# Nozzle Calculations

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```					        Chapter Three
Modeling and design
In this chapter, We are in interested in developing a designing
sequence with general parameters for most parts of monopropellant
thruster. The design procedure includes :
nozzle design, catalytic bed calculations, valve and regulator selection,
tank calculations , and injectors calculations.
As discussed in pervious section

3.1 Exhaust gas properties:-
I)Nitrous Oxide(            ) properties:

The equation of reaction for nitrous oxide
( 3.1.1)

The mole fraction can be estimated by
(3.1.2.a,b)

By using mass fraction one can get mole fraction
(3.1.3.a,b)

get gas constant

(3.1.4)

Then to determine chamber temperature(Tc), so apply balance of the energy equation
as follows:

(3.1.5)

Where for reactants:
(3.1.6)

And for products of reaction:

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(3.1.7)

by equating the above equations and integrating as shown
(3.1.8)

(3.1.9)

From above equations , temperature of reaction in catalytic bed Tc is
estimated , where Ts is standard or ambient atmosphere, the only
unknown will be Tc

other properties can be obtained now as follows:
(3.1.10)

(3.1.11)

(3.1.12)

(3.1.13)

(3.1.14)

Similarly, hydrogen peroxide properties could be gotten but with a slight
change as follows:

II)Hydrogen Peroxide properties:

(3.1.15)

(3.1.16)

(3.1.17.a,b)

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The same procedure can be followed to find properties .from eqn.(3.1.4)
gas constant can be calculated

(3.1.18)

(3.1.19)

(3.1.20)

(3.1.21)

(3.1.22)

After applying balance and getting chamber temperature the last
properties can be estimated from equations (3.1.10,11,12,13,14).

3.2Nozzle Design:-
A convergent-divergent nozzle is used in thrusters to accelerate
the exhaust gas thus producing the required thrust force. The nozzle is
designed to produce certain thrust force using certain propellant at a
known ambient condtions. Assuming the convergent and divergent nozzle
angles, the nozzle length can be computed using the estimated throat and
exit diameters. A theortical nozzle model based on a quasi-one-
dimensional and steady flow is used to compute the nozzle parameters
and geometry. The flow in the nozzle is assumed to be isentropic. The
thrust chamber pressure (Pc) is chosen to have a certain value. The
thruster is designed to produce certain thrust force F and the propellant
type is chosen.

As the chemical properities of propellant and the products              of
dissociations are calculted in section (3.1). After computing          the
dissociation temperature (Tc), and the exhaust gas properties,         the
characteristic velocity of the exhaust gases is calculated using       the
following equation:

 1
R * Tc       1     2*(  1)
C*                  *     
          2                                      (3.2.1)

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The conditions at the nozzle throat can be calculated from the following
equations:
.
2
Tth  Tc (
(3.2.2)
)
 1

 2   1
Pth  Pc * 
  1

     
(3.2.3)

th  Pth / R * Tth                                                (3.2.4)

uth   R Tth                                                        (3.2.5)

Assuming the nozzle is expanding ideally to the back pressure (i.e
pe=pb). Hence the exit conditions are estimated as follows:

 1
                                                       (3.2.6)
P    
*  c   1
2
Me 
  1  Pe 
         

           


Tc
Te 
  1       2
 1     * Me                                                  (3.2.7)
    2        
Pe
e                                                                  (3.2.8)
R * Te

ue  M e *  * R *Te                                                 (3.2.9)

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Then, the nozzle expansion ratio and thrust coefficient can be obtained
from the following equations:

 * 1                                
 1
Ae                                           1
                  1
Me     *        2
 1          2    *Me
2    2*  1 

Ath
(3.2.10)

              *  2 1 * 1                  * Pe  Pa
 1                           1

CF   *                                                   
2     1                     Pe
 1                   

Pc          

Pc

(3.2.11)
By calculating thrust coefficient, nozzle dimensions can be estimated as
follows:

Ath  CFF Pc
*                                                              (3.2.12)

Dth  2 *
Ath
                                                  (3.2.13)

From equation (3.3.10), exit area is obtained as follows:

Ae  Ath *                                                           (3.2.10a)

De  2 *              Aex
                                             (3.2.14)
Computing throat and exit diameters and assuming divergent angle
(Theta2), the length of divergent part of nozzle is calculated as follows:-

