Re: Sugar Cane Bagasse
What are the
1. Specific Heat of Bagasse at 50% moisture, 40% moisture
and 25% moisture?
2. What is bu;lk density of bagasse at 50% moisture, 40%
moisture and 25% moisture?
3. What is specific gravity of bagasse at 50% moisture, 40%
moisture and 25% moisture?
4. What is the specific heat of flue gases generated from
boiler fuelled by bagasse?
5. What is the specific gravity of flue gases generated
from boiler fuelled by bagasse?
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Question 50 - An aqueous solution with 2.5 g of a protein dissolved in 600 cubic centimeters of a solution at 20 degree Celsius was placed in a container that has a water-permeable membrane. Water permeated through the membrane until the h - level of the solution was 0.9 cm above the pure water. (a) Calculate the absolute temperature of the solution, T in Kelvin, where T (Kelvin) = T (degree Celsius) + 273.15. (b) Calculate the osmotic pressure, P of the solution by using the formula P = hrg where h is level of the solution, r is density of water with 1000 kg per cubic meter, g = 9.81 N / kg as gravitational acceleration. (c) Calculate the concentration of the protein solution, C in kg / cubic meter. (d) Calculate the molecular weight of the protein, (MW) = CRT / P where R = 8.314 Pa cubic meter / (mol K) as ideal gas constant.
Question 45 - According to Raoult’s law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
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i am going to appear vedanta written test so plz can any body suggest me for this on madhu.5393@gmail.com
FOOD ENGINEERING - QUESTION 23.2 : (a) A dryer reduces the moisture content of 100 kg of a potato product from 80 % to 10 % moisture. Find the mass of the water removed in such drying process. (b) During the drying process, the air is cooled from 80 °C to 71 °C in passing through the dryer. If the latent heat of vaporization corresponding to a saturation temperature of 71 °C is 2331 kJ / kg for water, find the heat energy required to evaporate the water only. (c) Assume potato enters at 24 °C, which is also the ambient air temperature, and leaves at the same temperature as the exit air. The specific heat of potato is 3.43 kJ / (kg °C). Find the minimum heat energy required to raise the temperature of the potatoes. (d) 250 kg of steam at 70 kPa gauge is used to heat 49,800 cubic metre of air to 80 °C, and the air is cooled to 71 °C in passing through the dryer. If the latent heat of steam at 70 kPa gauge is 2283 kJ / kg, find the heat energy in steam. (e) Calculate the efficiency of the dryer based heat input and output, in drying air. Use the formula (Ti - To) / (Ti - Ta) where Ti is the inlet (high) air temperature into the dryer, To is the outlet air temperature from the dryer, and Ta is the ambient air temperature.
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In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).