Answer / sr

Mass balance is also considered as material balance, but we
can do material balance by considering the molar flow rate
(kmol/hr)of the input and output material. When we do the
material balance considering the mass flow rate ( kg/hr)
then it is called mass balance.

Answer / zeeshan

WHAT EVER IS THE UNIT MOLAR OR KG ,WEIGHT ON EACH SIDE OF
THE EQUATION WILL BE UNDER THE LAW OF CONSERVATION OF MASS

Answer / viral

By the Material balance we know that how much amount of material c0‹2nsume,required,also by-product.while the mass balance we know only the quantity of material for particuler operation.

Answer / bharat

material balance and mass balance are one and the same.both are based on conservation of mass not in volume or moles.

ENGINEERING MATERIAL - EXAMPLE 12.3 : Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.

1 Answers  

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PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039 % carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air-fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.

1 Answers  

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What is minimum fluidization velocity?

9 Answers   Haldia,

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pls send me HPCL & reliance test paper for chemical engineering.....

4 Answers   HPCL, Reliance,

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What is a good approximation for the heat transfer coefficient of hydrocarbons inside 3/4" tubes?

0 Answers  

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What are the applications of screen analysis in the chemical industry?

0 Answers  

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we have 2 O2/Comb Analyzers Servomex Model 2700 in service for one of the furnaces in our chemical plant. It is giving High Temperature Alarm. Checked Temperature and found it to be 800 Deg C whereas Set Point is 700 Deg c. It is strange why is temperature not at SP. Also response is 23 % with 21 % Inst Air and with N2 reading is not coming down. please clarrify this issuses very urgent

1 Answers  

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How can we derive power factor equation p=vi cos phi?

0 Answers  

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Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt % methanol and the second contains 70.0 wt % methanol. If 200 kg of the first mixture is combined with 150 kg of the second, what are the mass and composition of the product? The symbol of weight percent is wt %.

2 Answers  

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Diffrence between centrifugal & recipocating pump?

8 Answers   Essar, IOCL,

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what are the precautions u are taking while starting HT motors?

2 Answers   HPCL, MRPL Mangalore Refinery and Petrochemicals Limited,

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ENGINEERING PHYSICS - EXAMPLE 30.4 : (a) Time evolution in Heisenberg picture, according to Ehrenfest theorem is m (d / dt) < r > = < p >, where m = mass, r = position, p = momentum of a particle. If v = velocity, prove that m < v > = < p >. (b) Lande g-factor is given by Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. If Gl = 1 and under approximation of Gs = 2, prove by calculation that Gj = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ].

1 Answers  

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