LEARN - MECHANICAL ENGINEERING

Problem: 1   A vessel of volume 0.04 m 3 contains a mixture of saturated water and steam at a temperature of 250°C The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and internal energy. Given data: V=0.04m 3 T= 250°C m 1 = 9 kg To find: p, m, v, h, S, and U   Solution: From Steam Tables corresponding to 250°C, v f =v 1 = 0.001251 m/kg V g = V s = 0.050037 m/kg p= 39.776 bar   Total volume occupied by the liquid, V 1 = m 1 v 1 = 9 x 0.001251 = 0.0113 m 3 Total volume of the vessel, V = Volume of liquid + Volume of steam V 1 +V s V s =0.0287 m 3   Mass of steam, m s = V s / V s   =0.0287 / 0.050037 =0.574 kg Mass of mixture of liquid and steam, m=m 1 +m s = 9+ 0.574 = 9.574 kg   Total specific volume of the mixture, v = V / m   = 0.04 /9.574 = 0.00418 m 3 /kg   We know that, v=v f +x V fg V fg = v g - v f 0.00418 = 0.001251 + x (0.050037 -0.001251) x=0.06   From Steam

  Properties of Pure Substances If any two independent intensive properties of a simple compressible system are defined, other properties automatically assume definite values.  These properties can be expressed in terms of charts, tables or equations. This chapter covers the charts and tables of properties of steam. Pure Substances Substances   of fixed chemical composition are known as pure substances    Example : Water, Helium, Nitrogen, Oxygen etc. Substances exist in any one of the three phases namely solid, liquid and gas. For example, H 2 O may exist in the form ice (solid), Water (Liquid) or Steam (Gaseous). In all these phases it will have the same chemical composition. A   Mixture of two or more phases of a pure substance should also be regarded as pure substance. If Water and Steam Co-exits in a container, the chemical composition of both the Vapour and liquid phases will be identical. Hence this heterogeneous system is also a pure substance. Phase Transform