Solution Manual For Principles of Nuclear Rocket Propulsion 2nd Edition By William J. Emrich, Jr

SOLUTION MANUAL For Principles of Nuclear Rocket Propulsion 2nd Edition By William J. Emrich, Jr Chapter 2-17 Chapter 2 A mission to Mars has been determined to require a total velocity increment of 14.2 km/sec. A vehicle incorporating a nuclear thermal rocket engine is to be used for the mission. Assume that the propellant is hydrogen ( ) and that its temperature upon leaving the reactor is 2844 K. Also assume that the rocket nozzle has an area ratio of 5 in its converging section and an area ratio of 300 in its diverging section. From this information, determine the engine specific impulse and the vehicle mass fraction.

Solution The first step in the calculation involves determining the subsonic Mach number of the hydrogen propellant in the core exit plenum. To accomplish this task, it is necessary to implicitly solve the Mach–Area relationship expressed by Equation 2.29. Note that this calculation and the ones that follow can be performed easily using the calculator found in Table 2.1.

√6

(

) 7

√6

(

) 7

With the Mach number known from Equation 1, the pressure ratio between the core exit plenum and the nozzle throat may be determined from Equation 2.30.

6

7 (

)

6

7 (

)

The supersonic Mach number of the hydrogen propellant as it leaves the nozzle assembly may also be determined from Equation 2.29 using the area ratio between the nozzle throat and the nozzle exit.

√6

(

)

√6

7

(

) 7

With the Mach number known from Equation 3, the pressure ratio between the nozzle exit and the nozzle throat may also be determined from Equation 2.30.

6

7 (

6

)

7 (

)

From Equations 2 and 4, it is now possible to calculate the pressure ratio between the core exit plenum and the nozzle exit, such that:

(

*(

*

Using the pressure ratio from Equation 5 in conjunction with Equation 2.19, the specific impulse for this nuclear rocket engine may be evaluated to yield:

√

6

√

( *

7

[

]

To determine the vehicle mass fraction, the specific impulse from Equation 6 is used in the rocket equation from Equation 2.8 which has been rearranged to yield:

(

*

Chapter 3 1. An NTR engine system is to be designed around a turbopump assembly which has a rated hydrogen flow rate of 15 kg/sec. It is estimated that because of shielding, cooling requirements, and strength of materials considerations, the weight of the engine system will grow with increasing hydrogen outlet temperatures according to the following equation:

(

)

Assume that a hot bleed cycle will be used for the engine cycle and that the bypass bleed flow can be neglected in your calculations. Also assume that the inlet hydrogen temperature to the reactor is:

For the stated conditions, determine the following parameters such that they yield an engine system which maximizes the engine thrust to weight ratio using only the information given.

- Reactor exit (i.e., chamber) temperature - Reactor power level - Specific impulse assuming a nozzle exit pressure of zero - Engine mass - Thrust level assuming a nozzle exit pressure of zero Note that Solution

The first step in solving this problem involves determining an appropriate relationship for the engine thrust to weight ratio. Since an expression for the engine mass was given in the problem statement, all that is required to determine the engine thrust to weight ratio is an appropriate expression for the engine thrust. This expression for the thrust may be

determined from the engine mass flow rate and the specific impulse. Using Equation 2.20 from the book, we find that:

√

Substituting in numerical values into Equation 1 then yields:

√

√

Using Equation 2 and the given relationship for the engine mass, the engine thrust to weight ratio becomes:

√ [

⁄

]

Taking the derivative of Equation 2 with respect to T4, setting the result equal to zero, and solving for T4 yield a chamber temperature which gives the maximum thrust to weight ratio for the engine.

(

) √

Using T4 in Equation 2 then yields for the maximum engine thrust to weight ratio:

√ [

⁄

]

Figure 1 shows the plotting of the thrust to weight ratio versus T4.

Figure 1

The thrust level may be determined from Equation 2 yielding:

√

√

Upon using the gravitational conversion constant, the engine specific impulse may be determined from Equation 1 such that:

The engine mass may be determined from the equation given in the problem statement yielding:

(

)

(

)

The reactor power level may be determined from the mass flow rate and the enthalpy change in the hydrogen propellant. Using the hydrogen property values from the appendix:

(9)

2. A Mars "hopper" is to be used to "hop" from place to place on the Martian surface using the atmosphere (primarily carbon dioxide) as the propellant in an NTR-based propulsion system using a cold bleed cycle. Assume the "hopper" is on the Martian surface and that carbon dioxide has been pumped into the propellant tanks. Based upon the given state conditions in the adjacent figure, determine: - The reactor inlet thermodynamic conditions - The minimum bleed power required - The mass flow rate through the turbopump turbine - The turbopump pumping power requirement - The mass flow rate through the reactor - The engine specific impulse - The Mach number at nozzle exit - The nozzle exit thermodynamic conditions - The engine thrust level

Given:

Assume pump is isentropic. Do not assume turbine is isentropic. Also, sketch the thermodynamic cycle on the carbon dioxide T–S diagram (see appendix) and be sure to show all work and justify any other assumptions you make.

Solution

The problem as written contains several unknowns (e.g. and statepoint 3) which must be determined in order to calculate the performance of the engine. To determine these unknowns, heat balances at various engine locations must be specified. For the bleed power it is found that:

Also noting that the power produced by the turbopump turbine must equal the power consumed by the turbopump pump it is found that:

Finally the heat balance across reactor may be represented by:

Solving Equations 1, 2, and 3 simultaneously allows us to eliminate and yielding an equation for the bleed power “ ” in terms of the enthalpy at unknown statepoint 3, “ ”:

The enthalpy values at various locations in the engine may now be calculated using the CO2 property calculator from the appendix, and the known statepoint data from the problem statement yield:

Assuming the pump is isentropic (5)

Assuming

Substituting the statepoint data from Equations 5 into Equation 4 yields for the bleed power:

Since the bleed power is to be minimized, the derivative of Equation 6 is taken with respect to and the result set equal to zero. Solving for then yields:

To determine the correct value for

,

is plotted versus

in Figure 2.

Figure 2 From the plot, it can be seen that

yields a minimum value for the bleed flow.

The other value for yields negative values for the turbine flow rate and is therefore discarded. Using in Equation 6 then yields for the minimum bleed power:

Using the CO2 property calculator from the appendix and the enthalpy and pressure at statepoint 3, the temperature at statepoint 3 may be found to be:

The total carbon dioxide mass flow rate may be found by using the value found for Equation 1, such that:

in

The turbopump pump power consumption can be found by using the total carbon dioxide mass flow rate from Equation 10 and the enthalpy change across the turbopump pump yielding:

The carbon dioxide mass flow rate through the turbopump turbine may now be found from Equation 2, such that:

Since the carbon dioxide flow rate through the reactor is simply the difference between the total carbon dioxide flow rate and the carbon dioxide flow rate diverted through the turbopump turbine, from Equations 10 and 12, it is found that:

To determine the engine specific impulse, we first use Equation 2.39 to calculate the Mach number at the core exit based upon the nozzle area ratio in the converging subsonic section of the nozzle, such that:

√6

(

) 7

√6

(

)

7

Using the Mach number determined from Equation 14, the pressure ratio in the converging subsonic section of the nozzle may be calculated using Equation 2.30 yielding:

6

7

6

7

Using the Mach number determined from Equation 14, the temperature ratio in the converging subsonic section of the nozzle may be calculated using Equation 2.31, such that:

(

)

(

)

