CALir'ORNIA STATE UNIVERSITY, NORTHRIDGE
SOLAR POWERED ABSDRP'l'ION
li
AIRCONDITIONER
A gradu~:te projc:ct submitted. in pa.rtial satisfaction of
the requirements for the degr~e of Master of Science in
E.hgineering
by
William Ronald Baxendale
lb)f;J
August 197-7
--~-@:------ --- -- -------- - ------------- --·-------------------
-------------------- ----·-- -·----------------------·--J
-----~---------------------------1
~---------~--------------------
I
!
~he
I
Graduate Project of William Ronald Baxendale is approveds
pr. William
Or.
I
fJ
Rivers
Dusan Zrnic
i
Dr. fL' Addis¥ockwood
(Chairman)-~~--
California State University, Northridge
-----·-------------- ----------------------------------------- - - - -
ii
TABLE OF CONTENTS
!SECTION
PAGE
:Table of Contents
iii
i
t
iLis t of Figures
v
.List of Symbols
vii
i
!Abstract
I
;chapter 1
X
Thermodynamic Calculations
·1
!
Section 1 - Evaporator
1
Section 2 - Absorber
.3
Section 3
Pump
5
Section 4
Heat Exchanger
6
Section 5 - Generator
8
Section 6
-
Condenser
9
Section 7
-
Conclusion
11
iChapter 2 - Absorber Experiment
12
I
i
Section 1
Experimental Setup
12
Section 2
Calibration
1.3
Section .3 - Procedure
15
Section 4 - Results
16
A - Pressure Drop
19
B ., Heat Transfer
25
Section 5 - Conclusions
30
i
I
i
Section 6 - Recommendations
.31
i
'I
L--~-------------~---------------------~----------------------------------___j
iii
~------------
-------···--------------··-·-·--·------,
---
i
:SECTION
PAGE
[References
33
!Table 1 - Properties of Ammonia and R-11
J4
•Appendix A - Thermodynamic Calculations
3.5
;
i
!Appendix B - Absorber Experiment
50
l
i
!__________________________
-----~--------------------------··--------j
iv
~-----------------------------------------
-------------------
1
LIST OF FIGURES
PAGE
I FIGURE
I
2
11 - Evapcrator
Absorber
Pump
5
~
i4 - Heat Exchanger
7
5 - Generator
8
i6 - Condenser
lG
'7 - Helix Tube Cross Section
25
:a - Refrigeration Circuit
9 - Expansion Valve
)6
37
'10 - Vapor Pressure vs Mole Fraction
.39
'11 - Vapor Pressures of NH
~·0
From Solutions of NaSCN
3
I
!12 - Integral Heat of Solution
45
113 - Densities of NaSCN and Liquid NH3 Solutions
46
•14 - Temperature - Enthalpy Diagram
49
1
I
i15
I
Experimental Setup
51
i16
Evaporator Setup
52
I
l
.17 - Condensor Setup
53
!18 - Calibration Curve for Flovmeter
54
:19
Change in Water Temp. for Constant Power Inputs .55
,20 -
Heat Transfer Coefficient (U)
I
'
:21
Experimental Results {P = JO Watts)
57
l
;
L_______________________ ---------------------·------------ ___ J
v
---------------------------- --- ----------------------
------~--------
------------------------- -------- ---!
PAGE
!FIGURE
.i2J -
Experimental Results (P = 40 Watts)
58
Experimental Results (P = 50 Watts)
59
24 -Experimental :Results (P
= 100,
125 Watts)
60
'
I
!2~
!
-
- Two-Phase Friction Multiplier (R)
Photograph of Evaporator
:26
l
'27 - Close-Up of Evaporator
61
62
6J
I
I
t-----~--- ---------~-------------·---- ------
vi
_j
I,IST OF SYMBOLS
Void fraction
A
Area
;
:coP
Coefficient of performance
Heat Capacity
Diameter
Equivalent diameter
.f
Friction factor
i
'G
Mass velocity
.h
I
:h
Heat transfer ccefficient
Specific enthalpy
Height
.k
Therme.l conductivity
iL
I
Length
!m
Mass flow rate
fM• W.
Molecular weight
Nu
Nusselt number
I
I
l
.Pressure
Pr
?randtl number
Heat transfer rate
Heat transfer per unit mass
r
radius
Two-phase friction multiplier
i
!_ ___________________________ _
---·--·---------------~-J
vii
(---------------·----------------
iRe
I
---------------------------------------~-------·-
Reynolds number
s
Specific entropy
t
Temperature
U
Total heat transfer coefficient
v
Specific volume
Wp
Pump work
Greek Letters
€
Heat exchanger
Q
Time
P
Dynamic viscosity
P
Dens.i ty
~
Quality
effectivenes~
Subscripts
i
io
I
Boiling portion
'
;e
Exit condition
:r
Friction
f
Evaluated at film condition
f
Property of saturated liquid
Difference in property for saturated vapor and
sc~turated
liquid
i
Inlet condition
ri
Inside condition
:g
Property of saturated vapor
I'
I
o
Nonboiling portion
i
:o
Outside condition
'
----------·-·-------------------------·---
viii
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _j
!s
Sensible
1SP
Single phase
I
,t
Tota_l
tp
Two phase
iv
Vapor
,w
Water
I
Superscript
Bar over symbol denotes property on· average
ba~~is
I
i
I
i
i
L__________________ _
- - - · - - - - - - - - - - - - - - - - - - - - - - - - - · - - - - - - - - - - - - · - - · - - - - - ·_ _ _ _ _ j
ix
..... ----
-·--· --·--- - - - - - -~- --------------- -----------;--
-- ---
-- ---~
-------------- --- ------
ABSTRACT
SOLAR
PO~~RED
ABSORPTION
AIRCONDITIONER
by
William Ronald Baxendale
Master of Science in Engineering
The object of this project is to analyze an existing
solar powered absorption air conditioning f;ystem built c.s
!
'a senicr
....
\,.,.._,._,.,,.,
d~~Jgn
~---""""-
project
and to do further study on thE:'
.,_,
___ ,·'·-""--~--'-,..~---
absorber ( 4}.
The solar airconditioner was originally designed. and
put together without a full investigation of the va.ricus
components due to lack of time to
!deadline.
m~et
the senior pro,ject
As a result, it was now necessary tc raanalyze
,various components, one of which that seemed to be ua.rtici
: ula.rly·
troublesome was the absorber.
'!'he purpose of the absorber is to lower the vapor pres.,
sure of the absorbent and refrigerant sclt:.tion.
As the
refrigerant is absorbed by the absorbent, heat has to be
~ -~-- - - - - - -
----- ----------------------·----------------·-.:--------·-
X
___ j
1removed from the system.
This is accomplished by ammonia
I
j<NHJ)
flowing through a helix within the absorber.
ithe flow of
r
NH.,
.J
Since
within the helix is two phase, i·t is im_por-
.tant to know just what type of
fl~w
is occurring there.
The
of flow in the helix is needed to predict the heat
~ype
I
I
:transfer from the solution to the ammonia i.n the helix.
An expel"iment was designed usin.g a glass helix with
Freon-11 to see what the flow characteristics were in the
'
helix and how they effected the heat transfer characteris-
I
:tics.
The first part.of this project consists of the thermo-
dynamic calculations for the entire system.
~light
There were only
variations in the results, which can be expected due
I
to various values
beir~
read from graphs, however since no
ca.lcula.tions appeared in the first report it is impossible
to
'
d~termine
where these differences occurred.
The second part of this project deals with the experii
mental setup, results for the eY.:perlment and some recom·men!
I
dations for improving the experimental data.
i
'
L----------------------------------------------------------------------·-----------------'I
xl.
~---
------------------- -------------------- ------------- ----------------
----------------
----------------- ------- ----------:
I
CHAPTER 1
THERr1:0DYNAIY1IC CALCULATIONS
In doir..g the
system~
this
calt~ulations
for the solar-refrigeration
only the r:dfrigera.tion circuit is considered in
r8p~·rt
as shown in E·igul"'e 8.
'I'he solar circui i; is not
. '
d an d the - _.;n
.:;t "-1 01.7 t:>C (_'>1... 5 °F.·);;:.
cons:!..vere•
_ c 0 m....~ nbo· wa.~t.o"'
...... .....
