to various degrees of superheat.

Rule

In order to define the condition of superheated steam it is necessary to state both the pressure and temperature of the steam. Thus, if a temperature is quoted for steam in a problem, check it against the tables and if it is higher than (Ts) for the corresponding pressure then superheated tables must be used.

The difference between the superheated temperature (T) and the saturation temperature (Ts) is called the degree of superheat.

Where exact values of the condition of steam are not listed in the tables, linear interpolation for both pressure and temperature may therefore be required. Exemplars and tutorials on the use of steam tables cover this aspect.

Specific Volume of Wet Ste

The specific volume of steam with a dryness fraction x is given by:

$v_x=v_f+x{v}_{fg}$ 

This can be illustrated by a graph of temperature against specific volume like the one shown below:

© A.Henderson, UHI

If we look at the graph we can see that v_fg­ = v_g – v­_f

_­This means that v_x = v_f + x(v_g – v­_f) = v_f + xv_g – xv­_f = (1-x) v_f + xv_g

_­Because v_{f ­}is so small compared to v­_g­, when can ignore the first term. This means that:

$v_x=x{v}_g$ 

Example

Determine the specific volume of wet steam having a pressure of 1.25 MN m^-2 and dryness fraction 0.9.

$1.25MN{m}^{-2}=1.25\text{bar}$ 

$v_x=x{v}_g$ 

$v_x=0.9\frac{0.1632\text{(12 bar)}+0.1512\text{(13 bar)}}{2}=0.04148m^3{kg}^{-1}$ 

Specific internal energy of wet steam

The specific internal energy of steam with a dryness fraction of x is given by:

$u_x=u_f=x{u}_{fg}$ 

This is illustrated in the graph of temperature against specific internal energy shown below:

© A.Henderson, UHI

If we look at the graph we can see that:

$u_{fg}=u_g-u_f$ 

Which means that

$u_x=u_f+x(u_g-u_f)=u_f+x{u}_g-x{u}_f$ 

Which in turn means that

$u_x=(1-x)u_f+x{u}_g$ 

Example

If we look at an example:

Determine the specific internal energy in wet steam at a pressure of 4 bar when it has a dryness fraction of 0.87

Specific internal energy

$u_x=(1-x)u_f=x{u}_g$ 

→ $u_x=(1-0.87)605+0.87\times 2554$ 

→ $u_x=78.6665+2221.98=2300.63kJ{kg}^{-1}$ 

Specific Enthalpy of Wet Steam

The specific enthalpy of steam with a dryness fraction x is given by:

$h_x=h_f+x{h}_{fg}$ 

This is illustrated on the graph of temperature against specific enthalpy shown below:

© A.Henderson, UHI

Example

So again if we look at an example calculation:

Determine the specific enthalpy of wet steam at a pressure of 70 kN m-2 and having a dryness fraction of 0.85.

Pressure of 70 kN m^-2 = 0.7 bar

Specific enthalpy:

$h_x=h_f+x{h}_{fg}$ 

$h_x=377+0.85\times 2283$ 

$h_x=377+1940.55=2317.55kJ{\text{kg}}^{-1}$ 

Using Steam Tables and Wet Steam Formulae

Example:

Determine the specific internal energy, specific enthalpy and specific volume of steam at a pressure of 3 bar (300 kN m-2) when it is .82 dry. What is the saturate temperature?

From steam tables at a pressure of 3 bar:-

Interpolation of Steam Tables

When quantities cannot be extracted directly from tables, intermediate values need to be interpolated between the nearest listed values above and below the required value.

Examples:

1. Determine the specific volume of wet steam at 68 bar:

At 65 bar: $v_g=0.02972m^3{\text{kg}}^{-1}$ 

At 70 bar:$v_g=0.02737m^3{\text{kg}}^{-1}$ 

Difference is therefore 0.00235

Therefore at 68 bar:

$v_g=0.02972-\left(\frac{3}{5}\times 0.00235\right)=0.02831m^3{\text{kg}}^{-1}$ 

Or

$v_g=0.02737+\left(\frac{2}{5}\times 0.00235\right)=0.02831m^3{\text{kg}}^{-1}$ 

Either method is perfectly acceptable

2. Determine the specific enthalpy of dry saturated steam at 52 bar.

At 50 bar: $h_g=2794kJ{\text{kg}}^{-1}$ 

At 55 bar: $h_g=2790kJ{\text{kg}}^{-1}$ 

Difference is therefore 4

Therefore at 52 bar:

$h_g=2794-\left(\frac{2}{5}\times 4\right)=2792.4kJ{\text{kg}}^{-1}$

Or

$h_g=2790+\left(\frac{3}{5}\times 4\right)=2792.4kJ{\text{kg}}^{-1}$

3. Determine the specific internal energy of superheated steam at 14 bar and 325°C.

At 10 bar and 325°C:

$u=\frac{2794+2875}{2}=2834.5kJ{\text{kg}}^{-1}$ 

At 15 bar and 325°C:

$u=\frac{2784+2868}{2}=2826.0kJ{\text{kg}}^{-1}$ 

Which gives a difference of 8.5

Therefore at 14 bar and 325°C:

$u=2834.5-\left(\frac{4}{5}\times 8.5\right)=2827kJ{\text{kg}}^{-1}$

Or

$u=2826+\left(\frac{1}{5}\times 8.5\right)=2827kJ{\text{kg}}^{-1}$