In the diverging section of the nozzle, Equation 2.29 is again used to calculate the Mach number, but this time based upon the nozzle area ratio in the diverging supersonic section of the nozzle, such that:

√6

(

) 7

√6

(

) 7

Using the Mach number determined from Equation 17, the pressure ratio in the diverging supersonic section of the nozzle may be calculated using Equation 2.30 yielding:

6

7

6

7

Using the Mach number determined from Equation 17, the temperature ratio in the diverging supersonic section of the nozzle may be calculated using Equation 2.31, such that:

(

)

(

)

The pressure ratio across the entire nozzle may now be determined from Equations 15 and 18, such that:

Using the pressure ratio across the entire nozzle determined from Equation 20 and the pressure at the reactor core exit, the pressure at the nozzle exit may be determined to be:

Using the temperature ratios from Equations 16 and 19 and the temperature at the reactor core exit, the temperature at the nozzle exit may be calculated to be:

Using Equation 2.19, the specific impulse may be determined, such that:

√

6

√

( *

7

[

]

Using the specific impulse from Equation 23 and the reactor mass flow rate from Equation 13, the engine thrust level may be calculated to be:

The rocket engine cycle may be illustrated on the T–S diagram presented in Figure 3.

\

Figure 3

3. A nuclear power system is being designed to energize an electric propulsion unit. Assume that an ideal Stirling cycle is to be used in the design. Determine the heat rejection temperature at which heat should be radiated to space in order to minimize the mass required to generate a given amount of electricity (e.g., maximum specific power or ).

Plot the relationship between and and determine the maximum system in at this optimal heat rejection temperature.

Note:

⏟

⏟

where the thermal to electric conversion efficiency of an ideal Stirling cycle is given by:

(

*

and heat is rejected to space according to the Stefan–Boltzmann law, where: where:

area of radiator fins electrical power

Parameter

Symbol

Value

Mechanical to Electrical Efficiency

0.9

Fin Emissivity

0.85

Temperature after Heat Addition

2000

Areal Density of Radiator

1.5

Reactor Thermal Power Density

1

Units

K

Stefan–Boltzmann Constant

Solution

To solve the problem, it is necessary to first determine as a function of the heat rejection temperature and then pick that temperature which maximizes . Beginning the analysis with a determination of the total heat output from the reactor yields:

(

* (

)

Since the waste heat which must be rejected by the radiators is just that heat which is not converted to electricity, using Equation 1, it is found that:

[ (

)

] (

)

The mass of the radiator can be calculated using the areal density of the radiator and the total radiator area, such that:

Incorporating Equation 3 into Equation 2 and solving for the mass of the radiator then yield:

[ (

]

[ (

)

] )

Rearranging Equation 1, the mass of the reactor may be determined to be:

(

)

Using Equations 4 and 5 and the definition of α from the problem statement, it is possible to determine α as a function of the heat rejection temperature alone, such that:

(

[ (

)

] )

[ (

)

] (

)

Substituting numerical values from the problem statement into Equation 6 then yields:

6 (

7 (

)

)

To determine the maximum α, the derivative of Equation 7 is taken with respect to result set to zero, and the resulting equation solved for to yield: (

, the

) (

Substituting the heat rejection temperature for the specific power from Equation 7 then yields

)

from Equation 8 into the relationship

Plotting Equation 7 in Figure 4 illustrates the relationship between

and .

Figure 4

Chapter 4 1. Determine the travel time to Mars if a spacecraft is launched from the Earth with a velocity vector that is in the same direction as the Earth's velocity vector around the Sun and a solar orbit eccentricity of 0.6. Assume that the distance for the Earth to the Sun is km and the distance from Mars to the Sun is km. Also assume that the solar gravitational constant is

.

Solution

The first step in the analysis is to determine the major radius of the vehicle transfer orbit based upon its orbital eccentricity. Using the fact that the spacecraft is launched from the Earth with a velocity vector that is in the same direction as the earth's velocity vector around the Sun (implying that the true anomaly at the Earth " " equals zero) and using Equation 4.27, it is found that:

[

]

[

]

Substituting numerical values into Equation 1, the major radius of the vehicle transfer orbit is found to be: [

]

[

]

When the spacecraft arrives at Mars, its orbital true anomaly may be determined to be:

0

1

0

1

Using the spacecraft true anomaly at Mars from Equation 3, it is now possible to calculate the spacecraft's eccentric anomaly at Mars through the use of Equation 4.66, such that:

0

1

0

1

Knowing the spacecraft's eccentric anomaly at Mars from Equation 3, the time required to travel between the Earth and Mars may now be calculated through the use of Equation 4.70 yielding:

√

[

]

√

[

]

2. Determine the engine specific impulse required for a one-way trip to Mars if the spacecraft described in Problem #1 is launched from a 200-km-high Earth orbit and captures into a 100km-high Mars orbit. Assume that the spacecraft has a vehicle mass fraction of 0.15 and that the radius of the Earth is 6378 km and the radius of Mars is 3393 km. Also assume that the Earth's gravitational constant is

and that the Mars's gravitational constant is

.

Solution

Using the major radius of the vehicle transfer orbit calculated from Problem 1 we first calculate the heliocentric velocity of the spacecraft after leaving the Earth orbit using Equation 4.31 yielding:

√

(

*

√

(

*

Calculating the velocity of the Earth around the Sun using Equation 4.34 gives:

√

√

Since it is known that the spacecraft's velocity vector and the Earth's heliocentric velocity vector are the same from the information given Problem 1 (implying that ), it is possible to calculate the hyperbolic excess velocity of the spacecraft from Equation 4.33 using the results from Equations 1 and 2, such that:

√(

)

√

The spacecraft velocity at engine cutoff may now be determined from Equation 4.36 using the hyperbolic excess velocity of the spacecraft calculated in Equation 3 yielding:

√

(

)

√

Next, the spacecraft earth orbital velocity is calculated by using Equation 4.37, such that:

√

√

The velocity increment required by the propulsion system to insert the spacecraft into the required Mars transfer orbit is the difference between the spacecraft velocity at engine cutoff and the spacecraft Earth orbital velocity, thus:

Upon arriving at Mars, it is necessary to insert the spacecraft the desired parking orbit. To determine the velocity increment needed to perform this maneuver we first must determine the heliocentric velocity of the spacecraft when it arrives at Mars. This velocity may be determined from Equation 4.45 and the results of Equation 1, such that:

√(

√

)

(

*

(

*

The angle at which the spacecraft arrives at Mars relative to the planet's orbital velocity vector may be calculated using Equation 4.47 and the results from Equations 1 and 7 yielding:

0

1

6

7

The orbital velocity of Mars around the Sun may be calculated using Equation 4.34 such that:

√

√

The spacecraft arrival velocity relative to Mars can now be calculated from Equation 4.48 using the results from Equations 7, 8, and 9, yielding:

√(

)

√

The spacecraft velocity at the time the Mars braking maneuver begins may be determined from Equation 4.52 using the results from Equation 10 such that:

√(

√

)

The orbital velocity of the spacecraft at the desired Mars capture height may be evaluated using Equation 4.37 yielding:

√

√

The velocity increment required to be delivered by the propulsion system to capture into its desired Mars parking orbit may be determined from Equations 11 and 12, such that:

The total velocity increment required to carry out the mission is the sum of the velocity increments required to leave the Earth and capture at Mars. Using Equations 6 and 13, the total required mission velocity increment which must be delivered by the propulsion system is thus:

To determine the minimum specific impulse required by the spacecraft engine to carry out the desired mission, the mission Δ from Equation 14 can be used with Equation 2.8 to yield:

Chapter 5 1. A nuclear rocket engine has a cylindrical reactor ( ) composed of fully enriched with hydrogen propellant flow holes (hole volume fraction = 0.3). The engine fires for a total of 100 min during a Mars mission and operates at 4000 MW.

a. What average neutron flux levels would you expect? b. What fractional fuel burnup would be expected during the mission?