.
......
!assumed to be correct as stated in the senior project
1
report.
The following analysis is a summary of the thermo-
i
jdynamic calculations, for a mare d.ete.i.led analysis see
[Appendix A.
I
I
1 - Evaporator
Tr.e first component an.alyzsd is the ev?.pora.tor, whi:::h
The
a required output of 2500 l):J/:t;;." { 2J70 BTU/hr).
ma_::dmtL"Tl ambient temperature is specified at J5. 0
and a r·com tam.perature of 2). 9 °C ( 7 5. 0 °F).
0-.
\,,
\95
An initial
'temperature difference of 16.7 °C (JO.O °F) is assumed
'
.between
:which is
th·~
refrigerant. (ammonia) and the room temperature
co~sistent
with the senior project report.
'l'h:ls
I
information is summarized in
Figur~
1.
The quality of the incoming refrigerant is de"termined
L______ ----- ·--~ -------.--- ___.... - -- ------- ---------~-~---------------------------·-··------- -----------~--------~------------:
.1
2
= 2500
qevap
pl =
...
"'o
p
2
-
= 2).9
At = 16.7
tl = t2
kJ/hr
(2370 BTU/lbm)
558.2 kPa
0(1
v
oc
= 7.2
(75
(80.96 psi a)
Op,)
(JO °F)
°C {45
OF)
Figure 1
i by
-
the conditions a.t the expansion valve and is equal to
I
:84.83 percent.
Since the enthalpy across the expansion
valve is constant, the enthalpy at point 13 is equal to the
enthalpy at point 1.
The refrigerant is assumed to leave
the evaporator as a vapor with a quality of 100 percent.
jFrom the above information the pressure a.nd
I
;known.
i
entha.lpi-~s
are
Fer the evaporatore the first law sln:plifies 'tos
!
Since everything is known except the mass flow rate,
this can now be determined.
!
:L___________
----
2e)8 kg/hr
(5.26 lbrn/hr)
-------------------------~----------------
2 - Absorber
The next component in the circuit to be analyzed is the
!absorber.
Again, the same assumptions as in the senior
project report are used.
The exit temperature of the solu-
ition of the absorber, sodium thyocyanate (NaSCN) ru1d of the,
i
irefrigerant, ammonia (NH ), were assumed to be at 50.0 °c
3
! (122 °F'). This gives a temperature difference between the
i
:absorber and the cooling refrigerant from the condenser,
iwhich is assumed to be at 46.1 °c (11.5 °F) ·of ).9 degrees
i
: Celsius ( 7 °F ) •
ABSORBER
!
From Heat
Exchanger
From
Condenser
_@
/...._
To Condenser
1----f- From Evaporator
'1'2'~'---'~-----t·~
'\B
l
1
-GsI
@
2
t
3
::: 50 °C ( 122 °Ii')
t9 = t12
To
Pump
-cs-
=r-·
-
46.1
oc (7.0
oc (115 OF)
OF')
At
:::.:
].9
p2
-·
PJ = p11 = 5.58.2 kPa
Figure 2
With a temperature of 50 °C (122 °F) at the absorber
!
exit and from the graph of vapor pressures versus mole
!
[
i fraction in Figure 10, the mole fraction of ammonia leaving;
'
'
~L·-----~----~ ------·-----·--·-------~----·--··--------------~·---··-------·--·------...1
4
1-th.e-;bsorberat -point J is-~pp~oxi;~t~ly -o~-8o-:--- si;:-~-~-th;--·
generator determines the mole fraction in the solution
entering the absorber at point 11. the temperature of
:97.2 °C (207 °F) is assumed at point 6 on the generator.
i
iThe pressure of 1834 kPa (266 psia), which is based on. the
vapor pressure of the refrigerant in the condenser, is used
in the graph of vapor pressures of NH
3 from solutions of
•NaSON L1 Figure 11 to determine ths mole fraction of NHJ
!which is approximately 0.777.
!
!
It can be shown from the mole .fractions of points 3 and.
!
i
;11 and conservation of mass relations, that the mass flow
,rate at point 3 is 40.52 kg/hr (89 . .33 lbm/hr) and at point
i11 is ]8.74 kg/hr (84.07 lbm/hr).
I
See Appendix A for
:details of this calculation.
The heat transferred to the cooling ammonia refrigerant
iin the helix can be described by the first law.
Since the ammonia entering at p·oint 2 enters as a vapor,
itt
1
dlta
hea.t o.f CC'Jndensation at tha"t polnt is described by the
i
!following.
!
qNH
.
::::
J
qNHJ
j
=
lb2 (hg - h f )
)005
kJ/hr
@
558.2 .lr..Pa
(2848
(80.96 psia)
BTU/hr)
The ammonia is then absorbed into the sodium thyocyanate
lin an exothermic reaction.
L - - - - - - - - - · - - - - - - · - - - - ··---·-----------··
From Figure 12 the data for
· - - - - - - ··------------ - - --------------·-----· __ J
.5
integral heats of solutions of NaSCN in liquid anu.11on.ia at
0
°c have been plotted.
For points 3 and 11, the moles of ammonia to moles of
.sodium thyocyar.ate are 3.998 and 3.483 respectively.
The
)heat generated in the exothermic absorption reaction can
I
l
now be found from Figure 12.
The heat generated in the
exothermic absorption reaction is 390 kJ/hr (370 BTU/hr).
The total heat generated in the absorber is the sum of
1
the heat .jf condensation of ammonia (3005 kJihr) and the
'
!
!heat generated in the exothermic absorption reaction (390
;
:kJ/hr).
!
The tota.l heat generated in the absorber is 3395
; .kJ/hr (3218 BTU/hr).
With the total heat transferred to the helix known and
with the assumption that.the cooling ammonia in the helix
i enters as. a liquid
2.. t
the saturation temperature and leaves
:<the helix as a vapor of 100 percent quality, the necessary
flow rate within the helix to accomplish this cooling is
found to be ).17 kg/hr (6.98 lb~/hr).
J ·-
Pump
Continuing aro-und the circuit from point
pump in the refrigeration
sy·~tem
3~
the only
is next analyztJd.
An ideal
pump is assumed, therefore the entropy is constant as shown
:in Figure 3.
PUMP
i
From Absorber
!
I
L____ _
.v-wp
®
To Heat Exchanger
s = s4
3
Figure
J
--;----------·-·--··--·-----·---··-----------------------------1
1
6
The work for the pump is given by the first law.
w.P =
=
From information already determined the following is knowna
(266 psia)
p4
=
1834 kPa
PJ
=
558.2 kPa
...
=
50.0 0 c
m~
_,
=
40.52 kg/hr (89 • .33 lbm/hr) with the per-
i .
I
'
".3
(80.96 psi a)
(122 Op)
:cent of am.monia equal to 45. 7 percent at point 3.
'"
From
.Figure 13, the density of NH.., - NaSCN solution can be found
,)
ia.nd converted into the specific volume which is 1.18 dmJ /kg
I
'co. o1ss
ftJ /lbm).
Substituting the above information into the first law
:and solving, the pump work is 60.68 kJ/hr (.57.12 BTU/hr).
From Figure 14, given the percent of ammonia at point J
!to be 45.7 percent and the temperature to be 50.0 °C (122
:°F), the enthalpy at point J is found to be 60.5 kJ/kg (26.o'
!BTU/lbm).
The first law gives h4 = 62.0 k.J/kg (26. 6 BTU/lbrrJ.
l~ -
Heat Exchanger
Solution from the generator enters the heat exchanger
at point 6 at an assumed temperature of 97.2 °C (207 °F)
:and is assumed to leave at 54.4 °C ( 130 °F).
Solution from
ithe absorber enters the heat exchanger at point 4 at 50.0 °C
i
!(122 °F).
L_ _____ ·--------·-·-------
!
I
- - - - · - - - - - - - - - - · - · - - - -..
7
----------------------- -----------------
HEAT--Exdli~~GER --------------------------~
j
To
Generator
From
Generator
+-@
0-t
J;
I<
I~ -~
I----
l ;f_ >·
cl -b®
OF)
=
97.2
t
=
.54.4 oc (130 oF)
10
t,_. --
.--·
oc
t6
p4
-
50.0
(207
oc (122 oF)
p5 = p6 -· Plo =
18~4
./
kPa
.
