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Bivouac Underground Thermal Model

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zoe_lithoi

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Mar 3, 2010, 11:54:41 AM3/3/10

THERMAL MODEL FOR AN UNDERGROUND BIVOUAC
3/2/2010
TOBY ANDERSON

-----------
Background
-----------

In World War I and II, German, French, and American soldiers lived in
temporary shelters, often underground, called Bivouac's. Their letters
reflected their apprciation of being able to stay warm in the winter.
Ground troops carried small collapsible shovels for digging fox holes
as well as these Bivouacs. SOme of their more advanced, larger, longer-
term, underground shelters had wood burning stoves.

Today, the underground Bivouac, is obsolete, except, perhaps for
hunters, living in remote off-grid locations, or emergency situations.
And perhaps, in this time of economical depression, some may be forced
to consider such quarters.

---------------------
Purpose & Conclusion
---------------------

This thermal model hopes to answer the questions:

1. what will be the temperature inside the Bivouac if there are 2
people inside in the winter time.
Answer: About 45degF when it is 0degF outside.

2. How much heat is required to heat the Bivouac to 70degF?
Answer: 1 gallon per day or 0.004 cords of wood a day

3. If the Bivouac ground is heated up over a period of time, then how
long will it continue to keep one warm?
Not yet answered...

A thermal model of a small thermal underground shelter might be
helpful in developing a thermal model for a larger underground
dwelling such as the earthships in Taos, New Mexico. Once the simple
model for the Bivouac is developed, it could be further developed by
adding more insulation, or/and adding a thermal solar heat collector
or/and by adding seasonal solar heat storage.

-----------------------------------------
Description of Bivouac:
-----------------------------------------

Dig a 4'x 4'x8' hole, make the roof of 1.5inch diameter or larger logs
(or 2x4's) and branches, tarp it, and cover it with 8inches of dirt.
Insulate the ceiling with thin, lightweight R7 "insultex" material, or
perhaps, a 3 inch layer of grass, weeds, leaves held in place by
placing them on a suspended tarp.

Description of Bivouac heater:

The idea here, would be to have a firepit outside, which would be used
to heat up rocks or firebricks. Then,this heated thermal mass would be
brought into the Bivouac to keep the place warm.

-----------------------------------------
General assumptions on the thermal model:
-----------------------------------------

1. assumes the ouTfide 'average monthly' temperature' of 0degF
2. Our goal is to have the bivouac temperature = 70F
3. It calculates the equivalent radiant thermal resistor
4. all 3 of these parameters affect the heat flowing from the house to
the ouTside, hence, this value
is also calculated, and, in fact, used in the calculation of the
floor surface Temperature, Tf.
5. It uses 8" thick sand layers
6. Thermal Analysis will initially be done using Nick Pine's method
called "Ohm's Law for heat flow". In the future, a computer-program
based program will hopefully be coded.

-----------------------------------------
Basic thermal parameters
-----------------------------------------

Cu = underground capacitance
Cu = 30 Btu/F/cuft * 8" x (1'/12")
* { 2*4x4 ***ends
+ 2x4x8 ***sides
+ 1x4x8 ***floor} ...i.e. 128sqft
Cu = 2560 Btu/F

Cr = roof capacitance
Cr = 30 Btu/F/cuft * 8" x (1'/12")
* {4x8} ***roof
Cr = 640 Btu/F
Rsc = thermal conductive bivouac sand resistor
= 0.083 Hr-sqft-F/Btu/inch * 8 inch / 128 sqft
= ~ 1/200 Hr-F/Btu
Rscr = thermal conductive bivouac sand resistor on the roof
= 0.083 Hr-sqft-F/Btu/inch * 8 inch / 32 sqft
= 1/48 Hr-F/Btu
Rwa = the warm air resistance over 128 sqft
= 2/3 / 128
= 1/200
Rwar = The warm air resistance on the roof
= 2/3 / 32 = ~1/50
Rrad = thermal radiant resistor
= Qrad/(Tf-Th) ---I may get to this later someday if the analysis
proves promising.
Tf = temperature of floor surface, deg F (assume Tf = 85 initially)
Rb = if insulation (likean R7 insultec is placed in the roof, Rh =7/
Area in sqft (32sqft)=7/32 = ~1/4
Tb = temperature of house
Tg = ground Temp = 40degF
Rg = if unsulation (like an R5 foam board) is placed in the ground,
Rg=5/Area in sqft
if no insulation is used, Rg = Rsc
Qp = heat from 2 people = 2*300Btu/hr
Qfb = heat from firebrick
Txi = the initial ground temperature,t 8 inches below the floor, Txi =
40F
Ta = the ambient outside average temperature = 0degF

-----------------------------------------
The Model using Nick Pine's Ohm's Law for Heat Flow
-----------------------------------------

NOTE: ***** Make this the "Lucida Console" font to make it appear
correctly

Qb--> Qa-->
Rwar Rscr Rb Rwar
continued.... -www-www-Tr-+-www-www-Ta(0F)
|
|
VQr
|
---
---Cr
|

-----------------------------------------
How deep into the earth should we make the model?
-------------------------------------------

We should look at 1 day's worth of heatflow

24hrs = R*C = the 'time constant' using conventional electronic
circuit theory

If we just consider dirt, but not the 'warm air resistance' nor any
other insulation resistance.