Note: Solution The fission power produced by the rocket engine may be determined from:

Rewriting Equation 1 to solve for the neutron flux then yields:

The fission rate in the engine may be determined from the power, such that:

The fuel volume in the engine nuclear reactor may be determined from:

The fuel atom density may be rewritten using more appropriate units yielding:

Using the numerical parameter values from Equations 3, 4, and 5 into Equation 2 then yields for the neutron flux:

The fractional burnup is simply the ratio of the number of atoms fissioned to the total number of uranium atoms in the reactor. The number of atoms fissioned may be determined from the reactor power and the time of operation yielding:

The total number of uranium atoms in the reactor may be determined from the atom density from Equation 5 and the reactor volume from Equation 4, such that:

The rocket engine fuel burnup for the Mars mission is just the total number of fissions from Equation 7 divided by the total number of U atoms from Equation 8 yielding:

2. The threshold for fission in is often taken to be 1.4 MeV since the fission cross section is small below this energy. Using this threshold, what fraction of fission neutrons are capable of producing fission in ?

Solution The neutron energy spectrum resulting from fission can be represented by the as given by Equation 5.6 and repeated below. (√

function

)

Since fission in requires neutron energies greater than 1.4 MeV, that portion of the distribution greater than 1.4 MeV represents that portion of fission neutrons which can initiate fission in as illustrated in Figure 1.

Figure 1 To determine the proportion of the χ(E) distribution represented by the shaded portion in Figure 1, the χ(E) distribution is integrated over the range 1.4 MeV to ∞ MeV, such that: ∫

∫ (

[

(√

)

√ ]

[

√ ]) [√

Equation 2 is illustrated in Figure 2 yielding:

]|

Figure 2

3. The prompt fission neutron spectrum is often represented by the function: √ where: , , and

are constants

- If the function χ(E) is normalized to 1.0, that is, if: ∫ Show that the constant “ ” is given by: - Find a transcendental function for the most probable energy.

- Show that the average energy of a prompt fission neutron is given by: ̅ Solution

To solve for the unknown parameter “ ”, the integral of the prompt fission neutron spectrum presented in the problem statement must first be evaluated. Carrying out the required integration of the prompt fission neutron spectrum yields: ∫

√

∫

⁄

8√

6

√ 5

4√

4√

√ 57

√

(√

√

)9 |

At E = ∞, Equation 1 evaluates to:

⁄

√

8√

6

[

√

√ 5

4√

(√

)]9

⁄

√ 57

4√

[

{√

]

√

√ [ ]}

At E = 0, Equation 1 evaluates to:

⁄

8√

6

4

√ 5

4 √ 57

√

9

Using Equations 2 and 3 to solve for “ ” then yields:

√

√

The most probable energy for the prompt fission neutron spectrum corresponds to the peak in the distribution function presented in the problem statement. This peak energy may be determined by that point where the derivative of the distribution function equals zero, therefore:

(√

0

)

(√

√

)1

Rearranging Equation 1 then yields a transcendental function of the form: √

(√

)

The average energy for the prompt fission neutron spectrum may be represented by energy weighting the distribution function presented in the problem statement such that: ∫

√ [

(

(√

)

√

)

⁄

] √

2 √ [

(

√ √ .

√

)

0 √ √

/13|

√

]

.

√

√ √

√

√

/

√

⁄

Incorporating the relationship for parameter “ ” from Equation 4 into Equation 7 then yields:

√

√

⁄

4. The absorption mean free path for 2200 m/sec (0.025 eV) neutrons in a material having a absorption cross section is 1.0 cm. The corresponding reaction rate is material has an atomic mass of 10 amu and a density of

. The

. Find:

a. The 2200 m/sec neutron flux b. The material's microscopic absorption cross section (

) for 10 eV neutrons in barns

Solution Recalling from Equation 5.11 that absorption reaction rates may be determined from:

Also, recalling from Equation 5.36 that the absorption mean free path for neutrons in a material may be represented, such that:

Substituting Equation 2 into Equation 1 and rearranging terms then yield for the neutron flux:

The atom density of the material in the problem statement may be calculated from Equation 5.9, such that:

Using the macroscopic cross section from Equation 2 and the atom density from Equation 4 along with the definitions for microscopic and macroscopic cross sections, it is found that:

Using the microscopic absorption cross section determined from Equation 2 and recalling from Equation 5.7 the energy-dependent behavior of cross sections in the 1/V region, it may be determined that:

√

Chapter 6

√

1. Suppose that during an inelastic scattering collision the target nucleus absorbs an amount of energy . Show that:

becomes

where

√ Solution In the derivations which follow, the Center of Mass coordinate system will be used where the total linear momentum of the particles is zero. In this coordinate system, the center of mass of the system of particles is stationary as illustrated in Figure 1.

Figure 1

where :

neutron mass mass of nucleus velocity

The derivations begin by stating the equations for the conservation of momentum of the particles such that:

where: The equation for the conservation of energy for particles having inelastic collisions is given by:

Using Equations 1, 2, and 3 to eliminate

and

yields:

Rearranging Equation 4 yields an expression of the form:

From the figure above illustrating the two coordinate system descriptions, note that scattering in the Center of Mass system can be related to scattering in the Laboratory system such that:

Incorporating Equation 1 in Equation 6 yields:

The vector relations between the neutron and the target nucleus after they collide with one another can be illustrated in Figure 2. From these vector relations all necessary scattering trigonometric relationships may be determined:

Figure 2

The effort at this point will now be to derive a correlation which relates the scattering angle between the neutron and the nucleus and the neutron kinetic energy before and after the collision. Using the Law of Cosines:

Using the definition for kinetic energy yields:

Using Equation 11 in Equation 5 yields an expression of the form:

(

*

τ =√

where:

Combining the results from Equations 1, 7, 10, and 12 yields:

.

/

(

*(

)

Incorporating the definition for kinetic energy into the results from Equation 13 yields the required expression for scattering interactions with inelastic collisions such that:

2. Show that the expression for the average loss in the logarithm of the energy during a collision:

may be approximated by: (

Hint: Let Solution

and

and then do a series expansion

*

Making the substitutions suggested in the hint, the equation for the average loss in the logarithm of the energy during a collision becomes:

(

)

It is desired to expand about (e.g., large A); however, since is not analytic, a traditional Taylor series expansion cannot be performed. Nevertheless, it is possible to perform a Laurent expansion wherein a new variable is defined such that:

Rearranging Equation 2 then yields: 0

1

For the zeroth order term, it is found from Equations 1 and 2 that:

(

)

For the first-order term, it is found from Equations 1 and 2 that: [ [ ( As

(

)

]

] )

[

(

)

]

approaches zero Equation 5 yields a value of: [

( [

) (

] )

]

Incorporating the results of Equations 4 and 6 into Equation 3 and cutting off the series after the first-order term then yield: (

*

Finally recalling the substitutionary hints from the problem statement and using them in Equation 7 then yield the desired result such that: (

*

Chapter 7 1. Calculate the macroscopic transport cross section (

) and the diffusion coefficient ( ) for

beryllium, given that the nuclide has an atomic weight of 9, a density of 1.85 microscopic absorption cross section ( section ( ) of 7 b.

