Posted by *Curbie* on January 2, 2011, 3:21 am

daestrom,

Thanks, I've read a bunch of Heat-X books and still haven't found what

I'm looking for.

Curbie

*>On 12/30/2010 17:57 PM, Curbie wrote:*

*>> daestrom,*

*>>*

*>> I'm sort of gummed up with some of my distillation calculations,*

*>> mainly for water to water and maybe oil to water heat exchangers*

*>> (re-biolers) and vapor to water condensers. Could you recommend a good*

*>> heat-X book with completed math examples in imperial units?*

*>>*

*>> Thanks,*

*>>*

*>> Curbie*

*>>*

*>I have one at work, will look up the biblio on it next week when I'm *

*>back at work.*

*>IIRC, condensers are *majorly* affected by even traces of *

*>non-condensable gasses (i.e. air). Tends to build up as a boundary *

*>layer on the cooling surface and then you're limited by diffusion of *

*>vapor through the film layer of air.*

*>This is why steam condensers have the air-removal system suction *

*>embedded in the middle of the tube bundle (sometimes called a 'dry pipe *

*>suction'). Taking suction from the coldest spot improves the *

*>air-fraction in the suction flow.*

*>daestrom*

Posted by *Morris Dovey* on January 5, 2011, 1:27 am

On 12/30/2010 3:29 PM, daestrom wrote:

*> I also have "Thermodynamics: 4th Edition" by Wark (1983). Don't know if*

*> it's exactly the same one, it doesn't mention a 'Richards'. I've got*

*> about five different thermo books and a couple of heat-transfer ones.*

"Thermodynamics" (6e) arrived this afternoon. Richards is an ME prof at

Rose-Hulman, and I gathered that Wark (at Purdue) brought Richards in to

broaden the presentation. It's interesting to me because I studied

(math) at Rose and my first full-time job was managing their computing

center.

--

Morris Dovey

http://www.iedu.com/DeSoto/

PGP Key ID EBB1E70E

Posted by *J. Clarke* on December 31, 2010, 1:37 am

*> *

*> On 12/30/2010 9:37 AM, daestrom wrote:*

*> *

*> > This is an old text book but still very relevant. I've always liked*

*> > stuff by Burghardt for two reasons. The math he presents is*

*> > straight-forward algebra and doesn't leave you glassy-eyed. He uses*

*> > 'English' units with the lb-force and lb-mass convention, which is what*

*> > I first started out with.*

*> *

*> Heh! Straight-forward algebra sounds good 'cause I'm not allowed to haze*

*> out on math.*

*> *

*> I'm doing my level best to stick to SI units, so what amounts to an "old*

*> friend" to you is sheer aggravation for me. I suppose I should learn to*

*> be comfortable with both, but doubt I can manage that _and_ actually*

*> solve the problems in front of me...*

*> *

*> > There's a good section on gas laws and various 'processes'. And lots of*

*> > information in the appendices. One area it's a bit 'light' in is the*

*> > heat-transfer stuff, but it does cover the basics of that as well.*

*> > *

*> >*

*
(Amazon.com product link shortened)*
*> *

*> I've bookmarked it. The books already en route are*

*> *

*> "Thermodynamics" by Wark and Richards, and*

*> *

*> "Steam Tables: Thermodynamic Properties of Water Including Vapor, Liquid*

*> and Solid Phases" by Hill, Keenan, Moore, and Keyes*

*> *

*> ...and I'm already at the point where putting another book on the shelf*

*> means that one of the books already there has to go. :(*

*> *

*> Wouldn't it be great if someone developed a gizmo that ate paper books*

*> and spit out e-books on flash memory sticks? (Even better if*

*> high-lighting and marginal notes were preserved!) :)*

*> *

*> > Happy New Year*

*> *

*> Peace, health, prosperity, and sunshine to you.*

You might want to check Pirate Bay for any titles you need as ebooks.

An amazing variety of engineering and scientific texts have been scanned

and torrented by various people. Most of them are image files, not

searchable text, though.

Posted by *Curbie* on December 24, 2010, 8:18 pm

Morris,

I'm bouncing between the kitchen and computer preparing for two Skype

parties, but here is the Gas Law and pressure conversion modules I use

in my spread-sheets, let me know if you don't find the answer there,

and if you want, I'll set-up a sheet to cross-check your math question

in a couple of days.

Curbie

' Gas Law Conversion Routines

' Between (V)olume, (T)emperature, (P)ressure, and (n)mol using the

Gas Law equation.