C, of dirt, is 30 Btu/F/cuft
R, of dirt, is 1/12 hr*sqft*degF/Btu/inch
D = # of inches deep
A = cross-sectional area of the dirt

R*Cr = 1/12 hr*sqft*degF/Btu/inch * Dinch/A sqft * 30 Btu/F/cuft * D
inch * (1ft/12inch) * A sqft = 24

Get rid of the unit labelling
R*Cr = 1/12 * D/A * 30 * D/12 * A = 24

The A cancells
R*Cr = 30 * D^2/(12^2) = 24

D = sqrt(24*12^2/30) = 11 inches

Let's consider the warm air resistance = Rwa = 2/3 / A

Now, RC = (Rdirt+Rwa)*C = 24
(2/3/A + 1/12 * D/A) * 30 * D/12 * A = 24

A's cancell
(2/3 + 1/12 * D) * 30 * D/12 = 24

Multiply by 30
(10 + 2.5D) * D/12 = 24

multiply both sides by 12
(10 + 2.5D) * D = 24*12

2.5D^2 + 10D - 24*12 = 0
divide by 2.5
D^2 + 4D - 24*12/2.5 = 0

Quadratic Equation
D = {-b +/-sqrt(b^2-4ac)}/2a
D = {-4 +/- sqrt(16-4*1*(-115.2))}/2
D = {-4 +/-21.8}/2
D = 8.9 inches

Hence, I've chosen to make my model looking at 8 inch layers.

==========================================
initial simplification evaluation
==========================================

Rrad = infinity
Rb = 0
Rg=0
elliminate several of the Rsc and Cu
Qfb = 0 (for now)
Txi=40

<--Qx Qb--> Qa-->
Rsc Rwa Rwar Rscr Rb Rwar
(40F)Txi-+-www--+-www-+-Tb-www-www-Tri-+-www-www-Ta(0F)
|
+-<-Qfb
|
^ Qp=600Btu/hr
|

Qp + Qfb = 600Btu/hr = (Tb-Txi)/(Rsc+Rwa) + (Tb-Ta)/(Rb+2Rwar+Rscr)
600 + Qfb = (Tb-40)/(1/200 + 1/200) + (Tb-0)/(0+2/50 + 1/50)
600 + Qfb = (Tb-40)/(1/100) + Tb/(3/50)
600 + Qfb = (Tb-40)*100 + Tb*50/3
600 + Qfb = 100Tb-4000 + 17Tb
4600 + Qfb = 117Tb
Tb = (4600+Qfb)/117
And
Qfb = 117Tb – 4600

If we don’t add any more heat into the Bivouac using heated
FireBricks, what will the temperature of the

Bivouac be?

Qfb=0 -? Tb = 39.3

If we want Tb=70degF, how much more heat do we need to put into the
Bivouac using heated FireBricks, Qfb?

Tb = 70, then Qfb = 3590 Btu/hr

Let's relate this to heating by propane and by firewood.

1 cord of wood and 220 gallons of propane both yield about 20,000,000
Btu/hr of heat

3590 Btu/hr * 220gallon of propane/20,000,000 Btu/hr = 0.04 gallons of
propane

In 1 day, how many gallons of propane?
Well, roughly 0.04 * 24hrs = 1 gal.

3590 Btu/hr * 1 cord of wood/20,000,000 Btu/hr * 24 hours = 0.004
cords of wood

--------------------------------
Let's do the same thing, except, lets use Rb insultec R7 material
--------------------------------

so Rb = 7hr-sqft-degF/BTU /32sqft = 7/32 = ~1/4
Qp = 600Btu/hr = (Tb-Txi)/(Rsc+Rwa) + (Tb-Ta)/(Rb+2Rwar+Rscr)
600 + Qfb = (Tb-40)/(1/200 + 1/200) + (Tb-0)/(7/32+2/50 + 1/50)
600 + Qfb = (Tb-40)/(1/100) + Tb/(1/3.5)
600 + Qfb = (Tb-40)*100 + Tb*3.5
600 + Qfb = 100Tb-4000 + 3.5Tb
4600 + Qfb = 103.5TB

If we don’t add any more heat into the Bivouac using heated
FireBricks, what will the temperature of the

Bivouac be?