,a

) of 0.0095 b, and a microscopic scattering cross

Solution The transport cross section is defined in Equation 7.32, such that:

To calculate a value for the transport cross section, it is first necessary to convert the data given in the problem statement into more appropriate units. Starting with the atom density for beryllium, it is found that:

Next, calculating the values for the macroscopic cross sections from the microscopic cross sections using the atom density from Equation 2 yields:

(3)

The cosine of the average scattering angle “

” is determined from Equation 6.15 wherein:

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

Incorporating Equations 3 and 4 into Equation 1 then yields for the transport cross section:

The diffusion coefficient is also defined in Equation 7.32 and may then be calculated from the transport cross section yielding:

Chapter 8 1. For a bare cylinder of a given constant volume " " containing fissionable material, determine the height to diameter ratio ( ) which will yield a maximum value for . Neglect extrapolation distances in the calculations.

Solution The one group, one region criticality equation may be written as:

From Table 8.2, the buckling in cylindrical geometry may be written as: ( )

(

*

0

( *

1

[

]

where: By using the H/D ratio in the equation for the volume of a cylinder, it is possible to eliminate the height variable from the equation such that:

( *

(

*

Substituting Equation 3 into Equation 2 yields an expression for the buckling which contains only the H/D ratio and the core volume yielding:

( )

(

*

0

1

( *

(

* 0( *

1

Substituting the buckling from Equation 0 into the criticality relationship from Equation 1 then yields for keff:

(

) [( )

]

To determine the H/D ratio that yields the maximum keff, the derivative of Equation 5 is taken with respect to the H/D ratio and the result set equal to zero such that: ⁄

⁄

(

)

⁄

0

(

)

1 √

2. A nuclear rocket engine is configured as shown in the figure below. Cold hydrogen enters the engine through the reflector region, is heated to high temperatures in the core region, and finally exits the core where it is expelled as exhaust. The reactor power level is 100 MW. Assumptions: - The sides of the core and reflector are bare.

- The length of the core and the diameter of the core are equal.

- The length of the reflector can be considered effectively infinite.

- Include extrapolation distances on both the core and reflector when calculating their lengths.

- Assume that the radial buckling is zero.

Parameter

Reflector 0.088 0.0033

Core 1.45 0.0044

-----

2.5 0.0032

- Derive the one group critical equation for this reactor system.

- Plot

as " " goes from 0 cm to 500 cm.

- At what length " " is the reactor exactly critical? Indicate on the above plot.

- Derive the value for

(e.g., as

) for the reactor? Indicate on the above plot.

- At what length " " is the reflector considered infinite.

- Plot the neutron flux in the reactor as a function of axial position.

- Plot the power density (in

) in the reactor as a function of axial position.

- What is the axial power peaking factor (

)?

Solution The first step in solving this problem is to write down the governing neutron diffusion equations for the core and reflector regions of the reactor. Assuming that no fissions occur in the reflector, the governing neutron diffusion equation in the reflector is found to be: .

/

In cylindrical geometry, Equation 1 becomes:

(

⏟ (

*

)

Since the core is bare in the radial direction, the radial buckling in Equation 2 may be determined from Table 8.2 yielding: 0( ⏟

1

*

By solving Equation 3, it is possible to obtain a relationship of the form: [

]

[

]

Assuming that the reflector has boundary conditions wherein the flux goes to zero at the extrapolation distance (e.g., at ), Equation 4 for the neutron flux becomes: [

]

Using Equation 5 the neutron current in the reflector may be determined to be: [

]

For the core region of the reactor where fissions are occurring, the governing neutron diffusion equation are found to be:

Again, since the core is bare in the radial direction, the radial buckling in Equation 7 may be determined from Table 8.2 yielding:

⏟ (

)

(

*

⏟ [

]

By solving Equation 8, it is possible to obtain a relationship of the form: [

]

[

]

Assuming that the core has boundary conditions wherein the flux goes to zero at the extrapolation distance (e.g., at ), Equation 9 for the neutron flux becomes: [

]

Using Equation 10 the neutron current in the core may be determined to be: [

]

To solve for the remaining undetermined coefficients, use will be made of the fact that the neutron flux and neutron current must be continuous at the core/reflector interface.

Therefore, equating the expression for the neutron flux in the reflector from Equation 5 and the neutron flux in the core from Equation 10 at the core/reflector interface it is found that: [

]

[

]

Likewise, equating the expression for the neutron current in the reflector from Equation 6 and the neutron current in the core from Equation 11 at the core/reflector interface yields: [

]

[

]

By dividing Equation 12 into 11 and incorporating the definition for extrapolation length, it is possible to eliminate all the unknown coefficients so as to yield after some rearrangement an equation of the form: [

]

[

]

Note that Equation 14 is the criticality equation for the reactor since it uniquely relates the reactor material properties to the reactor geometry.

Recalling that the reflector may be considered infinite when the relationship used by the hyperbolic tangent is equal to 4, it is found by using the definition for α from Equation 3 that:

√4

5

Inserting numerical values into the expressions for a, α, and β yields:

√4

√( 5

)

√4

√ 4

5

√(

*

√ (

*

5

Inserting the expressions for the buckling and reflector length from Equations 16 into the criticality equation as expressed by Equation 14 yields a criticality equation which is a function of only the core length “ ” and “ ”. Plotting this criticality equation assuming yields the graph illustrated in Figure 1.

Figure 1

From Figure 1, note that an infinite number of core lengths will satisfy the criticality equation, but only the smallest core length (e.g., ) produces neutron flux equations which yield non-negative neutron flux values at all core locations,

.

If some sort of numerical routine is to be used to calculate the core length, it is important to give the routine a good starting estimate to ensure that it converges to the correct core length.

A good estimate for this purpose would be a value for “ ” just to the right of the first vertical ( ). A good

asymptote in the criticality equation. This first asymptote occurs at starting estimate is, therefore:

(

)

.

/

(

[ (

)√ .

)

] /

The for the reactor can be calculated by recalling that all geometric effects are neglected (e.g., buckling is zero). As such, is simply the ratio of the core neutron production rate divided by the core neutron absorption rate yielding:

Plotting the criticality relationship from Equation 14 using the numerical values from Equations 16 yields the graph shown in Figure 2. Note that, as expected, asymptotically approaches as the core length and core width increase.

Figure 2

With the core length “ ” determined, all other reactor characteristics may be readily evaluated. The reflector length and the reactor bucklings may be calculated from Equation 16 yielding:

√(

√(

√(

)

*

√(

)

*

√ (

* √ (

*

The neutron flux in the core from Equation 10 may be determined by using Equation 12 to eliminate the arbitrary constant “ ” in the expression such that: [ [

] ] [ [

] ]

[

]

The arbitrary constant “ ” is determined from the core power level. By dividing the total core power by the volume of the core the average core power density may be determined, yielding:

The average core flux may now be determined by using the core power density calculated in Equation 21 and by recalling that:

The average core flux may also be determined by integrating the pointwise neutron flux over the core volume to yield: ∫

[ [

]∫ ]

[ [

] { ]

[

[

]

]

[

]}

Solving Equation 23 for the arbitrary constant “ ” and substituting in numerical values for the parameters from Equations 19 and 22 then give:

[ [

]{

[

] [

]

]}

Using the expressions for the neutron flux in the reflector and core from Equations 5 and 20, respectively, and the numerical values calculated earlier yields: [

] (25)

[

]

The flux expressions from Equation 25 in the core and reflector regions of the reactor are plotted in Figure 3.

