Answer / dhaval modh

unit operation means the operation in which only physical
change occur not chemical.ex.fluid operation,mechanical
operation,heat tranfer operation, mass tranfer
operation.ie.distillation,drying,adsorption etc......

Answer / darshan jagatap

MAIN CHARACTERISTICS OF UNIT OPERATIONS IS THERE IS ONLY
PHYSICAL CHANGE (PHASE CHANGE) AND NO REACTIONS

CHARACTERISTICS:

PHASE CHANGE IS TOBE OBTAINED BY MAINTAING THE PARAMETERS
SUCH AS TEMP, PRESSURE, FLOW, LEVEL ETC

Answer / alok kumar

basically fluid machanics ,heat transfer operation operation
,mass transfer comes in catagory of unit operations.
the basic thing is that during 1 or more unit operations
there is no chemical composition change occur only physical
changes such as temp. etc. change.

Answer / ritam

The maint terms in unit operations are Distillation,
crystallization, leaching, and drying etc.

Answer / praveen peda

main unit operations are DISTILLATION,
DRYING ,LEACHING,ABSORPTION, ADSORPTION,etc

Answer / anusha

the main terms of unit operations are-
drying
size reduction
distillation
evaporation
crystallisation

Answer / nimesh patel

Unit operatons means operation in which physical changes
takes place.
e.g. Drying, extraction

Answer / nisha sharma

basic laws, molecular units, mole fraction, gas laws,
primary and secondary quantities, dimensional formulas,
dimensions, dimensionless equations,dimensional
analysis,graphical integration, all expressions to express
concentration like molarity, molality,etc.

What is a surfactant?

0 Answers  

Explain some common piping materials used to transport slurries?

0 Answers  

which is the major cement producing state in western India ?

2 Answers   BNPM,

Explain how can you prevent bridging in a dilute phase pneumatic conveying system?

1 Answers  

Are you having any experience in antibody designing?

0 Answers  

name of the fraction at which benzene xylene and toulene is obtained during coal tar distillation

3 Answers   IFFCO,

POLYMER ENGINEERING - QUESTION 24.3 : The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18, with their respective quantities in N units having the ratio of Na : Nb : Nc = 2 : 3 : 5. (a) Find the numerical average molecular weight of the polymer by using the formula (Ma Na + Mb Nb + Mc Nc) / (Na + Nb + Nc). (b) Find the weighted average molecular weight of the polymer by using the formula (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) / (Ma Na + Mb Nb + Mc Nc). (c) Calculate the polydispersity Q by using the answer in (b) divided by answer in (a). (d) Find the volumetric average molecular weight of the polymer by using the formula (Ma Na Ma Ma + Mb Nb Mb Mb + Mc Nc Mc Mc) / (Ma Na Ma + Mb Nb Mb + Mc Nc Mc). (e) Estimate the polydispersity Q by using the answer in (d) divided by answer in (b).

1 Answers  

CHEMICAL MATERIAL BALANCE – EXAMPLE 2.6 : 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.

1 Answers  

DIFFERENTIAL EQUATIONS - EXAMPLE 20.1 : By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?

1 Answers  

I HAVE BEEN SELECTED FOR IOCL PANIPAT INTERVIEW,ACCTUALLY IAM DOING APPRENTIES TRAINEE IN CPCL.IAM GOING TO ATTEND THE IOCL PANIPAT INTERVIEW.SO PLEASE HELP ME HOW I PREPARE FOR INTERVIEW BY SUBJECT AND ALSO BY FIELD.

0 Answers   IOCL,

What is a good method of minimizing shell side pressure drop in a shell and tube exchanger?

0 Answers  

Question 73 - (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.

1 Answers