Qfb = 0, then Tb = 4600/103.5 = 44.4
Therefore, Using R7 insultex material increases the Bivouac's
temperature by 5degF (from 39.3 to 44.4degF).

And now, more heat flows from the Bivouac into the ground than
outside.

If we want Tb=70degF, how much more heat do we need to put into the
Bivouac using heated FireBricks, Qfb?

4600 + Qfb = 103.5*70
Qfb = 103.5*70 – 4600 = 2645Btu/hr

By adding R7 insulation to the roof, we need 1000 Btu/hr less to
heat the Bivouac to 70degF.

Over a 24hour period, the fire brick would have to supply (without
insultec insulation:
Qfb = 2645 BTU/hr * 24 hr = 63480 BTu

You would need this amount of 180degF firebrick cooling down to say
80degF

63480 BTu = 30 Btu/degF/cuft * (180-80)degF * Vcuft
Vcuft = 21cuft

The firebrick would need to cover an area 4’x4’x1.25’ inside the
Bivouac

Not to worry though, the situation is not that bleak.
You see, when the heat leaves the firebrick and is absorbed into
into the ground. As the ground warms up, less and less of the heat
from the firebrick goes into the ground and

thus you’ll need less and less firebricks to keep you warm.

The amount of heat that the ground absorbs is determined by the
thermal capacity of the ground. That is why we

need to look at those thermal capacitors in the above circuits.

==================
thermal capacitors
==================

<--Qx
Rsc Rwa
(40F) Tx-+-www--+-www--Tb-
|
VQcx
|
---
---Cu
|

The above is the classic RC circuit

In the steady state, there is no current flow. The temperature across
the capacitor is initially 0. Tx starts out at some minimum
temperature, Tx(min) = 40degF.

Then, as heat is added to the Bivouac, Tb rises to some maximum
temperature Tb(max). Current starts flowing into the capacitor, and Tx
rises exponentially from 40F to Tb(max). Hence, the equation for Tx
can be

expressed as:

Tx = Tx(min) + (Tb(max)-Tx(min))(1-e^-t/[Cu(Rsc+Rwa)])

Just to make this not so long, I define Tbm = Tb(max) and Txi =
Tx(min)

Tx = Txi + (Tbm-Txi)(1-e^-t/[Cu(Rsc+Rwa)])

==================
The new circuit with capacitors
==================

<--Qx Qb--> Qa-->
Rsc Rwa Rwar Rscr Rb Rwar
Tg (40F) Tx-+-www--+-www-+--Tb-www-www-Tr-+-www-www-Ta(0F)
| | |
VQcx +-<-Qfb VQcr
| ^ |
--- | ---
---Cu Qp ---Cr
| |

eqn 3). Qp + Qfb = Qx + Qb ---at 'Tb'
eqn 4). Qx = Qcx --- at 'Tx'
eqn 5). Qb = Qa + Qcr --- at 'Tr'
eqn 6). Qp + Qfb = Qx + Qa + Qcr ---combining eqn 3 and 5

We will look at Qx and Qb

------------------------
Qx
------------------------

From circuit theory:
Tx(t) = Txi + (Tb(t)-Txi)(1-e^(-t/25.6)) --Equation A

Txi = 40
(Tb-Txi)/(Rwa+Rsc) = Cu*dTx/dt

dTx/dt=(Tb-Txi)/[(Rwa+Rsc)*Cu]

G=(Rwa+Rsc)*Cu = 25.6 ----Equation G

G=(1/200+1/200)*2560 = 25.6

dTx/dt=(Tb-Txi)/[(1/200 + 1/200)* 2560]
dTx/dt=(Tb-Txi)/[(1/100)* 2560]

dTx/dt=(Tb-Txi)/G ----equation 1

Let: Tx = Txi + (Tb-Txi)(1-e^(-t/G)) --Equation A

The product rule: d[x(t)y(t)]/dt
dx(t)/dt * y(t) + x(t)*dy(t)/dt

d/dt[Tx(t)] = d(Tb-Txi)dt * (1-e^(-t/G)) + (Tb-Txi)*d(1-e^(-t/G))/dt
dTx/dt = dTb/dt * (1-e^(-t/G)) + (Tb-Txi)* [(1/G)*(e^(-t/G))]
dTx/dt = dTb/dt * (1-e^(-t/G)) + (Tb-Txi)/G*(e^(-t/G))
dTx/dt = dTb/dt * (1-e^(-t/G)) + Tb/G*e^(-t/G)) - Txi/G*(e^(-t/G))