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Dex  Dth
Ldiv nozzle 
2 * tanTheta 2                                 (3.2.15)
Then the nozzle mass flow rate could be estimated from the following
equation:

m
           P * Ath
c
C                                                    (3.2.16)
After getting catalytic bed area which is nozzle inlet area from equations
(3.3.3) , (3.3.11), the nozzle inlet diameter is estimated as in equations
(3.3.4), (3.3.12), The length of convergent part of nozzle is calculated as
follows:
Dc  Dth
Lconv nozzle      
2 * tanTheta1
(3.2.17)
The total length of nozzle equals:

(3.2.18)

3.3 Catalytic Bed Calculations:-
No one can deny that most of the catalytic bed calculations and
relations are empirical as each fuel or propellant has its special catalysts
and its relations. So we can use two methods for geometry calculations
one for hydrogen peroxide and another for nitrous oxide.

I)Hydrogen Peroxide:
This catalytic bed consist of screens which is packaged in
casing of rocket and supported as shown in picture

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Fig. 3.1 Catalytic bed screens.

This for silver catalyst activity=75 kg h2o2/ liter silver,minte.      (3.3.1)

It is required to calculate inlet nozzle area to get diameter of catalytic

(3.3.3)
Note that mass flow rate is gotten from equation (3.2.16)

Now catalytic bed diameter is gotten

(3.3.4)
From activity eqution(3.3.1.1) and flow rate equation(3.2.16) volume of
catalytic bed is calculated as follows:

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(3.3.5)

Then bed length

(3.3.6)

To have no of possible screens

(3.3.7)

(3.3.8)

To get pressure difference or pressure drop across catalytic bed

(3.3.9)

II)Nitrous Oxide:
From previous chapters we can say that catalyst bed efficiency
and ability to decomposition depend on length.
So catalyst bed length here is variable to obtain the most efficient
catalytic bed .so A design parameter to get suitable initial length which is
called characteristic length(L*)
For nitrous oxide from previous experience it can be let that
So bed area from

(3.3.11)
Where Ac is A_inlet nozzle
And diameter

(3.3.12)

As mentioned previously the nitrous oxide catalysts are similar to
hydrazine catalysts. So from others experiments it can be estimated that
the characteristic length is (L*)=10.75 m. As the length is changed

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experimentally to ensure the required length, the calculated length is only
initial length but the bed volume is obtained first using:

(3.3.13)
Then initial length is obtained as follows:

(3.3.14)

3.4Tank Design and Calculations:-
To design and manufacture pressure or propellant tank u
should determine its geometry and make selection to a suitable and
available materials .so we follow this simple method of design and
calculation by using simple equation of state.

I)Hydrogen peroxide tank
As mentioned in pervious chapter we should use pressurized tank for this
fuel .then our tank has two things inside first liquid hydrogen peroxide
down and inert gas which is nitrogen above liquid fuel.
For initial state fuel still in tank

(3.4.1)
From equation of state after finish of fueling
(3.4.2)

For spherical tank
(3.4.3)
To determine thickness of tank wall we should determine diameter and
make simple stress analysis for material selected.
(3.4.4)
For material density and
(3.4.5)

From above equation we can get thickness(t)

(3.4.6)

Then outer diameter of tank

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(3.4.7)

Then volume material of tank

(      -        )                                    (3.4.8)

Now we can get mass of material tank

(3.4.9)

So the total tank mass

(3.4.10)

If we assume that residual mass of propellant equal zero as no fuel
accumulation occur then

System with nitrogen pressurization

As we have two cases initial and final case .for final case the tank volume
is the nitrogen volume when fuel is finished so we can say

(3.4.11)

Where tank volume is constant in two cases then

(3.4.12)

Apply equation of state for fuel and nitrogen in two cases

(3.4.13.a,b,c)

From above five equations we can get five unknowns
Then we design for minimum mass of nitrogen by plotting it with
pressure. by plotting curve for pressure with tank mass we can choose
design point which have minimum mass with suitable tank pressure
which is design point .