Option Explicit ' force explicit

declaration of all variables

Public Function GL_R(P As Double, V As Double, T As Double)

' Returns Gas Law Constant (R) for the given Pressure, Volume, and

Temperature based on 1

' atmosphere of pressure, 22.4 liters of volume, and a temperature of

273.15°K or 492°R

' using the Gas Law equation.

' 1. Assume a basis. Assume gas is at standard conditions, that is, 1

g-mol gas at 1 atm

' (101.3 kPa) pressure at 0°C (273°K or 492°R) occupying a volume of

22.4 L.

' 2. Compute the gas constant Apply suitable conversion factors and

obtain the gas constant in

' various units. Use PV=RT; that is. R=PV/T. Thus:

' a. R =(1 atm)[22.4 L/(g-mol)](1000 cm3/L)/273°K = 82.05

(atm)(cm3)/(g-mol)(°K)

' b. R =(14.7 psia)(359 ft3/lb-mol/492°R = 10.73

(psia)(ft3)/(Ib-mol)(°R)

' c. R =(1 atm)[359ft3/(lb-mol)]/273°K = 1.315 (atm)(ft3)/(lb-mol)(°K)

' d. R =[10.73(psia)(ft3)/(lb-mol)(°R)](144 in2/ft2)[3.77x10-7

kWh/(ft-lbf)] = 5.83x10-4 kWh/(lb-mol)(°R)

' e. R =[5.83x10-4 kWh/(lb-mol)(°R)](1/0.746 hp-h/kWh) = 7.82x10-4

hp-h/(lb-mol)(°R)

' f. R =(101.325 kpa/atm)[22.4 L/(g-mol)][1000

g-mol/(kg-mol)]/(273°K)(1000 L/m3) = 8.31 (kpa)(m3)/(kg-mol)(°K)

' g. R= [7.82x10-4 hp-h[(lb-mol)(°R)][6.4162x10+5 cal/(hp-h)][1/453.6

lb-mol/(g-mol)](1.8 °R/K) = 1.99 cal/(g-mol)(°K)

GL_R = (P * V) / T

End Function

Public Function GL_V(P As Double, T As Double, n As Double) As Double

' Returns Volume in Liters for the given Pressure in (kiloPascals),

Temperature (in kelvin), and

' mol using the Gas Law equation.

GL_V = (n * GASR * T) / P

End Function

Public Function GL_P(V As Double, T As Double, n As Double) As Double

' Returns Pressure in KiloPascals for the given Volume (in liters),

Temperature in (kelvin), and

' mol using the Gas Law equation.

GL_P = (n * GASR * T) / V

End Function

Public Function GL_T(V As Double, P As Double, n As Double) As Double

' Returns Temperature in kelvin for the given Volume (in liters),

Pressure (kiloPascals), and

' mol using the Gas Law equation.

GL_T = (P * V) / (n * GASR)

End Function

Public Function GL_n(V As Double, P As Double, T As Double)

' Returns Moles for the given Volume (in liters), Pressure

(kiloPascals), and Temperature in (kelvin)

' using the Gas Law equation.

GL_n = (P * V) / (GASR * T)

End Function

' Pressure Conversion Routines

' Between (A)tmospheres, kilo(P)ascals, (T)orr, milli(B)ars and

(L)Pounds per Square Inch

Option Explicit ' force explicit

declaration of all variables

Public Function CP_A2P(a As Double) As Double

' Returns pressure in kiloPascals for the given pressure in

Atmospheres.

CP_A2P = a * 101.325

End Function

Public Function CP_A2B(a As Double) As Double

' Returns pressure in milliBars for the given pressure in Atmospheres.

CP_A2B = a * 1013.25

End Function

Public Function CP_A2L(a As Double) As Double

' Returns pressure in P.S.I. for the given pressure in Atmospheres.

CP_A2L = a * 14.69595

End Function

Public Function CP_A2T(a As Double) As Double

' Returns pressure in Torr for the given pressure in Atmospheres.

CP_A2T = a * 760

End Function

Public Function CP_P2A(P As Double) As Double

' Returns pressure in Atmospheres for the given pressure in

kiloPascals.

CP_P2A = P * 0.009869233

End Function

Public Function CP_P2B(P As Double) As Double

' Returns pressure in milliBars for the given pressure in kiloPascals.

CP_P2B = P * 10

End Function

Public Function CP_P2L(P As Double) As Double

' Returns pressure in P.S.I. for the given pressure in kiloPascals.

CP_P2L = P * 0.1450377

End Function

Public Function CP_P2T(P As Double) As Double

' Returns pressure in Torr for the given pressure in kiloPascals.

CP_P2T = P * 7.500617

End Function

Public Function CP_B2A(B As Double) As Double

' Returns pressure in Atmospheres for the given pressure in milliBars.