equate this to:
dTx/dt=(Tb-Txi)/G ----Equation 1

dTx/dt
=(Tb-Txi)/G = dTb/dt * (1-e^(-t/G)) + Tb/G*e^(-t/G)) - Txi/G*(e^(-t/
G))
= Tb/G -Txi/G = dTb/dt * (1-e^(-t/G)) + Tb/G*e^(-t/G)) - Txi/G*(e^(-t/
G))
Tb/G -Txi/G - Tb/G*e^(-t/G)) + Txi/G*(e^(-t/G)) = + dTb/dt * (1-e^(-t/
G))
Tb/G - Tb/G*e^(-t/G)) - Txi/G + Txi/G*(e^(-t/G)) = + dTb/dt * (1-e^(-t/
G))
Tb/G (1 - e^(-t/G)) - (Txi/G)*(1 - e^(-t/G)) = + dTb/dt * (1-e^(-t/
G))
Tb/G - Txi/G = dTb/dt
dTb/dt = (Tb-Txi)/G ----Equation A2

------------------------
Qb
------------------------

(Tb-Tri)/(Rwar+Rscr) = Cr*dTr/dt + (Tri-Ta)/(Rb+Rwar)
dTr/dt = (1/Cr) * [(Tb-Tri)/(Rwar+Rscr) - (Tri-Ta)/(Rb+Rwar)]
dTr/dt = (-Tri/Cr) * [(1/(Rwar+Rscr)+ 1/(Rb+Rwar)] + Tb/[Cr*(Rwar
+Rscr)] + Ta/[Cr*(Rb+Rwar)]

Let H = Cr/[(1/(Rwar+Rscr)+ 1/(Rb+Rwar)]

H = 640/[(1/(1/50 + 1/50)+ 1/(1/4 + 1/50)]
H = 640/[1/2/50 + 1/1/4]
H = 640/[25+4] = 22hrs
H= Cr/[(1/(Rwar+Rscr)+ 1/(Rb+Rwar)] = 22hours ---Equation H

dTr/dt = -Tri/H + Tb/[Cr*(Rwar+Rscr)]
+ Ta/[Cr*(Rb+Rwar)] ---Equation 2

Let Tr(t) = Tri + (Tb(t)-Tri)(1-e^(-t/H)) ---equation B

dTr(t)/dt = d{Tri + (Tb(t)-Tri)(1-e^(-t/H))}/dt
= dTri/dt + d{(Tb(t)-Tri)(1-e^(-t/H))}/dt
= 0 + d{(Tb(t)-Tri)(1-e^(-t/H))}/dt
= d{(Tb(t)-Tri)(1-e^(-t/H))}/dt

The product rule: d[x(t)y(t)]/dt
dx(t)/dt * y(t) + x(t)*dy(t)/dt

so:
dTr/dt = d(Tb-Tri)dt * (1-e^(-t/H)) + (Tb-Tri)*d(1-e^(-t/H))/dt
dTr/dt = dTb/dt * (1-e^(-t/H)) + (Tb-Tri)* [(1/H)*(e^(-t/H))]

equate this to:
dTr/dt = -Tri/H + Tb/[Cr*(Rwar+Rscr)]
+ Ta/[Cr*(Rb+Rwar)] ---Equation 2

-------------------------------
Solve for dTb/dt --- this part is superfluous for now...
-------------------------------

dTr/dt =
-Tri/H + Tb/[Cr*(Rwar+Rscr)] + Ta/[Cr*(Rb+Rwar)]
= dTb/dt * (1-e^(-t/H)) + (Tb-Tri)* [(1/H)*(e^(-t/H))]

Thus,

dTb/dt * (1-e^(-t/H)) =
-Tri/H + Tb/[Cr*(Rwar+Rscr)] + Ta/[Cr*(Rb+Rwar)] - ((Tb-Tri)/H)*(e^(-
t/H))

dTb/dt * (1-e^(-t/H)) =
-Tri/H + Tb/[Cr*(Rwar+Rscr)] + Ta/[Cr*(Rb+Rwar)] - (Tb/H)*e^(-t/H) +
(Tri/H)*e^(-t/H)

dTb/dt * (1-e^(-t/H)) =
-Tri/H + (Tri/H)*e^(-t/H) + Tb/[Cr*(Rwar+Rscr)] - (Tb/H)*e^(-t/H) +
Ta/[Cr*(Rb+Rwar)]

dTb/dt * (1-e^(-t/H)) =
(Tri/H)*(1 - e^(-t/H)) + Tb*{1/[Cr*(Rwar+Rscr)] - (Tb/H)*e^(-t/H)} +
Ta/[Cr*(Rb+Rwar)]

dTb/dt = Tri/H
+ {Tb*{1/[Cr*(Rwar+Rscr)] - (Tb/H)*e^(-t/H)} + Ta/[Cr*(Rb+Rwar)]}/(1-
e^(-t/H))
----Equation B2

