Question

An industrial boiler produces 68,000 kg/hr of saturated steam at 4.5 MPa using residual fuel oil (No. 6 fuel oil). The ambient air temperature is 255 K, which is also the temperature of the oil storage tank. There is no air preheater but the oil is preheated to 373 K. The oil is pressurized to 5 MPa. There is a 50 mm water pressure drop across the air ducts and wind box. An economizer heats the feed water to 400 K. The stack temperature is 470 K. The heat loss due to incomplete combustion is 0.5% of the total heat input, and the radiation heat transfer losses are 5% of the total heat input. The excess air is measured to be 5%. The efficiency of the pump is 90%, and the efficiency of the blower is 70%. Calculate the boiler efficiency. Assume that properties of the products are the same as the properties of air.

Question 2: An industrial biler produces 68,000 kg/hrof saturated steam at 4.5MIPa uwsing residal ftel oil CNo. 6 fuel oil). The ambient air temperature is 255 K, which is also the temperature of the oil storage tanlk. There is no air preheater but the oil is preheated to 373 K. The oil is pressurized to 5 MPa. There is a 50 mm water pressure drop across the air ducts and wind box. An economizer heats the feed water to 400 K. The stack temperature is 470 K. The heat loss due to incomplete combustion is 0.5% of the total heat input, and the radiation heat transfer losses are 5% of the total heat input. The excess air is measured to be 500. The efficiency of the pump is 90%, and the efficiency of the blower is 70%. Calculate the boiler efficiency. Assume that properties of the products are the same as the properties of air. Steam 4.5,MPa saturated Given: 373 K 5 MPa Fuel 470 K EA 5% Air, 255 K Watox, 68,000 kg/h p-0.5 kPa Qu 55%

Answer 1

Given that: - Industrial boiler: - m_s = 68,000 kg/hr, saturated stream pv = 4.5 mpa = 45 bav [stream flow rate] h_2 = 2797.95 kj/kg T_sat = 257.44 degree C Feed meter temperature = 400k = 127 degree C < T_sat p = 45 bav h_1 = 536.509 kj/kg man flow rate of fuel, nf = 1.14 kg/s = 4104 kg/hv calorie value of fuel oil oil no = 6 = 152400 /gallon (Ref: engineers ] = 152400 times 1055.065/3.785 times 10^-3 ms specific gravity between 0.95 - 1.03 let's value = 0.99 (mean value) = 4.248 times 10^10 /ms = 4.248 times 10^7 kg/ms Density of fuel oil no = 6 = 0.99 times 1000 = 990 kg/m^3 \r\n Therefore ev of fuel oil no: 6 = 4.248 times 10^7/990 = 42.909/kj/kg Therefore Boiler efficency (Direct method) = m_s times (h_2 - h_1)/m_f + c_v times 100% = 68,000 times (2797.95 - 536.509)/4104 times 42,909 % = 87.325%