Figure 3 To determine the power density as a function of position in the core, the neutron flux in the core from Equation 25 may be used such that:

[

]

[

]

The reactor power distribution from Equation 26 is plotted in Figure 4.

Figure 4 The peak reactor power may be determined from Equation 26 wherein: * + The axial peaking factor for the reactor may now be determined from Equations 21 and 27 yielding:

3. A thin-walled spherical container (the effects of which may be neglected in your calculations) contains a gaseous uranium compound being examined for possible use in a gas core nuclear rocket engine. Using a one energy group model and assuming the gaseous uranium compound behaves as an ideal gas: - Derive an expression relating the gas pressure, gas temperature, and container radius to .

- Derive an expression for the one group flux as a function of " ".

Hint: Note that

(

) and use the solution:

Solution The first step in solving this problem is to ascertain the appropriate criticality equation to use.

This relationship may be determined from Equation 8.21 for a one group, one region reactor where the is found to be:

Using the expression for the geometric buckling for a sphere from Table 8.2 in the criticality relationship from Equation 1 the criticality equation becomes:

( where:

)

(

)

Rs = radius of the fissioning sphere

Expressing the macroscopic cross sections in terms of microscopic cross sections times the atom density yields:

Substituting in the microscopic cross section expressions from Equations 3 into the current expression for the reactor criticality from Equation 2 then yields an equation of the form:

(

,

From the ideal gas law, it is found that:

Substituting the density derived from the ideal gas law from Equation 5 into the current expression for the reactor criticality from Equation 4 then yields for .

(

)

To solve for the neutron flux in the sphere, the one region, one energy group neutron diffusion equation is used such that:

⏟

In one-dimensional spherical coordinates, Equation 7 becomes: (

*

Taking the derivative of the neutron flux equation described by the hint in the problem statement yields an equation of the form:

Incorporating the derivative of the neutron flux from Equation 9 into the expression for spherical one-dimensional neutron diffusion from Equation 8 then yields: [

(

*]

[

]

Solving the differential equation for the neutron flux from Equation 10 yields an equation of the form:

Since the neutron flux must be finite at the center of the sphere, the arbitrary constant “ ” must be equal to zero, therefore:

To be critical, the material buckling must equal the geometric buckling, implying that:

Substituting the buckling relationship from Equation 13 into the expression for the neutron flux from Equation 12 then yields the final expression for the neutron flux in the gaseous spherical reactor such that: (

)

(

)

4. A fissionable gas, assumed to be ideal, is contained in a cylinder that is pressurized by a piston which exerts a constant 200 kPa pressure on the gas. The absorption cross section of the gas varies as while the fission cross section and the diffusion coefficient remain constant. At STP (300 K and 101 kPa), the height of the gas in the cylinder is 43 cm. At what height and at what temperature is maximum criticality (e.g., ) attained? What is the maximum ? What is the height and temperature of the gas in the cylinder when it is exactly critical?

Parameter

Value 2.5 600 b 300 b 0.9 cm

13,811

Note: From Kinetic Theory of Gases ...

Neglect extrapolation lengths in calculations Solution The unknowns in this problem are the density of the fissioning gas, the core neutron buckling, and the functional relationship for the absorption cross section. To solve the problem, it will be necessary to first get the variables in terms of the temperature of the gas and then plot the as a function of the temperature to determine the maximum value. Beginning with the criticality equation for a bare reactor, the may be written as:

Since the absorption cross section is a

√

√

absorber, it may be determined that:

√

√

Substituting the expression for the absorption cross section from Equation 2 into the criticality relationship from Equation 1 then yields:

√

The density of the fissioning gas may be determined as a function of temperature from the ideal gas law by noting that:

Substituting the expression for the gas density from Equation 4 into the current criticality relationship from Equation 3 then yields:

√

√

Turning now to the buckling term. Note that since the geometry of the reactor is that of a bare cylinder, the buckling from Table 8.2 may be used such that: ( )

(

*

( )

(

⁄

*

( )

To determine the height “ ” as a function of temperature, the ideal gas law may again be used yielding: (

)

Since the height is known at STP and also that pressure in the system always remains constant, the relationship of the height to the temperature may be determined to be:

Substituting the expression for the height of the fissioning gas column as a function of temperature from Equation 8 into the buckling relationship from Equation 6 then yields: (

)

Finally, substituting the expression for the buckling as a function of temperature from Equation 9 into the current criticality relationship from Equation 5 then yields:

[(

√

)

]

√

Equation 10 now yields as a function of temperature alone. By taking the derivative with respect to the temperature and setting the result equal to zero, it is possible to solve the resulting equation for the maximum temperature such that: * [

(

) (

√

⁄

+

) ]

Substituting the temperature yielding a maximum from Equation 11 into the expression for the height of the fissioning gas column from Equation 8 then yields:

To determine the maximum , the temperature from Equation 11 is substituted into the criticality expression from Equation 10 to yield a maximum of:

[(

)

]

√

A plot of from Equation 10 as a function of temperature and gas column height is illustrated in Figure 1.

Figure 1 Note from the plot that fissioning gas column can be exactly critical at two temperatures (534 K and 3828 K) corresponding to two different heights of the fissioning gas column (e.g., 39 cm and 277 cm, respectively).

5. A two-region symmetrical slab reactor is constructed as shown in the figure below. Derive expressions for the flux in each region of the reactor and . Sketch a plot of the expected flux shape. Use a one group model. Clearly state all assumptions. Assume the reactor is infinite in the " " and " " directions.

Solution The first step in solving this problem is to write down the governing neutron diffusion equations for the core and void regions of the reactor. For the void region, the governing one-dimensional neutron diffusion equation is found to be:

.

/ ⏟ (

)

Integrating Equation 1 then yields:

At the reactor center line symmetry considerations dictate that the derivative of the neutron flux should be equal to zero. For this to be true, the constant of integration " " from Equation 2 must be equal to zero. A zero derivative of the neutron flux also implies that the neutron current is equal to zero. | Integrating Equation 2 for a second time yields:

where:

= constant neutron flux value in the void region

For the fuel region, the governing one-dimensional neutron diffusion equation is found to be:

(

+ ⏟ (

where:

)

materials buckling

Solving the differential equation for the fuel region from Equation 5 yields: [

]

[

]

where: At the extrapolation boundary of the fuel region, the neutron flux goes to zero which requires that the arbitrary constant “ ” also be equal to zero.

Using the value for the arbitrary constant from Equation 7 in the expression for the neutron flux in the fuel from Equation 6 yields a fuel neutron flux equation of the form: [

]

Using the derivative of the neutron flux in the fuel from Equation 8, the equation for the neutron current in the fuel then becomes: [

]

At the interface between the fuel and void regions, the neutron currents must equal one another, therefore using Equations 3 and 9, it is found that:

For the equality condition expressed in Equation 10 to hold, it is necessary that the geometric buckling be of the form:

At the interface between the fuel and void regions, the neutron fluxes must also equal one another. Using the geometric buckling from Equation 11 with the expressions for the neutron flux from Equations 4 and 8, it is possible to determine a value for undetermined coefficient “ ” such that: ( ) Using the value determined for “ ” from Equation 12 in the expression for the neutron flux in the fuel from Equation 8, it is found that: [

]

The criticality relationship for the reactor may be determined by setting the materials buckling equal to the geometric buckling. Using Equations 5 and 11 the criticality equation for the reactor may thus be determined such that:

(

) (

)

Sketching the neutron flux distribution implied by Equations 4 and 13 then yields the graph presented in Figure 1.