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-real equation model ( van der waal eqn. of state)

Van der waal model usually used for high pressures because of the effect
of compressibility as the system of simple equation of state will have
some errors. so this model can be used here .
The eqn. of state

(3.4.14)
Where

,                                                            (3.4.15)

Where Tc,Pc are data of critical point of fuel
For van der waal model

C=0 ,d=0                      ,                                          (3.4.16)

From eqn.(3.4.16) a ,b can be calculated in eqn.(3.4.15)

After applying eqn. (3.4.16) in eqn.(3.4.14) become

(3.4.17)

Eqn.(3.4.17) non linear eqn. so that it need to be simplified .after
algebraic simplification the new eqn. become

(3.4.18)

Where A,B,C,D, constants depend on fuel and system properties
Eqn(3.4.18) is quadratic eqn.in specific volume

II)-Nitrous oxide tank

The big advantage of nitrous oxide that it has natural self
pressurization system .no need to built pressurization system so we
design for it only. Similarly by using the same procedure of hydrogen

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peroxide we can find mass of tank hence minimum mass for point of
design. The only difference for nitrogen system that we have cylindrical
tank then volume will change to First we have

(3.4.19)

(3.4.20)

Solve above equation for inner diameter (Din).Then apply the same
procedure to get curves and choosing design point

3.5 Regulator Selection:-
The two most important parameters to consider during regulator selection
and operation are droop and supply pressure effect. Droop is the
difference in delivery pressure between zero flow conditions and the
regulator’s maximum flow capacity. Supply pressure effect is the
variation in delivery pressure as supply pressure decreases while the
cylinder empties. Single-stage and two-stage regulators have different
droop characteristics and respond differently to changing supply pressure.
The single-stage regulator shows little droop with varying flow rates but a
relatively large supply pressure effect. Conversely, the two-stage
regulator shows a steeper slope in droop but only small supply pressure
effects.

The effect of these differences on performance can be illustrated with
some examples. For instance, when a centralized gas delivery system is
supplying a number of different chromatographs, flow rates are apt to be
fairly constant. Supply pressure variations, however, may be abrupt,
especially when automatic changeover manifolds are used. In this
scenario, a two-stage regulator with a narrow accuracy envelope (supply
pressure effect) and a relatively steep droop should be used to avoid a
baseline shift on the chromatographs. On the other hand, if gas is being
used for a short-duration instrument calibration, a singlestage regulator
with a wide accuracy envelope (supply pressure effect) but a
comparatively flat droop should be chosen. This will eliminate the need
to allow the gas to flow at a constant rate before the calibration can be
done.

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Delivery Pressure Range - Determining an appropriate delivery pressure
range for a regulator can be confusing but can be accomplished by
following these steps:

1. Determine the gas pressure needed.
2. Determine the maximum pressure the system might require
(this pressure and the gas pressure are often the same)
3. Select a delivery pressure range so that the required
pressures are in the 25% to 90% range of the regulator’s
delivery pressure (a regulator’s performance is at its best
within this range).

Relieving/Non-Relieving - A relieving regulator has a hole in the center
of the diaphragm. As long as the diaphragm is in contact with the poppet,
the regulator does not relieve. When the pressure under the diaphragm
increases as a result of back pressure from downstream, the diaphragm
will rise, allowing the pressure to relieve through the opening in the
diaphragm. While the internal gas is relieving through this opening, the
surrounding atmosphere (i.e. air) is diffusing into the gas stream. Oxygen
(a component of air) is a harmful contaminant, especially when a gas
stream is intended to be oxygen-free. It is well documented that oxygen
affects gas chromatographic results. Relieving regulators should not be
used in specialty gas applications.

Accuracy Envelopes for Single and Two-Stage Regulators at Two Supply
Pressures - The envelopes are bounded by inlet pressure curves of 2000
psig (138 bar) and 500 psig (35 bar). Each regulator was set to the shown
delivery pressure with 2000 psig (138 bar) inlet pressure and zero flow.
Once set, this delivery pressure was not manually changed during the
evaluation. The above curves generated are the result of increasing flow
through the regulator to its capacity, decreasing the flow rate through the
regulator to zero.

Linked Poppet/Tied Diaphragm - The poppet and diaphragm are
mechanically linked. An increase in pressure in the cavity below the
diaphragm will cause the diaphragm to move upward, pulling the poppet
to improve its seal against the seat. A tied diaphragm regulator is
effective in corrosive gas service, especially in the event that corrosive
particles form under the poppet or on the seat. Tied diaphragm or linked
poppet are terms used by manufacturers to describe this regulator feature.