(266 psia)
·ro
Pump
Absorber
Figure 4
;
.
I
i
Since the flow rates on both sides of the heat exchanger
are known as well as the percent of ammonia in solutions, the
:enthalpies
ca.n be determined for points 6 and 10 from Figure
i
14 with the above temperatures for these points.
j
enthalpies
known~
With these
the enthalpy at point 5 can be determi!'1ed ·
:from
the first law; see Appendix A for details of this cal;
i cu...at:1on.
i
,
•
h4
62.0 kJ/kg
-·
hs =
h6
::::
h 10 ·•
~c..
165 kJ/kg
(70.8 BTU/lbm)
172
( 74·. 0 BTU/lbm)
o/J
kJ/kg.
ir,T/kg
.,.,.
..
(27 BTU/lbm)
i
I
!
i
l____ ~~~-~-:::ure
(26.6 BTU/lbm)
14, the
point 5 is now found'
~-~~p~ratu~-e-at
- - ---·-------··-----------------1
!
8
heat exchanger is 91.4 percent (J, p.JJ6).
5 - Generator
The generator is considered next.
From the heat ex-
!char~er calculations the temperature at point 5 is 8?.8 °C
i
i
{190 °F).
The temperature at point 6 is assumed to be 97.2
•1 ~..., (207 °~-~- ).
10
1
i
mh
· de t
J.. e pressure ~s
·
erm~ne
db y ~·
t,;ne vapor pres-
sure of the re.frigerant in the condenser.
At a. temperature
of 46.1 °C (115 °F) the vapor pressure is 18)4 kPa (266
;
1
. )•
ps.ta
Figure 5 below shows the generator configuration with
the hot water entering
fr~m
the solar circuit at point A
. aml leaving at point B.
GENERATOR
To Solar
Circuit
From Heat
Exchanger
To Heat
Exchanger
Figure 5
J
The mass flow rates at points 5 and 6 are both known.
'-------------------·------------------------
______________________________ j
Q
----··------------ ---·-·--------------------··-···--·-- ------ -·------·--- -------- ----- ·---- ----------- -1
----·------- ----- - ----
The mass flow rate at point 7 is determined by "the original.
l
!COoling requirements of the evaporator and is equal to 2.J8
i
.
:kgI hr (5.26 lbm/hr).
I
At this point it is assumed. that the temperature at
point 7 is equal to 9J.J °C (200 °F) which gives a tempera-:
i
ture difference of J.9 °C (7 °F) between the exit temperature at point 6 and point 7.
With the temperature at point.
?, the enthalpy at that point is found from the ammonia
( 697.0 BTU/lbm)
1621 kJ/kg
=
'11 he overall heat transferred to the ammonia-sodium
thyocyanate solution in the generator is described by the
first law as followsc
q•
gen
= 3?59
kJ/hr
(3563 BTU/hr)
6 - Condenser
The final component in the circuit is i.:he condenser.
The high pressure side of the refrigeration circuit is
determined by the assumed temperature of the condenser
!which is 46.1 °C {115 °F).
i
1
This pressure is 18)4 kPa (266
ps.ia), the saturation pressure for the assumed temperature.
I
1
From previous calculations, the following information
lis known.
L
1
I
m,-
<s.26
= 2.:38 kg/hr
lbm/hr)
___________ ,_______ j _________________________________________________
,_______ _
I
1
10
h?
=
1621 kJ/kg {697 BTU/lbm)
h
~
1489 kJ/kg (640 BTU/lbm)
9
From conservation of mass, the mass flow rate leaving
'the condenser is 5.55 Jt"..g/hr (12.24 ll:>m/hr) at point 8.
CONDENSOR
From Generator
From Absorber Helix
To Evaporator
and Absorber Helix
Figure 6
· From the first law the total heat rejected by the condensor, which is also the total heat rejected by the refrig!
~eration
circuit,can be calculated.
]
4cond =
6347 kJ/hr
(6016 BTU/hr)
The coefficient of performance can now be determined as
i
lI
follows a
L_
11
COP
=
2.3?0 BTU/hr / .356.3 BTU/hr
COP
=
0. 665
7 - Conclusion
This completes the therm.odynar.lic analysis of the refrig..:.
·eration circuit.
The refrigerant that flows through the
I
ihelix is driven solely by a va.por syphon in the helix.
fO
1
I
I
calculate the flow rate, more information ls needed on the
.
i
iruode of flow within the helix.
The experimental results
:help to clarify this area.
, In addition, the heat transferred to the helix is also
ianother area which requil"'es more investigation.
The absor-
;ber - refrigerant solution entering the absorber from the
I
- :hea.t exchanger at point 11 is sprayed oYer the helix.
To
ide the heat transfer calculation an experiment should be
:
~set
up to determine how complete the coverage is over the
I helix,
At present there is no way of telling whether the
'
:S·:>lution
drips from one loop of the helix to the next, which
!
I
lis desirable,
Ol'
simply runs down the side of the absorber
!wall.
:
;
!
'
l____·----------------------· ·---------·-·-----------·---·--·-----·----·--·-----~
-----,
----------------------------~--
I
i
I
CHAPTER 2
. I
ABSORBER EXPERIMENT
The purpose of the absorber experiment is to determine
·what type of flow is occurring within the helix and to
I
l
!determine the heat transfer coefficient for the helix.
1 - EJcperimental Setup
The experimental setup consists of two glass helixes,
:one representing the absorber (in this case the evaporator)
I
!and the other the condenser, see Figure
15 for complete
·setup.
The evaporator (absorber) consists of a glass helix as
shown in Figure 16.
Freon 11, the refrigerant used in the
i
·experiment, enters at point A as a liquid and is heated as
it rises through the coils until it becomes a vapor and
I'
leaves at point B.
Thus a vapor syphon is formed and ere-
ates the desired flow.
The evaporator is bathed in distilled water which is
'electrically hea.ted through
8~-
feet of #16 Nichrome heating
'wire which is coiled around the outside of the helix.
Power
input is controlled using a variable transformer and moni1tored by a wattmeter.
A stirrer is also used to keep the
!water temperature uniform throughout.
Two thermometers are ·
_l __________ ·---------·----·-· ------------------------· ----------------- · - - - - - - - - -----------------------------------~
12
1)
~-----------------~-------------------------------
--------
-------
--- - - - - - - - - - - - - - - - -
iueed, one in the distilled water and the: other to monitor
;
lthe Freon vapor leaving the evaporator at point B.
!entire assembly is hc,used in
a glass
·
The
bottle and insulated
iwith fiber glass.
The condensor consists of' a simple glass helix and a
fan for cooling as shown in Figure 17.
Between the two helixes on the vapor side, the J/8 inch
Teflc'n tubing, which is used throughout, is. insulated wt th
J/8 inch thick rubber insulation and electrically heated tc1 .
:maintain
a vapor state until reaching the condenser.
I
Also
ion the vapor side there are a vacuum port and a Helicoid
!pressure
gage ( 0 to JO psi).
On the liquid side is a :fo'reon fill port and a flowmeter ..
:Figure 18 shows the calibration cur.ve for the Fischer and
!Porter
flowmeter, number FT-1/8n-zo-G-,5, for Freon i1.
;
;Also on the liquid side is a valve to restrict the flow if
i
:necessary.
2 - Cali"oration
'l'he calibration of the flowmeter is done as prescribed
i in the F'ischer and Porter Handbook, ( 6); Figure 18 is the
!calibration cu:t'Ve.
To determine the heat loss in the evaporatorD the flow
'Valve is shut off so that any heat loss is to the surround-
i
1
ings and not carried off by the Freon.
Using this method a'
const.9.nt power input is maintained for one hour and the rise
_in water temperature recorded.
Figure 19 is a graph of the·
'-------------~--------------------------·----------------------------·
14
I
!Change in water temperature for various power inputs.
