CALCULATED by the following formula\(\ETA = \FRAC{m\;\times\;Δ h}{m_{COAL}\;\times\;C.V}\)where, m = water supplied per hour, Δh = rise in enthalpy, mcoal = mass of fuel, CV = calorific valueCalculation:Given:m = 205 kg/hr, mcoal = 23 kg/hr, Δh = 145 kJ/kg, CV = 2050 kJ/kg\(\eta = \frac{m\;\times\;Δ h}{m_{coal}\;\times\;C.V}\) = \(\frac{205\;\times\;145}{2050\;\times\;23}\)η = 0.6304 or 63.04% \(\approx\) 63%

"> CALCULATED by the following formula\(\ETA = \FRAC{m\;\times\;Δ h}{m_{COAL}\;\times\;C.V}\)where, m = water supplied per hour, Δh = rise in enthalpy, mcoal = mass of fuel, CV = calorific valueCalculation:Given:m = 205 kg/hr, mcoal = 23 kg/hr, Δh = 145 kJ/kg, CV = 2050 kJ/kg\(\eta = \frac{m\;\times\;Δ h}{m_{coal}\;\times\;C.V}\) = \(\frac{205\;\times\;145}{2050\;\times\;23}\)η = 0.6304 or 63.04% \(\approx\) 63%

">
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In a boiler, feed water supplied per hour is 205 kg while coal fired per hour is 23 kg. The net enthalpy rise per kg of water is 145 kJ. If the calorific value of coal is 2050 kJ/kg, then boiler efficiency will be

Energy Engineering Function Boilers in Energy Engineering 8 months ago

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Concept:Boiler efficiency is CALCULATED by the following formula\(\ETA = \FRAC{m\;\times\;Δ h}{m_{COAL}\;\times\;C.V}\)where, m = water supplied per hour, Δh = rise in enthalpy, mcoal = mass of fuel, CV = calorific valueCalculation:Given:m = 205 kg/hr, mcoal = 23 kg/hr, Δh = 145 kJ/kg, CV = 2050 kJ/kg\(\eta = \frac{m\;\times\;Δ h}{m_{coal}\;\times\;C.V}\) = \(\frac{205\;\times\;145}{2050\;\times\;23}\)η = 0.6304 or 63.04% \(\approx\) 63%

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Posted on 26 Nov 2024, this text provides information on Energy Engineering related to Function Boilers in Energy Engineering. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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