# HVAC SMS #53



## svcetiquette (Sep 17, 2011)

Good evening fellas (and ladies),

I just had a quick question regarding #53 on the HVAC SMS. The questions gives a 100 boiler HP boiler operating at 15psig and 180 deg F feedwater, and asks what is the maximum possible steam generation rate.

I thought I would shortcut it and pulled hL of 180 deg water, and hV of 30psia steam from the steam tables to calculate the delta h per lbm. Divided that into the Btu/hr for the boiler for my maximum mass flow rate. This results in 4093 lbm/hr.

The book shows the right way to do it is to use cp delta T to get to saturation, then use the hLV just for the latent, which gives an answer of 3296.1

Just curious if someone could straighten me out on the source of the difference - I didn't think the change in sp volume would yield such a difference. Hope I'm not loosin it this late in the game - although after 6.5 hrs its now 11:30pm over here in UAE and I gotta work tomorrow. Thanks for the help anyone!


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## jamiecta (Sep 18, 2011)

svcetiquette said:


> Good evening fellas (and ladies),I just had a quick question regarding #53 on the HVAC SMS. The questions gives a 100 boiler HP boiler operating at 15psig and 180 deg F feedwater, and asks what is the maximum possible steam generation rate.
> 
> I thought I would shortcut it and pulled hL of 180 deg water, and hV of 30psia steam from the steam tables to calculate the delta h per lbm. Divided that into the Btu/hr for the boiler for my maximum mass flow rate. This results in 4093 lbm/hr.
> 
> ...


You have the idea/method right but you must be pulling the wrong numbers out of the steam tables or something. Either way you know the boiler needs to produce Q=3,347,500 BTU/hr of energy. So, to find the steam flow rate m in lbm per hour you do the following:

m=Q/(enthalpy change of water)

You can calculate the enthalpy change of water in different ways: 1) The way you did and 2) the way the 6MS did. In either scenario the rate of steam will be based on the full enthalpy change from liquid water to saturated steam at the operating pressure.

1) In this scenario you would look at the enthalpy of liquid feedwater at 180F. This enthalpy is 148 BTU/lbm. You would also look at the enthalpy of saturated vapor at 15psig (~30pisa). This enthalpy is 1164.1. The enthalpy change using this method is 1016 resulting in a steam flow rate of 3295 lbm/hr

2) In this scenario you do just as the book does. You take the h-fg from the steam tables at 15psig (~30psia) and then add the enthalpy change required to originally raise the temperature from 180 to the required 250 to boil. The enthalpy change using this method is 1016 resulting in a steam flow rate of 3295 lbm/hr

As you can see, both methods work here.


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## jamiecta (Oct 7, 2011)

did this help at all?


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## svcetiquette (Oct 10, 2011)

Thanks Jamie - actually I got up early the next morning and tried it again, got the right answer no problem. I think I may have pulled the value from the metric steam tables... definitely gave me a renewed appreciation for the endurance factor involved with this exam. Perhaps overkill but there is now a large blue-pen "X" across every metric steam table I have.


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