Gauges - Generally single and two-stage regulators are equipped with
two gauges - a cylinder or inlet pressure gauge and a delivery or outlet

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pressure gauge. The cylinder pressure gauge has the higher pressure
range and is located adjacent to the inlet port. The delivery pressure
gauge of the lower pressure range is located adjacent to the outlet port.
The actual pressure gauge range is usually greater than the pressure range
for which the regulator is rated. For example, a regulator that has a
delivery pressure range of 1-50 psig (0.1-3 bar) will typically be supplied
with a 0-60 psig (0-4 bar) delivery pressure gauge. This ensures that the
rise in delivery pressure as a result of the regulator’s supply pressure
effect will not exceed the gauge pressure range. Not all cylinder
regulators have two gauges. A line regulator is typically provided with a
single gauge that monitors the outlet pressure or reduced pressure. This
gauge is usually situated in the 12 o’clock position. Regulators designed
for liquefied gases may not have a cylinder pressure gauge because the
cylinder pressure varies only with temperature as long as liquid is present
in the cylinder.

Regulator Placement - Specialty gas regulator applications are divided
into two types. The first is when the regulator is fastened to a gas cylinder
using a CGA, DIN or BS fitting. The second application is when a
regulator is located in a gas line, providing a means to further reduce the
line pressure. A line regulator is identified by having the inlet and outlet
opposite of each other and by a single gauge as discussed above.

from    above the selection of pressure regulator estimated by:
1.   Calculation of Cv(flow ceoffient).
2.   Amount of flow rate.
3.   Minimum pressure difference (∆p).
4.   Maximum inlet pressure and flow rate.
5.   Type of flow to select material of regulator.
6.   Chemical compatibility of material to match propellant.

The goal in designing is to have suitable pressure difference in pressure
regulator.So it be considered that pressure drop vary between 10% and
20% of chamber pressure(Pc).
∆P=.17Pc                                                           (3.5.1)
To calculate Cv we need to define volume flow rate(Q)
(3.5.2)
Now we can determine regulator flow capacity(Cv)from

(3.5.3)

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Flow Performance:
The flow properties of a pressure regulator are illustrated by the flow
curve. The vertical axis indicates the delivery pressure at which the
regulator is set and the horizontal axis indicates the gas flow that the
regulator passes. The curves are made by setting the delivery pressure
while there is no gas flow and then slowly
opening the outlet valve downstream while measuring both the flow and
the delivery pressure. Typically, as flow increases, delivery pressure
drops. The portion of the curve to the far left is fairly flat and it is in this
range that the regulator demonstrates a
stable pressure regulation even though the flow is changing. For example,
increasing the flow from point “A’’ to point “B’’ shows only a slight
decrease in pressure. The portion of the curve to the right shows a rapid
drop in pressure with increasing flow rate,
indicating that the regulator valve seat is almost wide open. If flow is
increased from point “B’’ to point “C’’, there is a large drop in pressure
that is typical for all regulators.

Fig 3.2/ regulator flow curve

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3.6 Valve Sizing and Selection:
Sizing flow valves is a science with many rules of thumb that
few people agree on. In this article I'll try to define a more standard
procedure for sizing a valve as well as helping to select the appropriate
turbulent flow
STEP         #1:       Define        the        system
Key Variables: Total pressure drop, design flow, operating flow,
minimum flow, pipe diameter, specific gravity

STEP #2: Define a maximum allowable pressure drop for the valve
The usual rule of thumb is that a valve should be designed to use 10-15%
of the total pressure drop or 10 psi, whichever is greater.

STEP #3: Calculate the valve characteristic

At this point, some people would be tempted to go to the valve charts or
characteristic curves and select a valve. Don't make this mistake, instead,
proceed to Step #4!

STEP           #4:           Preliminary          valve       selection
Don't make the mistake of trying to match a valve with your calculated
Cv value. The Cv value should be used as a guide in the valve selection,
not a hard and fast rule. Some other considerations are:
a. Never use a valve that is less than half the pipe size
b. Avoid using the lower 10% and upper 20% of the valve stroke. The
valve is much easier to control in the 10-80% stroke range.

Before a valve can be selected, you have to decide what type of valve will
be used (See the list of valve types later in this article). For our case,
we'll assume we're using an equal percentage, globe valve (equal
percentage will be explained later). The valve chart for this type of valve
is shown below. This is a typical chart that will be supplied by the
manufacturer (as a matter of fact, it was!)