Figure 1 6. A nuclear reactor is being designed for a nuclear thermal propulsion system. Determine the core length and reflector length combination which will yield the minimum nuclear reactor mass. Assume that the reactor criticality equation is given by Equation 8.41 and that the reactor dimensions (other than along the length) are effectively infinite.

Parameter ( ) ( ) ( )

Core Reflector 0.00499 0 0.0042 0.0092 1.23 0.78 15 8

Solution Using Equation 8.41 as specified in the problem statement, the nuclear criticality equation may be represented by:

[

]

where:

one half the length of the core length of the reflector

Noting that the mass of the reactor may be given by:

It is possible to eliminate the core length “ ” by incorporating Equation 2 into Equation 1 and rearranging terms to yield: {

} ]

[

The core buckling used in Equation 3 may be determined from the buckling definition presented in Equation 8.30 such that: ⁄ √⏟

√

√

Likewise, the reflector buckling used in Equation 3 may be determined from the buckling definition presented in Equation 8.35 yielding:

√

√⏟

√

Solving for the areal mass in Equation 3 and incorporating the numerical data from the problem statement along with the bucklings from Equations 4 and 5 yield: {

[

} ] (6)

{

[

} ]

{

[

} ]

To find the reflector length which yields the minimum areal mass for the reactor, the derivative of Equation 6 is first taken with respect to “ ” and the result set equal to zero. The resulting equation is then numerically solved for “ ” to yield: {

[

[

]

}] (7)

[

]

[

]

To find the minimum areal mass for the reactor, the reflector length determined from Equation 7 is incorporated into Equation 6 such that: {

[

]

}

To find the total core length which yields the minimum mass for the reactor, the reflector length determined from Equation 7 and the areal mass determined from Equation 8 are incorporated into Equation 2 to yield:

Chapter 9 1. In a NERVA-type nuclear rocket engine, the fuel element propellant channel walls often must be clad with some type of material to protect the underlying uranium fuel material from being corrosively attacked by the hot hydrogen propellant. It is, therefore, important that this protective cladding maintain its structural integrity during operation. Should there be a mismatch in the thermal expansion coefficients between cladding and the underlying fuel material, the differential thermal expansion which would occur during operation could cause the cladding to fail, thus exposing the fuel material to the hot hydrogen propellant, potentially resulting in the failure of the fuel element. A likely place for this cladding failure to occur is where the channel wall surface temperature gradient is maximum. Given below are typical operating conditions for a NERVA-type fuel element.

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

Parameter

Value 140 cm 0.100 cm 0.150 cm 14.5 0.0035 0.00021 0.5 300 K 3.5 0.021

Assume the axial power profile varies according to the following equation (e.g., from Equation 9.32): [ ( (

*]

)

-

Derive an expression describing the location of the maximum channel wall temperature gradient

-

Plot the axial power distribution and the axial channel wall temperature distribution

-

Locate on the above plot, the point where the maximum channel wall temperature gradient occurs

-

Comment

Solution From Equation 9.36, the propellant channel wall temperature may be expressed as:

Formatted: Indent: Left: 0.25", Bulleted + Level: 1 + Aligned at: 0.5" + Indent at: 0.75"

8

* ( (

)+

9

) * ( (

)+ )

Likewise, from Equation 9.37, the propellant channel wall temperature gradient may be expressed as: ,

* (

)+ (

* (

)+-

)

Taking the derivative with respect to axial position of Equation 2 and setting the result equal to zero yield an expression which locates the position where the fuel channel surface temperature gradient is the highest: ,

* (

)+ (

* (

)+-

)

Solving Equation 3 for the location where the surface temperature gradient is the highest then yields: .

/

All information required to solve Equation 4 for the location where the surface temperature gradient is the highest is available directly from the problem statement except for the heat transfer coefficient. The heat transfer coefficient may be determined by using the Dittus– Boelter correlation which is a function of the Reynolds number and the Prandtl number where the Reynolds number is:

and the Prandtl number is:

The numerical value for the heat transfer coefficient can now be evaluated such that:

Incorporating all the numerical data into Equation 4 then yields for the location where the channel surface temperature gradient is the highest:

(

,

If is incorporated into Equation 1 along with the other numerical data from the problem statement, the channel wall surface temperature is found to be only a little over 1600 K. Incorporating all the numerical data into the power distribution given in the problem statement and the channel wall temperature distribution from Equation 1 and plotting the results in Figure 1 then yield:

Figure 1 It is interesting to note that the likely failure point for the cladding is not near the exit of the fuel channel where the wall temperature is the highest, but rather at a location between the middle of the fuel element and the fuel element propellant inlet. At this location, Equation 1 indicates that the channel wall surface temperature is only a little over 1600 K. Another reason why the cladding might fail at this location besides the high channel wall surface temperature gradient includes the fact that at 1600 K, the cladding is still cold enough that it remains somewhat brittle and thus prone to cracking when exposed to high thermal stresses.

Had the temperature been somewhat higher, the cladding would have been a bit more ductile and thus better able to resist cracking. Cladding failures such as described above were

observed during the NERVA program and were identified in the literature as the “midband corrosion problem.”1 2. Derive a core power distribution " " which will yield a constant fuel temperature along the length of the core. This distribution is optimal with regard to minimizing the size of a nuclear rocket core since all of the fuel would be able to operate at its maximum allowable temperature. In the derivation, assume that the fuel temperature is constant everywhere (e.g., the fuel temperature does not vary axially or radially) and neglect any calculations related to radial power variations. Do you think it is possible to practically achieve this power distribution within the core? Briefly discuss.

(

Assume: ) where:

a general conductance term = the heating rate per unit length

Solution Performing a heat balance with the propellant flow yields an equation of the form:

1

Lyon, L. L., "Performance of (U,Zr)C-Graphite (Composite) and of (U,Zr)C (Carbide) Fuel Elements in the Nuclear Furnace 1 Test Reactor", Los Alamos Informal Report, LA-5398-MS, (1973).

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

Since the fuel temperature is a constant, the propellant temperature can be taken to be a function of the fuel heating rate alone such that:

Taking the derivative of the propellant temperature from Equation 2 then yields:

Substituting Equation 3 into Equation 1 then yields:

Integrating both sides of Equation 4 yields an equation of the form:

∫

∫

( *

The shape of the power distribution described by Equation 5 is illustrated in Figure 1.

Figure 1

Equation 5 shows that an exponentially decreasing power distribution is required to yield a constant axial fuel temperature distribution. While this distribution cannot be obtained

exactly in a critical reactor, it can be closely approximated through the use of reflectors and by varying the axial fuel enrichment.

3. A nuclear rocket engine core was accidently misloaded such that fuel with an extremely high fission cross section was inserted at a particular axial location yielding the power distribution illustrated below. Based on your best engineering judgment, sketch in the blank plot below the expected axial propellant and fuel temperature distributions resulting from the misloaded core. Explain your reasoning.

Solution A sketch of the propellant and fuel temperatures is presented in Figure 1.

