26 PROPERTIES OF STEAM AND AMMONIA 



Each diagram gives several families of curves. Lines parallel to the 

 coordinate axes give, respectively, values of heat content and entropy as 

 read on the scales along the margin. There is a family of constant- 

 pressure curves, in the superheat region a family of constant-tempera- 

 ture curves, and in tin- mixture region a family of constant-quality curves. 

 Any point on the diagram represents a definite state of the fluid. If 

 the point lies in the region of superheat the heat content, entropy, pres- 

 sure, and tempi rat UK are read off directly; if it lies in the mixture 

 region the quality is given but the temperature must be obtained from 

 tin- pressure. Two important properties, the volume and energy, are 

 not given by the diagram as constructed. While it is possible to con- 

 struct constant- volume and constant-energy curves, the inclusion of so 

 many families of curves on a single diagram would lead to confusion. 

 Furthermore, at low pressures the volume changes so rapidly that it is 

 impossible to read volumes with any degree of accuracy. The volume and 

 ^y may, however, be easily obtained from the other properties 

 Thus to find the volume: If the point lies in the region of superheat 

 read the pressure and temperature from the diagram and simply look 

 up the corresponding value of v in Table 3; if it lies in the mixture region 

 read the pressure and quality from the diagram, look up the value of the 

 saturation volume v" for the pressure, and multiply this by the quality. 

 Having the specific volume, the energy is readily obtained from the 

 relation 



n = i - 144 Apv 

 = i- 0.1852 pv. (log o.i 852 = 1.26758.) 



If the pressure is given in inches of mercury the formula becomes 



u = i 0.091 pv. 



Illustrative Examples. --The following examples illustrate some of the 

 more important uses of the diagrams and tables. 



Example i.* Find the properties of steam at a pressure of 120 Ib. per sq. 

 in. and a temperature of 412 F. 



From the steam diagram the point that represents the state of the steam is 

 found at the intersection of the curves /> = 120 and / = 412. From the scales 

 are read the values i - 1231 B.t.u., s = 1.637. From Table 3 'the volume of 

 i Ib. is found to be 4.16 cu. ft. Therefore 



u - t - 0.1852 pv - 1232 - 0.1852 X f20 X 4.16 = 1138.5 B.t.u. 



Example 2. Steam in the initial state p = 120 Ib., / = 412 F. expands 

 adiabatically. At what pressure does it become dry and saturated? 



* In this example, the use of the diagram is superfluous, for the values of i and 5 are 

 obtained as easily and with greater accuracy from Table 3. When the problem involves a 

 change of state or a comparison between two states, as in Examples 2 and 3, the advantage of 

 the diagram becomes more apparent. 