From the plot of the change in water temperature with
1
I
:respect tc t:i.me, Figure 19, the overall heat transfer coef-:
I
:ticient "Un is determined as followss
Equation #t s
.
q
in =
+
The value of qin is equal to the power input of the
'heater as read off the wattmeter plu.s the power input into
t.he water by the stirring
action~
CP is equal to the heat
capacity of the system and L:.t 1... /e..9 is equal to the rate of
hchange in the water temperature with respect to time and is.
I
i
:equal to the slope of the line in the graph in Figure 19.
iThe value of
;
~t
2 is defined as the difference between the
!a.verage water temperature for a particular run at a given
I
I
:·power setting and the room ambient temperatureo
The power input by stirring is determined by noting the
:temperature rise in the water from the equilibrium temperai
>ture caused by only stirring.
:
From the calculated heat
I
:capacity of the system and the rate of change in the water
temperature the following calculation is made to determine
the power input by stirring •
Equation #2a
.
qstirrer
=
The total heat capacity, Cp• is computed using the vol1ume of water, mass of the glass container and that portion
I
.
of the steel stirrer that is in the water. The volume of •
I
:
l
--·------·-~---
--- --------------·~-------------·----------------------___;
'
15
I dis-i;i.1led
I
wa-t-er-ise_q_ual to52omii-ii1iters-;;h-i~h- ha~-;------~
.
I
! heat capacity ·::>f 2157.5
JjOc (1.1)68 BTU/°F).
The mass of
'
!the glass container is 3:36 grams which is equal to a heat
I
·capacity of 351.6
J;oc (0.1853
BTU/°F).
The mass of the
stirrer subrnerged in the water is approximately )8. 5 grams
iwhich has an equivalent heat capacity of 16.J.
BTU;oF).
;J·;oc
JjJc {0.0084-8
The total for the entire system is eq'lB.l to 2525
(1.JJO BTU;oF).
The value of At 1/AQ caused only by stirring is equal to.
?.1J x 10-J °C/minute. Substituting into equation #2, the
:}.'lower input by stirring is 0.300 Watts.
Solving equation#!
:for the total heat transfer coefficient, the following equa'
tion is obtained.
u
=
The values of the total heat transfer coefficients are
~shown
in Figure 19 and plotted as a function of temperature
iI
:difference (twater - tambient) in Figure 20.
From this plot
:the
heat less can he determined by knowing the water and
I
i
lambient temperatures by the following equation.
=
u
(twat
~.
t)
er - t amo~en
_,
3 - Procedure
For the initial start-up both the evaporator heater as
!well
as the tube heater are turned on to start the Freon to
'
I
!'------·---·-----------------------··--
16
i flow.
Once a flow is achieved the tube heate:;." can be turned
off.
Readings are takenevery five minutes cf water tempera,ture, refrigerant exit temperature on the evaporator, flow
rate, system pressure and power input to the heater.
ures 21 through
21}
are the graphical results for JO,
Fig40~
50, 100 and 125 watt power levels.
The 50 watt
readin~wer.e
taken with the flow valve open.
As a result of this, large fluctuations were encountered in
'the flowmeter which made making accurate readings impossible.
On all other runs the flow valve was partially closed
ito proYide stability to the system.
Also, after the 50
\watt run, which was the first run made, more Freon 11 was
I
'placed in the system which also helped to maintain-a stead-.
[ier flow rate.
All the recorded runs were made with the condensor 28
i inches above the level of the evaporator.
At the 100 and 125 watt power levels, as the temperaturc and pressure of the Freon 11 began to rise close to
i )0
psig 7 the flow rate
va1~1e
was opened slightly mere to
l
i a.llow more flow to occur and thereby reducing tha Freon
I
i
; temperature in the evaporator and the system pressure.
iCan
be seen in the resultant graphs in Figure 24 and
!for the fluctuations in the flow rate,
4 - Results
'rhis
acco~
1'1
!started on the lower tube surface and quickly forms a sepa·- ;
I
;
:rate region on the upper portion of the tube.
This predom·· :
jinately occurs on the third loop from the top of the helix .
·as can be seen in Figures 26 and.27.
The bubbles a.re initi-
ated in two distinct locations, possibly in surface :flaws
·at that point.
As the refrigerant leaves the helix, the va.por also car;ries over some liquid with it.
At this point the two phase
flow seems to resemble ar.u"1ular flow where the vapcr forms a
continuous phase, carrying only dispersed liquid droplets
;and travels up the channel core leaving an annulus of super'
!heated liquid adjacent to the walls {2, pp. 325 ~ 326).
I
i
!
!
'.rhe nonboiling length ·(L0
)
in which only the sensible
heat is added to the incoming subcooled coolant (Freon 11)
1
i
iis estimated to be approximately two feet.
At that point
the coolant reaches its saturation temperature.
The remain-
ing coolant path, 1.5 feet of the helix, consists of the
~boiling
'
length (Lb).
The following data is taken from Figure 21 for the power
input equal to JO watts at 45 minutes into the run.
i,.
L ---·------
q
=
30 Watts
P
=
8 psig
t wa t er
=
41.5 °C
l
I
I
_______ j
18
, - - - - - - - - - - - - - - - - - - - - - - - - - - ------------ --------------------
l
m
=
-------~
i
5.J lbm/hr
From Figure 20 the heat loss to the surroundingo can be
'determined.
With a water temperature of 41.5 °G
a:1d
an
ambient temperature of 23. 0 °C • the temperature dift'e:r ...:nce
is equal to 18 • .5 °C.
This corresponds to a. total heat
tz·ansfer coefficient of 0.13 Joules/sec. °C.
The heat loss
is equal to the followinga
0 1J
0
qloss --
Joules
-...-.. ---·sse °C
--~
2.405 Joules/sec
(2.405 Watts)
The actual total amount of energy (qt) input is equal
to
th~~.
measured input of JO watts minus the energy loss plus
the energy put into the water by the stirring which is equal
!to
o.;oo watts.
.
•
q."t
-
q
~.soI.
=
30 watts
+
qstir
+
•
qloss
0.)00 watts
2.405 watts
<1t
=
27.9 watts
(95.2 BTU/hr)
The total amount of energy input per pound mass of
coolant {qt) is equal to the total heat transfer rate <4t)
i
!divided by the mass flow rate of coolant (m).
i
[_ ______ - - - - - - - - - - - - - - - - - - - - - - - - -
·-----------------
19
r----------------------------- ------------ ---· ------- --:--·--·- ----------------------···-· -------..,
I
I
=
The mass flow rate in this case is equal to 5.3 lbm/hr
i
ifor the power input equal to 30 watts at 4.5 minutes into
'
I
)the run as stated on page 17.
f
I
J
t8.0 BTU/lbm (total hec.t added):
A - Pressure Drop
The total friction pressure drop for the helix is comi
!posed of friction due to single phase flow and to two phase
I
i
!flow.
I
The following procedure used in the calculation of
!the friction drop in the helix was developed for vertical
i
!channels ( 2, pp. 336 - .342).
For the nonboiling length (L 0 ) the friction pressure
!
;drop .is evaluated by the Darcy formula.
---2
v
'··a o
.10
=
2
i
The friction factor (f 0
)
gc
in the nonboiling length (L 0
)
!is dependent upon the average Reynolds' number and the waJ.l
!I
~oughness
I
in L0 •
L0 is the length of the glass tubing in
rhich single phase flow exists in feet.
'
:lent diameter of the tube in feet.
I
.
D9 is the equiva-
p 0 is the avera.ge
liquid density of Freon 11 in the single phase portion of
~he helix in lbm/ftJ and g 0 is a conversion factor equal to
!
i
8
2
!
~-L1_7__ x_lQ ______ lbm~f_t/~bf~hr. ___. ______ -----------~---------_j
20
~-------it-isas-sumed-th.a t
-t-te-Frion -11enters -the ev:aporato_r__
:at a temperature just slightly above room ambient temperai
:ture.
The inlet is assumed to be at JO °C.
Using the con-:
ditions a.s stated on the end of page 17. the average density
(1,) can be calculated from the inlet density
(,~Ji)
and the
density at the saturation temperature (l'r).
