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Tutorial solution of Thermal Power station (Principles of power system Vk-Mehta)

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Mohammad Usman Javed Sial

This document contains solutions to multiple problems related to principles of power systems and heat and energy calculations. Solution 1 calculates the heat produced by 0.75kg of coal and the heat equivalent of 1KWh to find the efficiency. Solution 2 calculates the overall thermal and electrical efficiency and the heat produced per hour by a 75MW plant. It then finds the coal consumption per hour. Solution 3 calculates the maximum units that can be generated according to a load factor and the corresponding coal consumption per day. It also calculates the overall efficiency. Solution 4 calculates the maximum energy that can be produced in a day by a 60,000KW plant and the corresponding coal consumption and specific fuel consumption per day.

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Tutorial solution Principles of power system by VK Mehta Solution#1 Heat produced by fuel by 0.75kg Coal= 0.75x Kcal Heat equivalent of 1KWh=860Kcal η=electrical output in heat units/heat of combustion 0.15=860/0.75x X=7644.44 Kcal/Kg Solution#2 η overall= ηthermal* ηelectrcial = 0.30*0.80=0.24 Heat produced/Hour =H= 75MW*860/0.24 =268750*106 Kcal Coal consumption per hour = H/Calorific value = (268750*106 Kcal)/ (6400 Kcal/Kg) =41.992188 tons= 42tons Solution #3 Units generated per Day: Maximum energy that can produced according to load factor: =65,000 kW*0.40*24 =624,000 KWh As coal consumption 1KWh >> 0.5Kg So, for 624,000kWh this can be 312,000Kg or 312tons Efficiency=electrical output in heat units/heat produced by fuel
Overall Efficiency= (624,000*860) / (312,000*15000) = 0.11466= 11.466% Solution #4 Maximum energy that can be produced in day: =plant Capacity * hour = 60,000KW*24h = 1440,000 KWh Calorific Values: 6950Kcal/Kg Electrical output in heat units = 1440, 000*860 = 1238400000 Kcal Coal consumption per day: =heat produced by fuel /calorific value = 1238400000/6950 =178187.0504 Kg = 178.187 tons Specific fuel consumption: =178187.0504Kg/1440000kWh =0.123Kg/kWh Solution#6 Calorific value??? 1Kg/kWh efficiency=15% 0.15=860/1x X=5733.33kcal/kg www.UsmanJavedSial.blogspot.com www.slideshare.net/ItsSial

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Tutorial solution of Thermal Power station (Principles of power system Vk-Mehta)

  • 1. Tutorial solution Principles of power system by VK Mehta Solution#1 Heat produced by fuel by 0.75kg Coal= 0.75x Kcal Heat equivalent of 1KWh=860Kcal η=electrical output in heat units/heat of combustion 0.15=860/0.75x X=7644.44 Kcal/Kg Solution#2 η overall= ηthermal* ηelectrcial = 0.30*0.80=0.24 Heat produced/Hour =H= 75MW*860/0.24 =268750*106 Kcal Coal consumption per hour = H/Calorific value = (268750*106 Kcal)/ (6400 Kcal/Kg) =41.992188 tons= 42tons Solution #3 Units generated per Day: Maximum energy that can produced according to load factor: =65,000 kW*0.40*24 =624,000 KWh As coal consumption 1KWh >> 0.5Kg So, for 624,000kWh this can be 312,000Kg or 312tons Efficiency=electrical output in heat units/heat produced by fuel
  • 2. Overall Efficiency= (624,000*860) / (312,000*15000) = 0.11466= 11.466% Solution #4 Maximum energy that can be produced in day: =plant Capacity * hour = 60,000KW*24h = 1440,000 KWh Calorific Values: 6950Kcal/Kg Electrical output in heat units = 1440, 000*860 = 1238400000 Kcal Coal consumption per day: =heat produced by fuel /calorific value = 1238400000/6950 =178187.0504 Kg = 178.187 tons Specific fuel consumption: =178187.0504Kg/1440000kWh =0.123Kg/kWh Solution#6 Calorific value??? 1Kg/kWh efficiency=15% 0.15=860/1x X=5733.33kcal/kg www.UsmanJavedSial.blogspot.com www.slideshare.net/ItsSial