----------------------
Solving for Tb
----------------------

Consider the following equations

Tx = Txi + (Tb-Txi)(1-e^(-t/G)) --Equation A
dTx/dt=(Tb-Txi)/G ----Equation 1
Tr(t) = Tri + (Tb-Tri)(1-e^(-t/H)) ---equation B
dTr/dt = -Tri/H + Tb/[Cr*(Rwar+Rscr)]
+ Ta/[Cr*(Rb+Rwar)] ---Equation 2

substitute eqn 4 into equation in 6.
Qp + Qfb = Qcx + Qa + Qcr

Note:
Qcx = Cu*dTx/dt
Qa=[Tr-Ta]/(Rb+Rwar)
Qcr = Cr*dTr(t)/dt

Substitute these 2 equations into the above equation

Qp + Qfb =
Cu*dTx(t)/dt
+ [Tr-Ta]/(Rb+Rwar)
+ Cr*dTr(t)/dt ---Equation C

Substitute Equation 1 and 2 into C

Qp + Qfb =
Cu*(Tb-Txi)/G
+ [Tr - Ta]/(Rb+Rwar)
+ Cr*{-Tri/H + Tb/[Cr*(Rwar+Rscr)] + Ta/[Cr*(Rb+Rwar)]}

Now, substitute Equation B into this

Qp + Qfb =
Cu*(Tb-Txi)/G
+ [{Tri + (Tb - Tri)(1-e^(-t/H))} - Ta]/(Rb+Rwar)
+ Cr*{-Tri/H + Tb/[Cr*(Rwar+Rscr)] + Ta/[Cr*(Rb+Rwar)]}

Now distribute everything.

Qp + Qfb =
Cu*Tb/G - Cu*Txi/G
+ Tri/(Rb+Rwar) + (Tb - Tri)(1-e^(-t/H))/(Rb+Rwar) - Ta/(Rb+Rwar)
- Cr*Tri/H + Cr*Tb/[Cr*(Rwar+Rscr)] + Cr*Ta/[Cr*(Rb+Rwar)]

Further distribute the 2nd row, and simplify 3rd row

Qp + Qfb =
Cu*Tb/G - Cu*Txi/G
+ Tri/(Rb+Rwar) - Ta/(Rb+Rwar)
+ (Tb/(Rb+Rwar) - Tri/(Rb+Rwar))(1-e^(-t/H))
- Cr*Tri/H + Tb/(Rwar+Rscr) + Ta/(Rb+Rwar)

Further distribute the 3rd row

Qp + Qfb =
Cu*Tb/G - Cu*Txi/G
+ Tri/(Rb+Rwar) - Ta/(Rb+Rwar)
+ (Tb/(Rb+Rwar))(1-e^(-t/H)) - (Tri/(Rb+Rwar))(1-e^(-t/H))
- Cr*Tri/H + Tb/(Rwar+Rscr) + Ta/(Rb+Rwar)

the Ta terms cancell out.

Qp + Qfb =
Cu*Tb/G - Cu*Txi/G
+ Tri/(Rb+Rwar)
+ (Tb/(Rb+Rwar))(1-e^(-t/H)) - (Tri/(Rb+Rwar))(1-e^(-t/H))
- Cr*Tri/H + Tb/(Rwar+Rscr)

move the Tb terms onto 1 side of the equation

Cu*Tb/G + (Tb/(Rb+Rwar))(1-e^(-t/H)) + Tb/(Rwar+Rscr)
= Qp + Qfb + Cu*Txi/G - Tri/(Rb+Rwar)
+ (Tri/(Rb+Rwar))(1-e^(-t/H)) + Cr*Tri/H

associate the Tb terms onto 1 side of the equation

Tb*{Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}
= Qp + Qfb + Cu*Txi/G - Tri/(Rb+Rwar)
+ (Tri/(Rb+Rwar))(1-e^(-t/H)) + Cr*Tri/H

Tb = { Qp + Qfb + Cu*Txi/G - Tri/(Rb+Rwar)
+ (Tri/(Rb+Rwar))(1-e^(-t/H)) + Cr*Tri/H
/{Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}

The Tri/(Rb+Rwar) terms cancell out

--------------------------------------------------------------------
Tb = { Qp + Qfb + Cu*Txi/G - (Tri/(Rb+Rwar))(e^(-t/H)) + Cr*Tri/H }
/{Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}
---Equation D
--------------------------------------------------------------------

Let's try some #'s

Cu = 2560 Btu/F
Cr = 640 Btu/F
Rsc = ~ 1/200 Hr-F/Btu
Rscr = 1/48 Hr-F/Btu
Rwa = 1/48
Rwar = ~1/50
Ta = 0
Tri = 40
Qp=600
H=22
G=25.6
Rb=1/4
Rwar=1/50
Rb+Rwar = ~1/4