~'i
-
jOf
= 90.87
Po=
91.88 J.bm/ftJ
t
(lei
lbm/ftJ
+ P.n}
J.
oc
@
.30. 0
@
33.5 oc
':{
-::::
91..38 lbm/ftJ
The averaee coolant fluid velocity
(Y~)
can be found
by
idividing the mass flow rate by the average density just
!
i
determined and the cross-sectional area of the helix tube.
The inside diameter of the helix tube (De) is 3/16 inch --
:with a corresponding area of 1.917 x 10 -4 ft 2 •
!
m/;:>0 A
513.6 ft/hr
The Reynolds' number is next determined as follows:
The va.lue of j-1~ is determined using the same method as
was used in determining t.he density above and is equal to
1
· 1. 017 lbm/ft·-hr.
Using the above values, the Reynolds number is found
!to be equal to ?.21 x 1.0 2 which indicates laminar i'low.
j
!
L---~---H---------·--·-----
---~---------~--------------)
21
JFroin-thfs-~tile- rrfct-i.~~n-iactor-(:t:~r---i.s--eCiuai-to--647Re~-.------i
64/Re 0
'Oe 0888
The friction pressure dr•:>p can .now be evalua.ted for the
!
inonboiling length
where the length of glass tubing in which
single phase flow exists (Lc.) is eq:ual to 2. 0 feet.
1
L
=
-·-2
/'o~
0
2
gc
0.002?.8 psi
The two phase friction pressure drop in the boiling
:length (Lb) is evaluated by first calculating the single
phase pressure drop in the boiling length a.ssuming that
only :sa:turated liquid of the same total mass flow rate exists
·in the channel.
·-2
--~. /'f Yfo
De
2
gc
The boiling length in the heli.x in which two phase flow ·
Ije Xl.S
• t S (Lb) is equal to 1.5 feet.
!
lfe:rence in (IJb) and (L 0
),
even
~he
th~u.gh
reason for the difthe two phase flow
ibegins in the middle of the helix 11 is that the value of (L0
1
)
includes the entry length of glass ·tubing running down the
I
·center of the helix as sho\vn in Figure 16.
/
I
vfo
= ~/,~f
A
=
S16.5 ft/hr
[_.~----------------~------------------~-----
J
22
II
I
Reb
=
0 el'f vfo
Reb
=
7 • 52.
!
fb
=
X
I f4 f
10 2
=
64/Re 0
0.0851
From the above information the f'riction pressure drop
lfor the boiling length can be determined as if only satu-
lrated liquid existed.
(aPsp )~
!
-
o. 237 lbr/f·t
2
=
o. 00165 psi
!
I
;
A two phase friction multiplier'(R) greater than one is
jnow multiplied by the above to give (APtp)Lb' the fr'i,ction
I
jpressure drop for the two phase flow.
I
I
R
i!
The values of the two phase friction multiplier (R) for :
i
.
~low
I
I
pt"'essures a11d high void fractions are plotted in Figure !
I\2 ...'\. where the mass velocity rate (G) is equal to 0.61+7 x 105
'
2 • The quality, (';(e) which is equal to the mass
llbm/hr-ft
iI
I
i
'flow rate of vapor divided by the mass flow ra.te of mixture,;
I
iis estimated by calculating the mass of vapor that would be
i
!achieved if just vapor was flowing.
i
=
I
_l _ _ _
'
I
J
_________H____
!------------------
p_VG----~
=
{AP)
--. ·-------
f -- --De 2 gc
The actual length of the channel (L) was represented by the
,.,.,
'1-f)
: na~gn~.
\~ ..
i·.
In this case, L was substituted since a helix
!is inYolved and the height is obviously not the total length
'
:of the fluid path.
To obtain a. value for the two phase friction mu.l tiplier~
!
1
the values in the original graph were extrapolated to a
I
iquali"ty of 20.2 percent to obtain an approximate value of R.
lrn addition, the graph is plotted for a mass velocity rate
2 as opposed to the ?.ctual
i (G) equal to o. 647 x 105 lbm/hr-ft
.
2
~
.value of 0.469 x 10.:> lbm/hr-ft •
The following is the Lottes - Flinn correlation for the .
:,two
phase friction multiplier (2, p.)l-}2).
1
R
1
2
+ { - - - ) + (---·-)
=
1 - a
e
J
l - ae
~
i
:The void fraction {ae) at the exit is foU-'I"ld by the following
method a
=
-x-9
:::;
0.202
vf
-·
0.01109 ft3/lbm
vfg
=
a~
..:::
"Xe
vg /
(v f +·'Xe vf g )
@
JJ.S oc
@
i
-----------------------·------]
25
~-----------·-
=
vg
1
------:)------ -·-··- ··-·:·---·------ -·-··· -~o· ·---·---···-·-· -----
1.765 ft /lbm
@
JJ.5 C
1
i
I
!
Substituting into the equation for the void fraction, a :
i
value of
o. 976
is obtained.
rrhis value of void fraction
gives a corresponding value of the two phase friction multi'
'plier (R) of 59J.
Using this value the total friction pres-
sure drop in the helix is 0. 981 psi.
'l'he other value for
the friction drop using: F'igure 25 is 0. 687 psi.
i
!midpoint
value of
I
i
iof head.
o. 8J4
Taking the
psi, this is equivalent to L 32 feet
The actual observed head is 1.83 feet.
This leaves
i
!a difference of 0.51 feet of head which can be attributed to
'
!line losses.
B - Heat IJ.1ransfer
For the heat transfer coeff'icient for the nonboiling
i
:portion o:f- the
helix-~-
the- glass helix is considered to be
;straight and heated through forced convection at a water
:velocity (uc;c.) of 1 ft/sec.
HELIX TUBE CROSS SECTION
Ucc
Figure
7
i
I
I
l
'
The bulk temperature of the coolant at the point of en-
!
1
I
\:tr:y_J.!fi-)---~ 6--~-~~~Il!~g__j:g __l}_~y-~--~_i~_rr_!P~-~~~-~!'.~__ Q'f_l_Q_.__Q_ °C ~ __1'h_e_j
26
,-------------------~--------------·-·---·-----------.:..
---------------------------------------
/heat transferred to the nonboiling length is therefore the
!
;sensible heat required to re:ach the saturation temperature.,
The rnass flow rutr:t (m) is t)qual to
5.3 lbm/hr, hfi is
the enthalpy at tile point of entry and is equal to 25.J
B'L'U/lbm and hf is the enthalpy at the saturation temperature
and is equal to 27.8 BTU/lbm.
Substituting into the above
eq.uationil the energy transferred to the nonboiling length
: (q 0 ) is
13.2.5 BTU/hr.
For the geometry, namely a long tube, the following is
i
! the he~a.t transfer equation.
1
h. A.
~
1.
This equation will be solved :for the heat transfer
1
coefficient for the inner surface of the helix (hi).
ii
A. ls
l.
;the surface area of the inner surface of the helix with
I
radius r 1 and length L0 (nonboiling length), and A0 is the
outer surface area with radius r 0 and length L0 • The fol-
:lowing is a list of the known values.
I
L _ __
r0
~~
3/16"
(1.562.5 x 10- 2 ft)
____________________________________ __j
27
~-------··
kg lass
::
0.45 BT'U/hr-ft OF
Lo
=
2.0 ft
Ao
=
0.216 --f'J-""
Ai
-· 0.108
ry
?
ft ....
The heat transfer coefficient f'or forced
convt~ction
on
the outer tube wall (h 0 ) with a water velocity (u..,..) of
roughly 1 ft/sec is determined using the following correls.-
tion for the average heat transfer coefficient in crossflow
(3, p.186).
=
The properties for the water are evaluated at the film
i
!temperature.
twall +_ twater
2
The wall temperature is assumed to be at 85 °F which
gives a :film temperature of' 88.65
0
F.
With this temperature
i
lth.e properties are as follows:
=
The value for the outside tube diameter (d 0
)
is equal to
I
[
----------------------------------------·----------·----------_j
28
,--------------------------------·------------~-----~------------
!
jJ.125 x 10
-2
I
I
feet.
=
.The conductive heat transfer coef.:f.icient (k) for the water
lis equal to 0.]65 BTU/hr-ft
The correlation for the
°F.
heat transfer
avera~e
cosffic~rt
in crossflow in the middle of the previous page can now be
'solved for (h0
)
and evaluated.