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

Figure 1

The propellant temperature is essentially proportional to the integral of the local power density in the reactor as a result of the fact that the heat generated in the core is progressively absorbed by the propellant as it travels from the core inlet to the core outlet. As a consequence, the temperature of the propellant as it enters the reactor, initially, begins to rise proportionally to the integral of the energy it absorbs from the core. When the propellant reaches the core location containing the fuel with the high fission cross section, the local power density spike causes the propellant temperature to suddenly increase at that location.

Since the propellant continues to absorb energy from the fuel downstream of the power spike, the propellant temperature cannot decrease, but rather continues to increase proportional to the integral of the local power density until the propellant finally exits the core.

The fuel temperature in comparison depends both upon the propellant temperature and the power density at a particular axial location. In other words, the fuel temperature depends both upon the local power density in the core and the integral of the local power density in the core. As a consequence, the temperature distribution is a composite of the local power density and the propellant temperature. At the location of the fuel with the high fission cross section, the fuel undergoes a temperature spike due to the increased propellant temperature and the power density spike. Downstream of the power density spike, the fuel temperature drops somewhat due to the decrease in the local power density; however, the fuel

temperature remains elevated due to the higher propellant temperatures to which the core power downstream of the power density spike must be transferred.

Chapter 10

1. A hydrogen turbopump for a nuclear rocket engine employing an expander cycle is to be designed. The inlet pressure to the pump from the propellant feed system will be 0.5 MPa and the liquid hydrogen is to be delivered to the engine system at a flow rate of 10 kg/sec. The pump impeller and turbine rotor should have the same diameter and each should have the same rotational speed.

Formatted: Indent: Left: 0", Hanging: 0.24", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

For the stated conditions, determine the following turbopump parameters.

-

Turbopump turbine inlet hydrogen temperature Reactor inlet hydrogen temperature Pumping power Turbopump turbine efficiency Turbopump pump efficiency Main shaft rotational speed Diameter of the pump impeller and turbine rotor

Solution The first step in the pump design is to determine the NPSH required by the pump to suppress cavitation. Knowing that the hydrogen to be pumped is subcooled at a pressure of 0.5 MPa and a temperature of 20 K, the liquid density can be determined to be 71.3 kg/m3. The required NPSH from Equation 10.2 is then:

The total required pressure head to be produced by the pump can be determined from Equation 10.3 such that:

Formatted: Indent: Left: 0.3", Hanging: 0.19", Bulleted + Level: 2 + Aligned at: 0.75" + Indent at: 1"

The Thoma parameter can now be calculated from Equations 1 and 2 yielding:

Using Figure 10.4 and the Thoma parameter calculated from Equation 3, Figure 1 illustrates that an upper value for the pump specific speed may be determined to be approximately ns = 0.3.

Figure 1

The volume flow rate of hydrogen through the pump may be determined from the specified mass flow and the hydrogen density such that:

From the definition of the pump specific speed described by Equation 10.1 and the volume flow rate of hydrogen from Equation 4, the impeller shaft maximum rotational velocity which can be tolerated without inducing cavitation is then found to be: ⁄

⁄

√

√

The pump specific speed can also be used in conjunction with Figure 10.5 to determine a pump specific diameter which will maximize the pump efficiency. An examination of Figure 2 reveals that a pump efficiency of about ηs = 67% is achievable in a radial flow pump having a specific diameter of ds = 8.6.

Figure 2

Using the definition of specific diameter from Equation 10.1 and the previously calculated values for the hydrogen volume flow rate and the pump pressure head, it is possible to determine that the impeller diameter should be: √

√ ⁄

(

)

⁄

The power required to drive the pump may now be determined by using the definition of efficiency from Equation 10.6 such that:

At this point, the pump section is completely specified, and the turbine characteristics must now be determined. Since the turbine inlet temperature is unknown, it will be necessary to guess a temperature to serve as a starting point in the subsequent calculations. Guessing 300 K as a start, the hydrogen state conditions are:

For the turbine outlet, assuming that the turbine is isentropic, the hydrogen state conditions are:

(9)

Using the inlet and outlet enthalpies from Equations 8 and 9, the pressure head across the turbine may be calculated from Equation 10.8 yielding:

The volume flow rate of hydrogen through the turbine may also be calculated using the hydrogen density at the turbine exit such that:

Using the pressure head from Equation 10, the hydrogen volume flow rate from Equation 11, rotational speed from Equation 5, and the impeller diameter from Equation 6, it is possible to calculate the turbine specific speed and specific diameter using Equation 10.1 yielding: √

√ ⁄

and ⁄

√

√

The specific speed and specific diameter calculated from Equations 12 and 13, respectively, may now be used in conjunction with the turbine efficiency chart from Figure 10.10 to determine a turbine efficiency of about 80% as illustrated in Figure 3.

Figure 3

Using this turbine efficiency, the work output of the turbine is found to be: (

*

Note that the turbine output from Equation 0 may also be represented as:

Using the actual turbine outlet enthalpy from Equation 15 and the turbine outlet pressure, the turbine outlet temperature and gas density may be determined to be:

Comparing the work output of the turbine from Equation 14 to the required drive power of the pump from Equation 7 ( ), it is apparent that the power being supplied by the turbine significantly exceeds that required by the pump. As a result, it can be seen that the turbine inlet temperature guess of 300 K which was used to calculate the output power of the turbine was too high. Guessing other turbine inlet temperatures and repeating Equations 8 through 14 then yield the results in Table 1.

Table 1 Turbine Turbine Pressure Hydrogen Volumetric Turbine Inlet Outlet Specific Specific Density Flow Rate Head Efficiency Temperature Temperature Speed Diameter ( ) ( ) ( ) ( ) ( ) ( )

300 100 142 130 132

252 83 116 106 108

92,143 29,286 43,367 39,490 40,102

6.748 23.20 15.06 16.72 16.42

1.482 0.4310 0.6640 0.5981 0.6090

0.3188 0.4062 0.3755 0.3823 0.3814

3.849 5.359 4.763 4.902 4.877

80 67 80 77 77

Turbine Power Output ( )

7.22 1.92 3.40 2.97 3.03

From Table 1, it can be seen that the turbine output power and the required pump drive power are equal at a turbine inlet temperature of 132 K and a turbine outlet temperature (reactor inlet temperature) of 108 K. Under these conditions Figure 10.10 indicates a turbine efficiency of about 77% as illustrated in Figure 4.

Figure 4

Chapter 11 1. A nuclear rocket has been firing for 2 hours in route to Mars and is now to be shut down for a long coast period. Assume that the engine had been operating at 1500 MW and a reactivity

insertion of -$1.00 was used to shut down the engine. How much total decay energy must be dissipated during the vehicle coast phase?

