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Hello,
I am working on a evaporator being used in the evaporation of 10 m3 sodium chlorate solution. The heating fluid is steam.
The present mass flow rate of steam is 1000 kg/hour and pressure = 5 kg/cm2. At present this steam flows through 64 1-inch Ti tubes inside the evaporator. Length of tube = 2.8m
We are about to add new tubes into the evaporator thus increasing the heat transfer area (96 tubes).
I need the new mass flow rate of steam keeping the present pressure constant (5 kg/cm2) . I would greatly appreciate it if someone helped me and suggest how to go about solving this problem.
The inlet temperature of steam and outlet temperature of water are known. Hence I know the amount of heat transfer occurring. Each batch process (presently) occurs in about 7 hours.
The inlet and outlet concentrations of the solution are also known.
It would also help if someone told me how to go about finding the times that is saved upon addition of the new tubes.
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Steam Pressure And Mass Flow Rate
Started by Guest_cartman_*, Jul 01 2008 07:24 AM
2 replies to this topic
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#2
Zauberberg
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Posted 01 July 2008 - 08:08 AM
Maximum steam flowrate in the new case is determined by the criterion of maintaining steam velocities within the limits which are preferred for condensation service. You will decrease the time required for evaporation of the NaClO3 volume from the base case (if inlet and outlet concentrations are as in the first case):
t1, s = (heat of evaporation of required volume of solution, kJ)/(Duty 1, kW)
t2, s = (heat of evaporation of required volume of solution, kJ)/(Duty 2, kW)
Duties are calculated by U*A*MTD, and the only factor which changes is surface area, A - if you maintain the same heat transfer coefficient in both cases.
t1, s = (heat of evaporation of required volume of solution, kJ)/(Duty 1, kW)
t2, s = (heat of evaporation of required volume of solution, kJ)/(Duty 2, kW)
Duties are calculated by U*A*MTD, and the only factor which changes is surface area, A - if you maintain the same heat transfer coefficient in both cases.
#3
Guest_cartman_*
Guest_cartman_*
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- guestGuests
Posted 01 July 2008 - 10:11 AM
Thanks a lot Zauberberg.
But the steam condensation details are not actually needed. Very less steam is taken away for condensation in a fall-film evaporator but most can be safely assumed to escape to the atmosphere (It is an open-pan evaporator). Can Hagen-poiseulle equations be applied?
But the steam condensation details are not actually needed. Very less steam is taken away for condensation in a fall-film evaporator but most can be safely assumed to escape to the atmosphere (It is an open-pan evaporator). Can Hagen-poiseulle equations be applied?
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