=
=
. h0
798. 4 BTU/hr-ft 2 °F'
This value of the average heat transfer coefficient
(h0
)
is for the entire helix, not only the nonboiling
!length.
!
The heat transfer equation on page 26 can now be solved
; for the heat transfer coefficien·t for the inner surface- of
the helix (hi) and evaluated.
The equation is repeated
hare for convenience.
.
qo
=
-
{tw
1
hi A.~
tfi)
ln(r /r.)
1
+ __
0~.....!_
+---
2'n"k Lo
ho Ao
Solving and evaluating for (hi)=
56.73 BTU/hr-:ft 2 °F
(nonboiling 1ength)
I
L____________·___
. _,__________________j
I
29
For the boiling length, the heat transferred to the cool:ant is the remainder of the energy not used to heat ti1e coolj
;ant to the saturation
temperature~
In this case, after
.accounting for the losses to the surroundings, the actual
:F!nergy transferred is 95.2 BTU/hr.
Therefore, the energy
:transferred to the boiling length is 95.2 BTU/hr minus the
i
t
!
energy used to rai.se the coolant tcJ the saturation tempera-
:tur& which 113 1) •.25 BTU/hr.
I
This difference leaves 82.0
.
!BlJ.lU/llr transferred to the boiling length.
!
Since the geometry is the same for· the boiling length
:as it is for the nonboiling length, the basic heat transfer
;equation remains the same, only the subscripts are changed.
fThe
following is the heat transfer equation for the boiling
I
!length.
!
'
'
.
qb
(t
=
1
-·h. A.
J.
i
J.
+
w
tsat)
ln(r0 /ri)
2 1i'- k Lb
1
+
ho Ao
This time (hi) is the heat transfer coefficient for the
,:.~nner surface of the helix for the boiling length, (t sa,t) is:
ithe coolant temperature at saturation and is equal to 41.5 °c
I
!
:
j(106.7
0
I
~ooiling
:
!
F) and (L,0 ) is the length of glass tubing f'or the
leng·th and .is equal to 1. 5 :feet.
Solving the above
.
~quation
for (hi) and evaluating&
(boiling length)
i
i
· - - - ·----·-----------i
30
r--------------------------------------·-----------------···------,
1
I
The two convective heat transfer coefficients for the
:
:glass
helix - Freon 11 interface are now 56~ 73 BTU/hr-ft °F;
!
:for the nonboiliP.g portion and 734 BTU/hr-ft 2 °F for the
2
boiling portion.
5 - Conclusions
By knowing the heat transfer coefficients for the absorber, both the boiling and ncnboili.ng SE.\Ctions, the size of
the helii'.: required can now be .::stimated.
From the thermodynamic calculations for the absorber,
the total heat needed to be removed by the absorber helix
is 3395 ktT/hr (3218 BTU/hr), see page .5.
Assuming that the
coolant enters at the saturation temperature, the entire
helix would be in the two phase fl.ow regime.
'l'his was the
origind·assumption in the calculations.
.
q
=
h·~ ·A.l. (tabsorber
- tNH3 )
The following is a list of known values.
q"
=
3218 BTU/hr
734 BTU/hr-:ft 2 OF
h.t.L
~
A.l.
=
(
d.~
=
1 • .562.5
tabsorber
i
di.) L
=
X
10-2 ft
122 OF
Q
l--------------------tNH
= 11.5 F
3--------------------------·-------I
.31
i - · - - - - - - - - - - - - - - - - - - - - - - - · ---1
---- ---- ---·- --·-··
........... - - ----. .. ----- ·--
------ ---- ·-- ···-- -------- - .. --------- -····-----
The value for the helix length· (L) can now be determineq.
L
=
12.76 f t
Using the existing helix with an overall outside dia-·
:
meter of 2i inches, a length of 12.76 feet would correspond!
i
to a helix with approximately 26 1oops.
i
The corresponding ·
,pressure drop for Freon 11 in this helix would be approxiI
'
·mat ely 3, '14 :psi. ( 5. 93 feet of head).
6 - Recommendations
To determine the amount of subcooling occurring, another
thermomttter should be placed. in the incoming side of the
evaporator.
Presently there is no certain way of
ing this temperature.
determin~
·
Also., the addition of several thermoJ
couples ,on the glass surface of the helix would give some
indication of the temperature of the glass and whether the
temperature is uniform over the entire helix as is the
assumption.
A more accurate method of determining the temperatures
with a recording capability would also be helpftll, particu.larly in the determination of the overall heat transfer
coefficient for the insulation surrounding the evaporator.
A liquid trap on the vapor side of the circuit would
allow measurement of the amount of liquid being carried over
Ito the condenser by the vapor leaving the evaporator.
With the above improvements i.n the setup, a more accu;..
;
I
'
'
!
j_:~t-~-~::aly~_is-~-oul~~~~-~-com~~~s~d ·--~it~- t~e~e i-~~-=-~~~_-___ j
32
jments~astud;y~- of
l
the- -iiiiec ts-. ()f--c ooiS.-nt___mas_s_f-low--ra teon_____
lthe heat transfer coefficient would be worth investigating. ·
I
I
I
l_______
I
- - - · · · - - - - - - · - - - - - - - - - _ _ _ _ _ _j
r-- ----------- ---- ------------------------ ---!
REFERENCES
·~
' J. •
Solutions of NaSCN in Liquid Ammonia."
Journal of the American Chemical Society. 11 April,
·~conc-entrated
1962.
'2~
El-Wakil 7 M.Ivl. Nuclear Heat Transport. Londoru
International Textbook Company, 1971.
New Jerseya
).
Holman, J.P. Heat Transfer. Jrd ed.
Prentice-Hall, Inc., 1971.
4.
eProject Report for Energy Resource Alternative Competition." De:partment of Thermal lt'luid Systems, .Sc.hool
of Engineering, Calife State Univ., Northridge~
January 1976.
i
; 5.
6.
"Theoretical Performance of an Ammonia Sodium Thyocyanate Intermittent Absorber Refrigeration Cycle."
Solar Energy. vol 12, April 1968.
"Tri·-Flat Variable-Area Flowmeter."
Warminster, Penna
Fischer and Porter Co., Handbook 1029010.
i
i
I
l___________._____.______.________________ __j
:-- ---------PROPERT!ES--OF -AI'JINIONIA- AND
___________ -----,
rt~fl
i
AT 1 ATM, 300 K, (except as noted)
Ammonia
{NH
3
!Density {kg/m3)
Specific Heat {kJ/kg-°K)
I
2
; Visco·si ty {N-s/m )
Therm51 Conductivity
(W/m
)
82J.5
3•• 476
4.38
0.870 @ 297'1<
I
0.01018@ 273 OK
0.00042
0.353
0.09.3 @ 29'lK
I
K)
Boili.ng Point (°K)
I
'
240
297
'
5.40
Prandtl Number
{cp ;V/ k)
IL________.
@
Table 1
------~------------------------·------
. 34·
273 °K
I
_j
--
APPENDIX A
THERMODYNAl'HC CALCULATIONS
i
I
i
I
L _____
----------------------
______________]I
3 .<
\~
r-----~
i
-----------l
REFRIGERATION CIRCUIT
To Solar
Circuit
i
Condensor
t
7
= 9J.J°C
l ..r=-
-
-)~,
.../
.,
I
I
______ !
.---1~·----~
_____.
'0
,1;8-
1°~h
(.J..; . . .k'D"'. . . ..
tg ::; J..i.6 .1 °c
8
r.:pansion
Valve
37
~-
----------------1-~-Evaporat~r------------------------,
The quality of the incoming refrigerant is determined
i
iby
the conditions across the expansion valve.
EXPANSION
VALVE
~(!)
h
hl
-_lJ =
Figure
9
= 266
psia
P13
Therefore&
hf'
= h13 = h1 = 17J
BTU/lbm
Equation #1a
80.96 psia
=
=
..
62·'+.1 BTU/l"brn
.531.8 BTU/lbm
Substituting into equa:tion #1 and solving for the qual-:
iitya
?(
·--
84.83 %
Since a temperature difference of JO °F is assumed between the refrigerant and the room temperature, t 1 and t 2
are fixed at 45 °F. Therefore P the pressure is f.ound in the
38
: ammonia tables.