Tb = { 600 + Qfb + 2560*40/25.6
- (40/(1/4))(e^(-t/22)) + 640*40/22 }
/{2560/25.6 + (1-e^(-t/22))/(1/4) + 1/(1/25))}

Multiply and divide

Tb = { 600 + Qfb + 4000
- 160(e^(-t/22)) + 1164 }
/{100 + 4*(1-e^(-t/22)) + 25)}

Add

Tb = { 5764 + Qfb - 160e^(-t/22) }
/{129 - 4e^(-t/22)}

If Qfg = 0 and t=0

Tb = { 5764 + 0 - 160e^(-0/22)) }
/{129 - 4e^(-0/22))}

Tb = {5764 - 160}/{125} = 44.8

If Qfg = 0 and t=infinity

Tb = {5764}/{129}

Tb = { 5764 + 0 - 160e^(-large#/22) }
/{129 - 4e^(-Large#/22)} = 44.7

Let's say, that we want the Bivouac Temperature at 70degF,
How much heat do we need to add via the firebricks, Qfb
at t=0

70 = { 5764 + Qfb - 160e^(-0/22) }
/{129 - 4e^(-0/22)}

70 = {5764 + Qfb - 160}/{129 - 4}
70 = {5604 + Qfb}/{125}

Qfb = 70*125-5064 = 3686 Btu/hr

3686 Btu/hr * 220gallon of propane/20,000,000 Btu/hr = 0.04 gallons of
propane

In 1 day, how many gallons of propane?
Well, roughly 0.04 * 24hrs = 1 gal.

But let's get a better ideafor Qfb at the end of t = 24hrs

Tb = { 5764 + Qfb - 160e^(-24/22) }
/{129 - 4e^(-24/22)}

Tb = { 5764 + Qfb - 160*0.34 }
/{129 - 4*0.34}

Tb = { 5710 + Qfb }
/127.6

Qfb = 127.6Tb - 5710

So, if Tb=70, then
Qfb = 3222 Btu/hr

We need about 450 Btu/hr less from the firebricks after 1 day, than at
the beginning.

----------------------
an easier way to calculate Heat flow
----------------------

Consider equation D

Tb = { Qp + Qfb + Cu*Txi/G - (Tri/(Rb+Rwar))(e^(-t/H)) + Cr*Tri/H }
/{Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}

multiply both sides by the Denominator

Tb* {Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}
= { Qp + Qfb + Cu*Txi/G - (Tri/(Rb+Rwar))(e^(-t/H)) + Cr*Tri/H }

Rearranging:

Qp + Qfb
= Tb* {Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}
- Cu*Txi/G + (Tri/(Rb+Rwar))(e^(-t/H)) - Cr*Tri/H

Distribute the Tb term

Qp + Qfb
= Tb*Cu/G + Tb*(1-e^(-t/H))/(Rb+Rwar)) + Tb*/(Rwar+Rscr)
- Cu*Txi/G + (Tri/(Rb+Rwar))(e^(-t/H)) - Cr*Tri/H

Rearrange the 'Cu' terms and the 'Cr' terms together and associate:
Qp + Qfb
= (Cu/G)*(Tb-Txi)
- Cr*Tri/H
+ Tb*(1-e^(-t/H))/(Rb+Rwar))
+ Tb/(Rwar+Rscr)
+ (Tri/(Rb+Rwar))(e^(-t/H))

Distribute the Tb term
Qp + Qfb
= (Cu/G)*(Tb-Txi)
- Cr*Tri/H
+ Tb/(Rb+Rwar) - (Tb/(Rb+Rwar)*e^(-t/H))
+ Tb/(Rwar+Rscr)
+ (Tri/(Rb+Rwar))(e^(-t/H))

associate the e^ terms
Qp + Qfb
= (Cu/G)*(Tb-Txi)
- Cr*Tri/H
+ Tb/(Rb+Rwar)
+ Tb/(Rwar+Rscr)
+ (Tri-Tb)/(Rb+Rwar))(e^(-t/H))

associate the Tb terms and the '-' from the e^ term
Qp + Qfb
= (Cu/G)*(Tb-Txi)
- Cr*Tri/H
+ Tb*[1/(Rb+Rwar)+1/(Rwar+Rscr)]
- (Tb-Tri)/(Rb+Rwar))(e^(-t/H))

Recall that
H = Cr/[(1/(Rwar+Rscr)+ 1/(Rb+Rwar)]
So, Cr/H = [(1/(Rwar+Rscr)+ 1/(Rb+Rwar)]

Substitute this into the above

Qp + Qfb
= (Cu/G)*(Tb-Txi)
- Cr*Tri/H
+ Tb*Cr/H
- (Tb-Tri)/(Rb+Rwar))(e^(-t/H))

Combine CR terms:

Qp + Qfb
= (Cu/G)*(Tb-Txi)
+ (Cr/H)*(Tb-Tri)
- (Tb-Tri)/(Rb+Rwar))(e^(-t/H))

Now, compare that to this:
eqn 5b). Qp + Qfb = Qx + Qa + Qcr

We can conclude, that

eqn 7). Qx = (Cu/G)*(Tb-Txi)
eqn 8). Qcr = (Cr/H)*(Tb-Tri)
eqn 9). Qa = - (Tb-Tri)/(Rb+Rwar))(e^(-t/H))

------------------------
Summary of equations so far:
------------------------

<--Qx Qb--> Qa-->
Rsc Rwa Rwar Rscr Rb Rwar
Tg (40F) Tx-+-www--+-www-+--Tb-www-www-Tr-+-www-www-Ta(0F)
| | |
VQcx +-<-Qfb VQcr
| ^ |
--- | ---
---Cu Qp ---Cr
| |

eqn A). Tx = Txi + (Tb-Txi)(1-e^(-t/G))
eqn A1). dTx/dt=(Tb-Txi)/G
eqn A2). dTb(t)/dt = (Tb-Txi)/G
eqn B). Tr = Tri + (Tb-Tri)(1-e^(-t/H))
eqn B1). dTr/dt=(Tb-Tri)/H
eqn B2). dTb/dt = (Tri/H)
+ {Tb*[1/[Cr*(Rwar+Rscr)] - (1/H)*e^(-t/H)] + Ta/[Cr*(Rb+Rwar)]}/(1-
e^(-t/H))
eqn C). Qp + Qfb = Cu*dTx(t)/dt + [Tr-Ta]/(Rb+Rwar) + Cr*dTr(t)/dt
eqn D). Tb = { Qp + Qfb + Cu*Txi/G - (Tri/(Rb+Rwar))(e^(-t/H)) +
Cr*Tri/H }
/{Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}
eqn G). G=(Rwa+Rsc)*Cu = 25.6 ----Equation G
eqn H). H= Cr/[(1/(Rwar+Rscr)+ 1/(Rb+Rwar)] = 22hours
eqn 1). dTx/dt=(Tb-Txi)/G ----equation 1
eqn 2). dTr/dt = -Tri/H + Tb/[Cr*(Rwar+Rscr)] + Ta/[Cr*(Rb+Rwar)]
eqn 3). Qp + Qfb = Qx + Qb ---at 'Tb'
eqn 4). Qx = Qcx --- at 'Tx'
eqn 5). Qb = Qa + Qcr --- at 'Tr'
eqn 6). Qp + Qfb = Qx + Qa + Qcr
eqn 7). Qx = (Cu/G)*(Tb-Txi)
eqn 8). Qcr = (Cr/H)*(Tb-Tri)
eqn 9). Qa = - (Tb-Tri)/(Rb+Rwar))(e^(-t/H))

=============================
The circuit for the 2nd day
=============================

The above circuit covers heat flow for day 1, at the end of which, Tx
(the ground temperature 8 inches underground, has risen from Txi=40F
to some new value. Let's figure this out, recall that Tb=70degF, and
use

eqn A).

Tx = Txi + (Tb-Txi)(1-e^(-t/G))
= 40 + (70-40)(1-e^(-24/25.6))
=58 degF

This is really:
Txi=58 for the second day

Do the same thing but for Tr

eqn B). Tr = 40 + (70-40)(1-e^(-24/22)) = 60
This is really:
Tri=60 for the second day

Heat will now flow from Tx, 8 inches underground to T2, which is 16
inches underground and which is 40degF.

The new circuit looks something like this:

Q2<-- Qx<-- Qb--> Qa-->
Rsc Rsc Rwa Rwar Rscr Rb Rwar
T2(40F)-+-www-Tx-+-www-+-www-+--Tb-www-www-Tr-+-www-www-Ta(0F)
| | |Rrad | |
| | +-www-+-<-Qfb VQcr
| | ^ |
VQc2 VQcx | ---
| | Qp ---Cr
--- --- |
---Cu ---Cu
| |

The left side of the above model is equivalent to the following.

Basically, all the equations above, can still be used with some
modification.

For instance, take the following equation from above:

eqn D). Tb = { Qp + Qfb + Cu*Txi/G - (Tri/(Rb+Rwar))(e^(-t/H)) +
Cr*Tri/H }
/{Cu/G + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}

Replace

Txi with T2i
Cu with 2Cu
Rsc with 2Rsc
G with G2

Where
eqn G). G=(Rwa+Rsc)*Cu = 25.6

should be replaced with:

eqn G2). G2=(Rwa+2Rsc)*2Cu = (1/200 + 2(1/200))*2*2560 = 76.8

Tb = { Qp + Qfb + 2Cu*T2i/G2 - (T2i/(Rb+Rwar))(e^(-t/H)) + Cr*Tri/H }
/{2Cu/G2 + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}

Lets substitute G2 in there..

