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# Calculating Mass Of Steam By Energy Balance

thermodynamics

7 replies to this topic

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Posted 06 March 2013 - 08:12 AM

how much steam of 145C and 14 psi is needed to raise the temperature of mixture of corn and water from 85C to 120C?????

### #2 Art Montemayor

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Posted 06 March 2013 - 05:05 PM

Use the heat transfer equation:

Q = W CP (T2- T1)

where,

Q = total heat transferred, btu

W = mass of the Corn + water

CP = heat capacity of the Corn + water

T2 = 120 oC

T1 = 85 oC

Since we don't do homework assignments for others, you should take it from here and work out the proper units and logic to arrive at the amount of saturated steam you need.

Good luck.

### #3 Raj Mehta

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Posted 07 March 2013 - 06:15 AM

Its quite simple and Mr. Art is right, you should figure out on your own, such simple things to calculate. He already has done half of your work and it would be better If I don't reveal the later half. Find out yourself and enjoy the happiness and excitement of finding solution to your problem.

Take care about the units or else you will surely mess it.

Thanks.

### #4 breizh

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Posted 07 March 2013 - 06:33 AM

Consider this resource to support your query , it's related to the food industry .

Chapters 5 & 6 in particular

http://www.nzifst.or...s/matlenerg.htm

Breizh

Edited by breizh, 09 March 2013 - 01:19 AM.

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Posted 08 March 2013 - 01:52 PM

its not an homework or assignment

and admin what you are telling is also wrong

its a general question for a heat exchanger to find amount of heat required to raise the temperature of a mixture ot a single component to a certain value.

### #6 Raj Mehta

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Posted 08 March 2013 - 03:29 PM

I guess he should be able to figure out the solution to this problem very easily (only one step, after Post # 2) if he has a steam table or he  takes some time to read basic heat exchanger problems or google about it. Moreover he also has not mentioned the mass flow of water + corn, how can any one give him the answer ? Mr. Art though has wonderfully explained the basic of calculating the heat required for the process.

Next thing is that, he needs the amount of steam required and not the heat required.

We will definitely be happy to help him, if he mentions what efforts he has taken and where he got stuck.

Thanks.

### #7 thorium90

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Posted 09 March 2013 - 12:27 AM

its not an homework or assignment

and admin what you are telling is also wrong

its a general question for a heat exchanger to find amount of heat required to raise the temperature of a mixture ot a single component to a certain value.

Oh, so how is the reply by Art wrong? Do explain?

### #8 kkala

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Posted 09 March 2013 - 07:01 AM

1. Can you please give additional data, preferably with a sketch? Is it a continuous or batch operation? I had supposed a batch operation with (insulated) autoclave receiving live steam, post no 5 indicates condensing steam in tubes. Is it a steam coil? Post no     2 expresses a correct principle (simplified), but you seem to want something more. So more data is needed.
2. As post no 6 notes, steam tables can be more convenient than heat capacities; especially in case of autoclave, where an extra amount of water will be evaporated (remaining in the closed volume) during heating from 85 to 120 oC.
3. Concerning corn, <http://www.tpl.fpv.u...2006/pdf/08.pdf> (*) indicates following
3.1 heat capacity of completely dry corn is about 600 J/kg/oC at 25 oC (figure 3)
3.2 mixing rule could be used for humid corn (equation (1)); e.g. for humidity=0.65 (kg H2O/kg dry corn), heat capacity is calculated as (600 + 4186*0.65)/(1+0.65) = 2013 J/kg/oC, versus ~ 2000 J/kg/oC from figure 3. Water heat capacity = 4186 J/kg/oC
3.3 Consequently corn can be considered as a mixture of completely dry corn and water. For want of more precise data, assume 600 J/kg/oC for dry corn, even at 120 oC.
4. Neglecting any water evaporation (per para 2), let us have an example: 10 kg of humid corn (humidity=0.65 per para 3.2) is mixed with 20 kg of water, at temperature 85 oC. The mixture has to be heated at 120 oC through steam condensing in tubes. Its condensing temperature has to be higher than 120 oC (**), say 135 oC. Thus heating steam can be saturated, 3.13 bara - 135 oC (assumed, it can be otherwise).
Dry corn in the mixture: 10/(1+0.65) = 6.06 kg
Water in the mixture: 20 + 10 - 6.06 = 23.94 kg
Heat to be supplied to water: 23.94*(503.7 - 355.9) = 3538 kJ (from steam tables)
Heat to be supplied to dry corn: 6.06*600*(120-85)=127260 J = 127 kJ (rather insignificant compared to that of water)
Heat supplied by condensing steam: Q= 3538+127= 3665 kJ
Heat of condensation (steam, 3.13 bara - 135 oC): 2185.9 kJ/kg (from steam tables)
Steam required: 3665/2185.9 = 1.68 kg, assuming no heat (or steam) losses.

Note (**): mentioned 14 psig steam (assumed saturated) corresponds to 120 oC condensing temperature.
5. Hopefully above example can help actual calculations (with adjustments / corrections). Conditions can be different, and so can assumptions.

#### Attached Files

Edited by kkala, 09 March 2013 - 07:25 AM.