I
=
=
80.96 psia
=
624. 1 BTU/lbm
2 - Absorber
The generator exit temperatura at point 6 determines the
mole fracticn in the solution entering the absorber at point
;11.
The temperature of 97.2 °C (207 °F), from the assumed
;temperature at point 6 on the
generator~
arid the pressure of
:266 psia which is based on the vapor pressure of the refrig'
!'erant in the condensorg is used in. the graph of vapor pressures of NH
3
from solution of NaSCN in Figure 11 to deter-
imine the mole fraction of ammonia which is approximately
'0.777
(NHJ) moles
-
0.777 at point 11
(NaSCN + NHJ) moles
With a. temperature of 50 °C (122 °F) at point 3 on the
absorber exit and from the graph o.f vapor pressures versus
mole fraction in Figure 10, the mole fraction of ammonia
!leaving the absorber at point J is approximately 0.80.
(NH ) moles
3
(NaSCN + NHJ) moles
=
0.80 at point .3
I
I
[_, ____________________________________ - - - - - - - - - - - - - - - - - - - - -
____ _j
39
~----- V;POH
JOO
'
----~-----------~-·-·-----------~-----------·---!
PRESSURE
vs MOLE FRACTION (1,
.
p~1078)
i
250
2.00
-"'
50 °C
1.50
orf
ra
-s..
A
4)
:::f
fll
tn
a>
~
~
100
50
1
r-_____ -- _-_-~
I
I
t
'
0
1.00
~----~--
.70
.80
.90
Mole Fraction of NHJ
lI- - - - - - - - - - - - - - - - - · - -
Figure 10
- __________j
40
-·-------------------------~----~----:-~--·------·--··
·--~---·--------·
-- --------------- ----·-!
VAPOR PRESSURES OF NH':) FROM
.}
600
SOLUTIONS OF NaSCN
(4)
500
400
-.....
·'UJ
tO
Pt
.300
266
·------------~-
.85
·~---r--·
.so
.70
Mole Fraction of NHJ
Figuz_·_~_t!___ _
I
_______________ j
41
,.-~
·---~-----------·------.,----
·-·
-------------
...
-
----
-----------·
-
Since a mole is equal to the mass divided by the mole1
cular weight, and the molecular weight of ammonia and sodium
.thyocyanate are 17.03 and 81.07 respectively, the equations
for points 3 and 11 are as follows.
mass (NH..,)
..J
Eqn /111
17.0.3
------
m.as~ (NH
3)
17.03
-
mass (NHJ + NaSCN)
17. OJ
+
0.777 at point 11
81. 07
Rewriting the equation for point
· - - @)
17.03
0.80
=
point .3
+ 81. 07
17. OJ
Eqn #2a
= o. 80 "~.t
mass (NHJ + NaSCN)
mia.ss
NH
Jo (Eqn #1)¥
l @3
17.0)
+
mass NaSCN
0.80 @3
81.07
Rearranging:
mass NaSCN
- - - - · - @.3
81.03
1.
=
mass NH.3
- - - - - - - @)
0.80 17.03
- - - - - - @)
17. OJ
Similarly with point 11, the identical procedure is used
starting this time with equation #2.
i
shown an the top of the next page.
The results of this is
,-------·mass NaSCN
· - - - - · ------
mass .NHJ
1
1
@11
81.03
=
17.03
0.777
mass NH
J
@11
@11
17.03
Since the mass of NaSCN enter.ing at point 11 must equal,
ithe mass of NaSCN leaving at point J, the following relation
!
iis obtained.
mass NaSCN
- @.3
81. OJ
mass NaSCN
-------·- @11
81. 0.3
=
Substituting the two equations just obtained for points.
/J
and 11 into the above equationp the following is obtained.
mass NH
1
mass NH.J ·
- - - - - - - - @11
0.777 17.03
1
=
17.03
mass
NHJ
0.80 17.03
j_
3
_____._.___: @11
@3
mass NHJ
·---- @J
17.03
This equation reduces to3
Rearranging&
=
0.287
- - - (mass NH ) @11
3
0.250
From conservation of mass consideration of the absorber
)the
I
following relationship is true.
,
J.-----··------------~----------------·------·--------~··--·------·-1
=
(mass NHJ) @.3
Since the mass of NHJ at point 2 is 5.26 lbro/hr from
the evaporator calculation, this can be substituted into
.the above equation •
.5.26 lbm/hr
(mass NH ) @11
3
+
=
(mass NHJ) @.3
From the last equation on the previous page and the
above equation. the masses for points 3 and 11 can be deter;mined.
(mass NH ) @11
3
=
J5.54 lbm/hr
(mass NHJ) @.3
=
40.80 lbm/hr
-Substituting into equations 1 a.nd 2 on page 41 and solving
for the mass of NaSCN at points .3 and 11p the following
.results are obtained.
(mass NaSCN) @11
=
48.5.3 lbm/hr
(mass NaSCN)
=
48.53 lbm/hr
@_)
The total mass flow rates at points J and 11 are the
sums of the mass flow of
Jh3
=
NH~,
.)
and NaSCN.
(NHJ) @)
+
(NaSCN)
@3
•
·- 89 • .3.3 lbm/hr with a concentration of
mJ
/NH
l__
3 of 45.7 percent.
\
_________________________________\
.·-·----------------------.,.......-···------~----~------------------··-·----------·---~--------~-------,
I
Similarly for point 11 a
!
'
!
=
I
84.07 lbm/hr
with a concentration of
J of 42.) percent.
NH
From Figure 12, the heat generated in the exothermic
I
iabsorption reaction is calculated as follows•
moles of NHJ
~
3.998
at point J
=
).483
at point 11
moles of NaSCN
moles of NH
3
moles of NaSCN
As shown on Figure 12 ~- ..AH is equal to 0. J,:. Kcal/g mole
;
:NasCN.
AH
6.H
A P.
=
=
=
400 cal
0.508 l.b mole NaSCN
g mole NaSCN
92252.8 cal
hr
454 g
hr
_).9685
X
1 lb
10-J BTU
1 cal
J70 BTU/hr
--
The tctal h(1at. genarated in the abst.')rber (qt), (3218
:BTU/lbw), is the sum of the heat of' condensation of amrnonie.
'and the heat generated in the exothermic absorption reaction
1
just determined above.
The necessary flow rate within the
helix to accomplish this cooling is found as followss
I
.
~
~
(h - h )
l------------------~----~-_g-----~---------------
q
-------
__ ____ __
.
:
--·---~--
INTEGRAL HEAT OF SOLUTION OF NaSCN IN
SOLUTIONS OF N~SCN IN I,IQUID NHJ AT 0°C ( 1, p.1081)
10
8
z
0
~
:z;
H = 0.4 Kcal/g mole of
- -- - - - - - - - - --I
6
ascN
I
I
I
l
l
jPOint J
Point 11
I
I
2
I
0
I
~---~--------~~-~~·--*~~~
0
i
~~-----
1
2
J
·~-~---
4
5
Moles NHJ/Mo1es NaSCN
_ _ _ F_;_§;ure__!2____ _
_,
0
7
8
I
I
________j
46
------------~---
-----~--
-----·-·----·- ···---~-~-
DENSITIES OF NaSCN AND LIQUID~NHJ -------~
. I
!
VERSUS TEMPERATURE (1, p.1076)
SOLUTIONS
0.9
~
.
--....._
45.7%
NH,.
o.s~t·
---~----~
.
.
I
-~
I.
0.8
-~........
~._
I
~ 7'7. 86%
-------
0.7
NHJ
-----
0.
40~~-----.
•
10
20
.--------.r
.......
30
--r·---+:-~~---------40
50
60
70
80
Temperature (
Figur~
--·-·--------
13
0
c )
J
47
qt
.
·---------------------------~---·--·-------------------------·,
1-,
=
Ih0
6.98 lbm/hr
=
/
,
The values of hg and hf are looked up in the ammonia
1
tables at the saturation temperature of 46.1 °C (115 °F).
J - Pump
To determine the enthalpy at point 4, the first law in
!
!the following form is used.
i
h •. · =
26.6 BTU/lbm
"''
4 - Heat Exchanger
The following is the procedure used to determine the
enthalpies for the heat exchanger.