-------------------------------------------------------------
eqn D2). Tb = { Qp + Qfb + T2i/(Rwa+2Rsc) - (T2i/(Rb+Rwar))(e^(-t/H))
+ Cr*Tri/H }
/{1/(Rwa+2Rsc) + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}
-------------------------------------------------------------

Tb = { Qp + Qfb + T2i/(Rwa+2Rsc) - (T2i/(Rb+Rwar))(e^(-t/H)) + Cr*Tri/
H }
/{1/(Rwa+2Rsc) + (1-e^(-t/H))/(Rb+Rwar)) + 1/(Rwar+Rscr)}

Now let's plug some numbers in there....

Tb = { 600 + Qfb + 40/(1/200+2*(1/200)) - (40/(1/4+1/50))(e^(-t/22)) +
640*40/22 }
/{1/(1/200+2*(1/200)) + (1-e^(-t/22))/(1/4+1/50) + 1/(1/50+1/50)}

Tb = { 600 + Qfb + 2667 - 160e^(-t/22) + 1164 }
/{67 + 4(1-e^(-t/22)) + 25}

Tb = { 4431 + Qfb - 160e^(-t/22) }
/{96 + -4e^(-t/22) }

If t = 0, and Tb=70, what does Qfb=?

70 = { 4431 + Qfb - 160 }
/{96 + -4 }

6440 = 4271 + Qfb
Qfb = 2169

If t = 24, and Qfb=2169 what does Tb=?

Tb = { 4431 + 2169 - 160e^(-24/22) }
/{96 + -4e^(-24/22) }

Tb = { 4431 + 2169 - 53 }
/{95 }

Tb = 69degF

-----------------------------
Discharging the ground heat into the Bivouac
-----------------------------

This part is not really started yet....

How long does it take for Cu to charge up to 80degF?
How long will it take for it to discharge such that Tb=60F?

daestrom

unread,

Mar 3, 2010, 6:41:59 PM3/3/10

zoe_lithoi wrote:
> THERMAL MODEL FOR AN UNDERGROUND BIVOUAC
> 3/2/2010
> TOBY ANDERSON
>
>

<snip>

> If we just consider dirt, but not the 'warm air resistance' nor any
> other insulation resistance.
>
> C, of dirt, is 30 Btu/F/cuft
> R, of dirt, is 1/12 hr*sqft*degF/Btu/inch
> D = # of inches deep
> A = cross-sectional area of the dirt
>
> R*Cr = 1/12 hr*sqft*degF/Btu/inch * Dinch/A sqft * 30 Btu/F/cuft * D
> inch * (1ft/12inch) * A sqft = 24
>
> Get rid of the unit labelling
> R*Cr = 1/12 * D/A * 30 * D/12 * A = 24
>
> The A cancells
> R*Cr = 30 * D^2/(12^2) = 24
>
> D = sqrt(24*12^2/30) = 11 inches
>
> Let's consider the warm air resistance = Rwa = 2/3 / A
>
> Now, RC = (Rdirt+Rwa)*C = 24
> (2/3/A + 1/12 * D/A) * 30 * D/12 * A = 24
>
> A's cancell
> (2/3 + 1/12 * D) * 30 * D/12 = 24
>
> Multiply by 30
> (10 + 2.5D) * D/12 = 24

^^
I think this should be
(20 + 2.5D) * D/12 = 24 (30*2/3=20 not 10)

>
> multiply both sides by 12
> (10 + 2.5D) * D = 24*12
>
> 2.5D^2 + 10D - 24*12 = 0
> divide by 2.5
> D^2 + 4D - 24*12/2.5 = 0
>
> Quadratic Equation
> D = {-b +/-sqrt(b^2-4ac)}/2a
> D = {-4 +/- sqrt(16-4*1*(-115.2))}/2
> D = {-4 +/-21.8}/2
> D = 8.9 inches
>

I get 14.9 inches with the change I mentioned above.

> Hence, I've chosen to make my model looking at 8 inch layers.
>
>

I didn't wade through the rest of all of it, but I think you omitted air
exchange. This can be a significant heat loss unless you plan on
sealing it very well. But with going in/out to get firebrick, you are
going to lose a lot of heat through the air exchange.

A primitive technique used in igloos is that since they are above ground
they have a raised floor and so the warm air is trapped inside above the
entrance/exit.

good luck,

daestrom

zoe_lithoi

unread,

Mar 4, 2010, 5:02:57 PM3/4/10

Hi Daestrom,

Thanks for catching that mistake and other observations.

Toby

> daestrom- Hide quoted text -
>
> - Show quoted text -

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