CP_B2A = B * 0.0009869233

End Function

Public Function CP_B2P(B As Double) As Double

' Returns pressure in kiloPascals for the given pressure in milliBars.

CP_B2P = B * 0.1

End Function

Public Function CP_B2L(B As Double) As Double

' Returns pressure in P.S.I. for the given pressure in milliBars.

CP_B2L = B * 0.01450377

End Function

Public Function CP_B2T(B As Double) As Double

' Returns pressure in Torr for the given pressure in milliBars.

CP_B2T = B * 750.0617

End Function

Public Function CP_L2A(l As Double) As Double

' Returns pressure in Atmospheres for the given pressure in P.S.I.

CP_L2A = l * 0.06804596

End Function

Public Function CP_L2B(l As Double) As Double

' Returns pressure in milliBars for the given pressure in P.S.I.

CP_L2B = l * 68.94757

End Function

Public Function CP_L2P(l As Double) As Double

' Returns pressure in kiloPascals for the given pressure in P.S.I.

CP_L2P = l * 6.894757

End Function

Public Function CP_L2T(l As Double) As Double

' Returns pressure in Torr for the given pressure in P.S.I.

CP_L2T = l * 51.71493

End Function

Public Function CP_T2A(T As Double) As Double

' Returns pressure in Atmospheres for the given pressure in Torr.

CP_T2A = T * 0.001315789

End Function

Public Function CP_T2B(T As Double) As Double

' Returns pressure in milliBars for the given pressure in Torr.

CP_T2B = T * 1.333224

End Function

Public Function CP_T2P(T As Double) As Double

' Returns pressure in kiloPascals for the given pressure in Torr.

CP_T2P = T * 0.1333224

End Function

Public Function CP_T2L(T As Double) As Double

' Returns pressure in P.S.I. for the given pressure in Torr.

CP_T2L = T * 0.01933677

End Function

' Global Constants

Public Const BASK = 273.15 ' Base of Kelvin scale °

Public Const BASR = 459.67 ' Base of Rankine scale

°

Public Const GASR = 8.314 ' Gas Law constant

Public Const MF_W = 18.016 ' Atomic Mass of water

equaling One Mol

Public Const MF_M = 32.042 ' Atomic Mass of

methanol equaling One Mol

Public Const MF_E = 46.068 ' Atomic Mass of ehanol

equaling One Mol

Public Const MF_P = 60.094 ' Atomic Mass of

propanol equaling One Mol

Public Const MF_B = 74.12 ' Atomic Mass of butanol

equaling One Mol

Public Const LHVW = 40.639 ' Latent Heat of

Vaporizrion for Water (kJm)

Public Const LHVM = 45.3 ' Latent Heat of

Vaporizrion for Methanol (kJm)

Public Const LHVE = 39.22 ' Latent Heat of

Vaporizrion for Ehanol (kJm)

Public Const LHVP = 40.3 ' Latent Heat of

Vaporizrion for Propanol (kJm)

Public Const LHVB = 4

Posted by *Morris Dovey* on December 25, 2010, 4:50 pm

On 12/24/2010 2:18 PM, Curbie wrote:

*> Morris,*

*> *

*> I'm bouncing between the kitchen and computer preparing for two Skype*

*> parties, but here is the Gas Law and pressure conversion modules I use*

*> in my spread-sheets, let me know if you don't find the answer there,*

*> and if you want, I'll set-up a sheet to cross-check your math question*

*> in a couple of days.*

<major snip to file CurbieCalc>

Mmmm - yummy. Some of these I already had but all appear useful (even

though I'm working hard to convert /myself/ to metric only). :)

I think I'm going to transliterate the entire collection to C language

modules in a (growing) C thermodynamics library.

Thanks!

--

Morris Dovey

http://www.iedu.com/DeSoto/

PGP Key ID EBB1E70E

>On 12/30/2010 17:57 PM, Curbie wrote:>> daestrom,>>>> I'm sort of gummed up with some of my distillation calculations,>> mainly for water to water and maybe oil to water heat exchangers>> (re-biolers) and vapor to water condensers. Could you recommend a good>> heat-X book with completed math examples in imperial units?>>>> Thanks,>>>> Curbie>>>I have one at work, will look up the biblio on it next week when I'm>back at work.>IIRC, condensers are *majorly* affected by even traces of>non-condensable gasses (i.e. air). Tends to build up as a boundary>layer on the cooling surface and then you're limited by diffusion of>vapor through the film layer of air.>This is why steam condensers have the air-removal system suction>embedded in the middle of the tube bundle (sometimes called a 'dry pipe>suction'). Taking suction from the coldest spot improves the>air-fraction in the suction flow.>daestrom