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STEP #5: Check the Cv and stroke percentage at the minimum flow
If the stroke percentage falls below 10% at our minimum flow, a
smaller valve may have to be used in some cases. Judgements plays role
in many cases.

STEP #6: Check the gain across applicable flow rates

Gain is defined as:

OTHER NOTES:

Another valve characteristic that can be examined is called the choked
flow. The relation uses the FL value found on the valve chart. I
recommend checking the choked flow for vastly different maximum and
minimum flow rates. For example if the difference between the
maximum and minimum flows is above 90% of the maximum flow, you
may want to check the choked flow. Usually, the rule of thumb for
determining the maximum pressure drop across the valve also helps to
avoid choking flow.

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3.7 Injectors design :-

For our system we use the showerhead injector may be designed
as following by using hydrazine monopropellant model of design .
The number of orifices is calculated from:

(3.7.1)

The injector pressure drop is calculated from:

(3.7.2)

Now we can get orifice diameter from:

(3.7.3)

For a good and safe design we should have space between orifices which
is calculated from the equation

(3.8.4)

Therefore ,the diameter of each orifice row is:

(3.8.5)

The number of orifice in each row is calculated from:

(3.8.6)

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3.8 Valve selection procedure for gas
application(N2O):
Cv:flow discharge coefficient

Q:volume flow rat in ft/m3

G: specific gravity

T: gas temperature in F0

P1: maximum inlet pressure in psi

P2:maximum outlet pressure in psi

m.:mass flow rate ibm/s

ρ:gas density in kg/m3

if o.53 p1 < p2

Cv=(Q/1349)*(√(460+T)*G/∆P*P2)…………………(3.8.1)

If o.53 p1 >p2

Cv=(Q/704 P1)*(√(460+T)*G)………………………(3.8.2)

∆P=pressure drop

Givens:

m.=0.015

T=68 F0

ρ air=1.225 kg/m3

ρ n2o=1.775 kg/m

p1=15*14.5 psi

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p2=13*14.5 psi

∆P=2*14.5 psi

Calculations:

From equation of state:
ρ =P1/R T……………………(3.8.3)
also
To calculate Cv we need to define volume flow rate(Q)

Q= m./ ρ……………..(3.8.4)
gas constant calculated from:
R = Ru/Mw……………………………(3.8.5)
=8341/44

then            R=188.96 J/kg—K

G = ρ n2o/ ρ air …………(3.8.6)
=1.775/1.225=1.45

from equ 3 after substituting about T ,R

ρ =27.092 kg/ m3

from equ 4 by substituting about m., ρ

then         Q=(0.015/27.092)*(127132.8002)=70.389

Q=70.389 ft3/hr

From equ 1
Cv=(70.389/1349)* (√ (460+68)*1.45)/(29*13*14.5))

Cv=0.02

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3.9 Valve selection procedure for liquid application
H2O2:
Givens:
m.=0.015
T=68 F0
ρ water=1000 kg/m3
ρ h2o2=1.391*1000 kg/m3
p1=15*14.5 psi
p2=13*14.5 psi
∆P=2*14.5 psi

Calculations
specific gravity calculated from:
G = ρ h2o2/ ρ water

G=1.391/1=1.391
volumetric flow rate calculated from:

Q= (m./ ρ h2o2 )*(1/6.3088*10^(-3)) in gal/min

=(0.015/1.391*1000)*( 1/6.3088*10^(-3))

Q=1.7092956 gal/min

flow discharge coefficient calculated from:

Cv=Q*(√ G/ ∆P)

=1.7092956*(1.391/29)

Cv =0.004

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.
3.10 HEATING SYSTEM
Required
Evaluation the time required to heating the catalytic bed up to 523 Kelvin
The available power(electrical power ) used to this process taken as : Ẅ which related
with temperature in the following relation:

Ẅ=m' cp ∆T ……………..(1)
=W/t……………………(2)
Given:
Ẅ=4000 watt
Dcb: catalytic bed diameter =0.048225 m
Dc:coil of heater diameter =0.008 m

Where;
W:heat transfer from heater coil to the catalytic bed
t: total time used in heating
Also;
W=m cp ∆T……………...(3)
= m cp [Tf-Ti ]…………(4).
m: mass of catalytic bed
cp: specific heat of catalytic bed material
ρ: density of catalytic bed material
V: volume of catalytic bed
m=ρV……………..(5)
1- FOR SHELL 405:
(iron..36 wt,γ_Al2 o3)