The percent of NHJ in
1
!solution at point 6 and 10 are the same as determined for
I
;
ipo:lnt 11 and equal to 42.3 percent.
With the assumed tern-
!
peratures at points 6 and 10 of 207 °F and 130 . °F respec-
1
'
:tivaly, the enthalpies for t.hese points can be found from
!
iF'igure 1/.,'-.
I
n10
.
i
The enthalpy at point
\tions.
I
=
74 BTU/lbm
=
27 BTU/lbm
1}
is known from the pump calc.ula-
From first law considerations the enthalpy at point
[_ _________._________
i
-----·------·---·--------J
48
point.
=
89.JJ lbm/hr (h
5
- 26.6 BTU/lbm)
=
84.07 lbm/hr {74 BTU/lbm - 27 BTU/lbm)
The percent of ammonia at point 5 is the same as at
I
ipoint J. From Figure 14, with a 45.7 percent concentration
'
tof NH
3
, the temperature at point 5 can now be determined.
· The maximum effectiveness for the heat exchanger is
;exr)ressed
by
the following (J, p.)J6) a
"t
E:
6
X
t6
€.
,
I
;__ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
- t10
:l:
:;:
100
- t4
91.4 %
I
I
· - - - _____________________________j
,
TEMPERNFUHE - ENTHALPY DIAGRAM
FOR NaSCN -
NHJ
SOLUTIONS (5, p.142)
50% NHJ
120
45.7%
NH
J
i-}2.
J%
NH.,.
J
100
i-
~-=_27
'i
::
- ·_uLlE~/
207 oF'
190 oF
zo[i;.3-:; 26OF
::::
I
I
I
I
' l
0
50
100
........
150
200
I
~...--.4
2.50
Degrees Fahrenheit
I
I
.l-------------
Fi~ure
14
--~~---------------- .. --~-------------.--
!
~-----------
1
I
•
APPENDIX B
ABSORBER EXPERINffiNT
'.
I
;
I
Il______________________________________________________________________________ _[i
50
··-·--·····
~·-------
1
--
.....
-------···
EXPERIMENTAL SETUP
i
I
~001~
I
p
j
i)
::;:?'
"· ~
Flowme oe-
~II
~~"~
:11
11
\
.
Fill
Port
Flow
Valve
~\
-~
.~
l1
•
.....-:;?
iP.
-:::;;;..::-:ffw11=s--6=
'~gU"'r:.
F .....
l,.-
1.5
____ j
\_n
......
52
~
I
EVAPORATOR SETUP
1
'
. To He-ater
r[J
Wattmeter
1
I
L---·
Variac
~--~--·---·----___j
I
l_______ -----------·-
Figure 16
- - - -----------'
53
CONDENSOR -
·--------
SETUP
Inlet
Fan
I
i
i
I
L----~-
-------
Figure
1?r
~
______ j
1'4
~)
,-i
-~-----------------------------
1
CALIBRATION CURVE FOR FISCKER
'
----"]
& PORTER
FLOWMETER #F•r-1/8-20·~G-5 FOR FREON- l.l
(Stainless-Steel Float)
20
18
16
6
4
2
oo
I
2
4
6
8
10
Flow rate (lbm/hr)
,
;
12
14
i
I
I
:_____________________________ _l_i_g1!_1;:?_1~-------------··-·-· __________ ]
_.,r::s
,----·-·-···--------·-----------------------,------------·---·-------------- ---1
.
I,
.
I
CHANGE IN WATER TEMPERATURE
I
1
'
FOR CONSTANT POWER INPUTS
P
~
10 wa·tts
0
U =- 0.1.'710 J/sec
P
5 Watts
-
r---------.. . . .-=:::::-_: =
U
60·
./ P
----- _.,/
-
0
0
U
U
:i=
= 0.1732
~
15
c:
J/sec °C
Watts
= 0.1297
J/sec °C
0 Watts
0. 1622 J/sec
~ - 10 Watts
,U;;; 0,090 J/sec
~
0
0
c
c
~/
/
~--p-= 5 Watts
30
~
20
t:::::::_-t-______,___,______.,. . _,_____.
0
10
20
30
40
50
60
Time {Minutes)
Figure 19
'---------------------------------
·-~··---~---~----'i
56
~-----------·-----
----- ---------·--
-----------------------------·------c----------~
HEAT TRANSFER COEFFICIENT FOR
j
'
J
I
EVAPORATOR INSULATION
I
40~
I
I
0
0
I
J5l
•
20
1.5
I
I
10 ~--+------ -_j-t--0. 08
0.10
'0;:12
~+-:
0.14
- - -
-1-1- . - - - -
0.16
Total Heat Transfer Coefficient
U (Joules/sec °C)
IL _____
I
_____Figur_e_z_g______________ __:_____ _ j
57
--------
------------------------------------------------------------
EXPERiblliNTAL RESULTS
1
P
17!
16
48'
16
15
46
15
1.4
44
~ 141
13
42r
~ 13!
12
40
~
a
,.0
~-12
...,
-
.......
0
.~11 -38
.
P,;
~ 11 cv 10
s...
::s
0
~ 3.0
U'J
m
.«!
OS:
"""'
oj
s...
Q)
9 §t34
Q)
8
9
8
32
8
7
30
6
28
51
26-
41
24
7
6
5
Ma.s s Flow Rate
t ambient --
I
i
L _________ _
Freon---·-
___./'"
.E 36
0}
Q)
t
s...
In
r.t
p..
/
4'
.._
~
WATTS
I
en
Q)
= JO
0
_·-----"~-~
10
2) • 0
°c
20
30
40
-----r---70
-~--+-!
50
60
Time (Minutes)
I
Figure 21
----------"~-
________ j
II
58
----------------------- - - - - - - - - - - - - - - - - - · - - -
I
EXPERIMENTAL RESULTS
. !
= 40
P
14
20
52.
13
19
50
12-
18
48
WATTS
I''
11
-
~10
17
46/
16
-·
-9
"Sf
~
0
...;;42
,.
Pressure
fl.t
1'1.1
-~ 6
12
.36
5
11
34
10
32
9
JO
I
81
Flow
--+--~---~-----~---~---l41---
28
10
0
15
20
25
JO
J5
Time (Minutes)
t ambient
= 23 · o
0
c
I
I
!
L____________________F~g~~~-__?~---·--------·-----------_j
59
~-------------···------
--,
---
!
EXPERIIv1ENTAL RESULTS
P
I
= 50 WATTS
5.
52
50
48
16
-
4
12
38
3
11
J6
2 .
10
3
1
Ol
JO ~-----+---~----·t---·t-~0
10
20
)0
40
' - ._.,_________;;
60
50
Time (Minutes)
tambicnt = ZJ;o-Oc----------
I
I
L_
Fi~~~e 2J
_I_
I
I
- - ___________ _J
60
EXPERI~lliNTAL
RESULTS
74
P :: 125 WATTS
6
28
l
26
66-
64
Flow
6
20
Mass
Flow
Rate
52
50
18
4
I
I
I .
48
46
i
2-
16
I
44
-t
10
0
.
---t>----t-------J.--+--r--·
20
30
40
50
60
70
Ti.me (Minutes)
t ambient
i
L___________________.__
= 2~.0
J
°c
i
__j
61
TWO-PHASE FRICTION MULTIPLIER {R)
k~D
AT LOW PRESSURE
HIGH VOID FRACTION (2, p.J40)
I .
'500
r--
Extrapolated Values
I
I
-
-----1
I
I
I
I
I
- --- - - 1- -
- -
I
I
I
I
/
I
, r::
.....0
,625
·r-i
·~
1%-t
:Q)
I
I
I
I
I
I
I
i
I
I
I
I
G = 0. 64'? x 105
lbm/hr-f't~
2
~m
'cd
~1.50
, I
~
E-t
100
502:----,T---r__..--:::8~--;1~0--l2- !LJ.
Quality ('X)
l6::-.-~1=8---2~0-22
(percent)
Figure 25
6)
CLOSE-UP OF EVAPORATOR
Figure 27
62
PHOTOGRAPH OF EVAPORATOR
Figure 26