Question

The following data collected from a steam turbine of 25 MW capacity thermal power plant:

Assume R for air as 287 kJ/kg C

Steam condensed = 40000 kh/hr

Temperaure of steam in condenser =40 degree

Dryness of steam enetring condenser = 0.85

Air leakage in the condenser = 140 kg/hr

Temperature of codenser =35 degree

Temperature at the suction of air pump = 32 degree

Borometer reading = 76 cm of Hg

Maximum rise in cooling water temperature =12 degree

Find : 1. Vacuum gauge reading in condenser , 2.Capacity of dry air pump in cume meter per hour 3.Quantity of cooling water passes through condenser in tons per hour

Answer 1

1. Volume of steam per hour = m_s v = m_sxv_g   where m_s= mass steam per hour

                                                                             x= dryness fraction of stem

                                                                             v_g = specific volume of dry steam

                                        V_s= 40000 x 0.85 x 19.52     ( from steam tables at 40^oC v_g= 19.52 m³/kg)

                                             =663680 m³/hr

By Daltons partial pressures law the volume of lekage air is also 663680 m³/hr

Partial pressure of air pa =

                                               = (140 x 0.287 x 313) / 663680

                                                = 0.0189 kPa

                                                = 0.000189 bar

At 40^oC steam pressure in the condenser p_s = 0.0738 bar

Total condenser pressue = p_a + p_s = 0.07398 bar

                                    = 5.44 cm of Hg

2. At air pump suction temperature is 32^oC, and saturation pressure of steam p_s = 0.047596 bar

partial pressure of air p_a= p - p_s = 0.07398-0.047596 = 0.02638 bar

                                                                            = 2.638 kPa

                                 V_a =

                                     = (140 x 0.287 x 305) / 2.638

                                    = 4645.52 m³/ hr

3. For heat balance

         Heat lost by steam = Heat gained by cooling water

         m_s (xh_fg + C <t_g - t_c>) = m_w C (t₂ - t₁)    where h_fg = latent heat at 40^oC from steam tables= 2406.5 kJ/kg

        40000( 0.85 x 2406.5 + 4.186 <40 - 35>) = m_w x 4.186 x 12

                                                            m_w = 1645528.74 kg/ hr

                                                                  = 1645. 52 tons/ hr