ρ av= ( ρ iron*.36+ ρ Al2o3*.64 )/2…………………………(6)
Where;
ρ iron =7.86 g/cm3
ρ Al2o3=3.96 g/cm3
after substituting in Equ…(6) then
ρ av= 2595.6 kg/m3 ……………………………………..(ans 1)
Cp av=( Cp iron*.36+ Cp Al2o3*.64)/2………………………….(7)
Where;
Cp iron=448 J/kg-k
Cp Al2o3=880 J/kg-k
after substituting in Equ…(7) then
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Cp av=362.24 J/kg-k…………………………………….(ans 2)

Now we can calculate the volume from the following formula
V =Л/4(D2)*L…………………………………………..(8)
D=Dcb- Dc………………………………………………(9)
Where;
Dcb: catalytic bed diameter
Dc: coil of heater diameter

Also after substituting in Equ…(5,8,9 ) then

D:net catalytic bed diameter =0.040225 m…………….(ans3)
..                  V=1.051315*10^(-4) m3…………………… ………(ans4)
….
m= 0.27288 kg ………………………………………….(ans5)

Now we can calculate the( W )from the Equ (3) ;

W=22240.8 J/kg—k ……………………………………(ans6)

Finally and from Equ 2

the required time for heating is

where;

t=W/ Ẅ ……………………………………………..(equ 2a)

t=111.204 sec……………………………………………(ans7)

required a battery of life time =5.56*1000
L_t=1.5444 hr………………………..(ans 8)

2-FOR IRON—AL:
(iron..95 wt, AL0.05 wt)

ρ av= ( ρ iron*.95+ ρ Al2o3*.05 )/2…………………………(6--a)

Where;
ρ iron =7.86 g/cm3
ρ Al2o3=3.96 g/cm3
after substituting in Equ…(6--a) then

52
ρ av= 4409kg/m3 ……………………………………..(ans 9)
Cp av=( Cp iron*.95+ Cp Al2o3*.05)/2………………………….(7—a)
Where;
Cp iron=448 J/kg-k
Cp Al2o3=880 J/kg-k
after substituting in Equ…(7--a) then

Cp av=437.8J/kg-k…………………………………….(ans 10)

Now we can calculate the volume from the following formula
V =Л/4(D2)*L…………………………………………..(8)
D=Dcb- Dc………………………………………………(9)
Where;
Dcb: catalytic bed diameter
Dc: coil of heater diameter

Also after substituting in Equ…(5,8,9 ) then

D:net catalytic bed diameter =0.040225 m…………….(ans3)
V=1.051315*10^(-4) m3…………………… ………(ans4)
….
m= 0.4659kg ………………………………………….(ans5)

Now we can calculate the( W )from the Equ (3) ;

W=45659.5 J/kg—k ……………………………………(ans6)

Finally and from Equ 2

the required time for heating is

where;

t=W/ Ẅ ……………………………………………..(equ 2a)

t=11.41sec……………………………………………(ans7)

required a battery of life time =11.41*1000
L_t=3.169444 hr………………………..(ans 8)

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3-FOR IRON—Ag) (IRON 0.95.0.05Ag)
The same steps of calculations so these final results

Where;
ρ iron =7.86 g/cm3
ρ Al2o3=3.96 g/cm3
after substituting in Equ…(6--b) then

ρ av= 6381kg/m3 …

also
Cp iron=448 J/kg-k
Cp Al2o3=880 J/kg-k
after substituting in Equ…(7--b) then

Cp av=270.3/kg-k…………………………………….
Also after substituting in Equ…(5,8,9 ) then

m= 0.670841kg

Now we can calculate the( W )from the Equ (3) ;

W=22240.8 J/kg—k
Finally and from Equ 2

the required time for heating is

where;

t=W/ Ẅ ……………………………………………..(equ 2a)

t=10.19976 sec……………………………………………(ans7)

required a battery of life time =10.19976*1000
L_t=2.833266hr………………………..(ans 8)

4- (FOR IRON—Mg o) (IRON0.95. Mg o0.05)

ρ av= 46285kg/m

Cp av=432 .05 J/kg-k

m= 0.48660 kg

54
W=4730 .105J/kg—k

t=11.824sec…

L_t=3.28444 hr

55

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