Assume: Solution The total energy which must be dissipated during shutdown is given by Equation 11.38 which states that: [

]

Changing the stated reactivity insertion from the problem statement into consistent units yields:

Substituting in the other data from the problem statement and the reactivity insertion from Equation 2 into Equation 1 then yields: [

[

]

]

2. For super prompt critical transients, it is usually satisfactory to assume that in the point kinetics equations remains constant during the transient. With this assumption, show that if the reactor is initially critical and at steady state with and a reactivity insertion occurs such that , then the reactor response is approximately: 2

√

[

(

√

*

(

√

*]3

Solution Inserting a linear reactivity increase into the point kinetics relationships from Equations 11.11 and 11.12 at that point where the reactor is exactly critical yields:

and

where:

initial neutron density

Substituting Equation 2 into Equation 1 yields a single equation for the neutron density as a function of time such that:

To solve Equation 3 the method of “integrating factors” will be used. The integrating factor for Equation 3 may be calculated by solving an equation of the form: ∫

where:

integrating factor

The solution to Equation 3 is then: ∫ Noting that:

∫

√

∫

[

(

√

*

(

√

*]

√

[

(

√

*

( √

*]

Substituting the results from Equations 4 and 6 into Equation 5 then yields:

√

[

(

√

*

(

√

*]

To solve for the arbitrary constant “ ”, the initial condition that the neutron density at time zero is is again used such that Equation 7 becomes:

√

[

(

√

*

( √

*]

Substituting the value for the arbitrary constant from Equation 8 into Equation 7 and simplifying then finally yield an equation for the time-dependent neutron density of the form:

√

[

(

*

√ 2

( √ √

*]

[

(

*

√

( √

*]3

3. A nuclear rocket has been firing for 1 hour in route to Mars and is now to be shut down for a long coast period. Assume that the engine had been operating at a power density of 1500 W/cm3 and a reactivity insertion of -$1.00 is used to shut down the engine. Also assume that 2 W/cm3 of auxiliary reactor cooling (i.e., independent of the cooling resulting from the flow of propellant through the engine) is available to take care of engine decay heat after shutdown.

,

where:

-

How long will some level of propellant flow need to be maintained in order that the reactor temperatures remain equal to or below their steady state values?

-

If propellant flow is terminated at the time of engine shutdown, how much does the reactor temperature rise over the steady state reactor temperature?

Solution To prevent core overheating due to decay heat after shutdown, the reactor is cooled for a time through the use of an auxiliary cooling system and shutdown propellant flow. To determine the amount of time required during which shutdown propellant flow is required, use is made of the point kinetics relationship from Equation 11.36 such that: 2

*

Substituting numerical values into Equation 1 then yields: {

(

)

+3

(2) [

]} [

]

To prevent overheating, some cooling using shutdown propellant will be required until drops below 2 W/cm3. By examining the plot of illustrated in Figure 1, it can be seen that 2 W/cm3 will not be reached for over 4600 seconds which is a little over an hour and 15 minutes. Also note from the plot that if 20 W/cm3 of auxiliary cooling were available, shutdown propellant flow would be required for only about 143 seconds or about 2 minutes and 23 seconds.

Figure 1

If no heat loss from the engine occurs other than that from the auxiliary cooling system, the engine will begin to heat up due to delayed fissions and fission product decay until the heat loss from the auxiliary cooling exceeds that from the delayed fissions and fission products.

That point occurs at a little over 4600 seconds as determined previously. To determine the maximum temperature which the fuel would encounter if no shutdown propellant flow were available, a heat balance is performed such that:

where:

total heat generated through delayed fissions and fission products change in temperature of the fuel during the course of the shutdown transient

The total heat absorbed by the fuel as a function of time may be found by integrating the power produced through delayed fissions and fission products as in Equation 11.37 minus the heat which is removed through the auxiliary cooling system yielding: ∫[

]

∫[

] (4)

2

0

1

[

]

*

(

)

+3

Substituting the results from Equation 4 into Equation 3 and incorporating the numerical data from the problem statement yield an equation of the form:

(5) {

[

]}

Plotting the fuel temperature rise over time, it can be seen in Figure 2 that quite high fuel temperatures can result if adequate cooling is not supplied to the reactor core. The peak temperature in the fuel without shutdown propellant cooling is more than 17,000 K over the fuel's temperature at shutdown, a temperature well over the melting point of any conceivable fuel form.

Figure 2

4. The point kinetics equations can be simplified considerably by employing what is called the "prompt jump approximation." This approximation assumes that the initial prompt jump which takes place immediately after a reactivity insertion occurs instantaneously and thereafter only the stable reactor period controls the rate of change of the neutron population.

The approximation assumes that the rate of change of the neutron population is small when compared to the other terms in the point kinetics equation, such that:

Using the prompt jump approximation in the point kinetics equations, determine an expression which yields the time-dependent behavior of the neutron population.

Solution Solving the point kinetics equation given in the problem statement for the time-dependent neutron population yields:

The time-dependent neutron population from Equation 1 is now used in the point kinetics equation for the time-dependent precursor population as stated in Equation 11.12 to yield an equation of the form:

Integrating Equation 2 yields an expression which gives the time-dependent behavior of the neutron precursors such that:

∫

∫

Substituting the expression for the time-dependent neutron precursor behavior from Equation 3 into the expression for the time-dependent neutron population from Equation 1 yields:

At the beginning of the power transient, the derivative of the neutron precursor population is zero; therefore, from Equation 11.12, it is found that:

Finally, substituting the expression for the initial neutron precursor density from Equation 5 into the expression for the time-dependent neutron population from Equation 4 yields the “prompt jump approximation” for the neutron population such that: (

*

Comparing Equation 6 with the expression derived earlier for the time-dependent neutron population from Equation 11.33, it can be noted that Equation 6 is simply Equation 11.33 with the first term in brackets missing. In Figure 1 Equations 6 and 11.33 are plotted for some typical kinetics parameters so as to illustrate the difference between the two equations.

Figure 1 Note from the plot that a $0.35 reactivity insertion results in over an almost instantaneous a 50% increase in the reactor power level. Such fast power increases can highly stress the reactor components; therefore, provisions must be made in the design of the reactor to allow for such rapid power transients. Otherwise, it will be necessary to limit the amount of reactivity inserted at any one time to mitigate the negative effects of the prompt jump. It is possible to estimate the instantaneous relative power increase due to the prompt jump by evaluating Equation 5 at time zero yielding:

5. The reactor in a nuclear rocket must reach full power in 30 sec. where: -

What step increase in reactivity (in $) is needed to induce the required period?

-

If the power must increase by a factor of 105 during this time, estimate the stable period required.

Solution The point kinetics expression for the time-dependent neutron population from Equation 11.33 is used to solve for the reactivity required to achieve the specified power increase:

Incorporating the data from the problem statement into Equation 1 yields:

Plotting the power versus reactivity relationship from Equation 2 as illustrated in Figure 1 indicates that a reactivity insertion of 0.00603 will result in the required power transient.

Figure 1 Converting the calculated reactivity to dollars yields:

The required stable period can be determined from the exponent of the second term in the expression for time-dependent neutron population from Equation 2. The first term in the time-dependent neutron population equation is the prompt jump term such that:

6. A nuclear rocket engine is configured as shown in the figure below. The engine is to be started up, run at 500 MW for 30 minutes, and then shut down. During startup, the control drums are rotated so as to achieve maximum positive reactivity (e.g., absorber plates are positioned nearest the outside of the reflector). This causes the reactor to become supercritical and results in the fastest possible temperature rise. Once the operating temperature has been reached, the control drums are rotated again so as to allow the reactor to become exactly critical. Because the reactor is exactly critical, the power and temperature rise ceases and the core temperature stabilizes at its operating temperature. At shutdown, the control drums are rotated so as to achieve maximum negative reactivity (e.g., absorber plates are positioned nearest the outside of the core). This causes the reactor to become subcritical and results in the fastest possible engine shutdown.

Assumptions: 

Control drum rotation varies the reactivity in the reactor by effectively changing the thickness of the reflector. For this problem, assume that the heterogeneity in the reflector may be ignored and that the reflector may be treated simply as a homogeneous region of varying thickness



At maximum core reactivity assume that the reflector is completely present



At minimum core reactivity assume that the reflector is completely absent



The control drum movement is instantaneous



Neglect extrapolation distances



There are no reflectors in the axial direction



The axial power distribution is constant at any radial location



The shutdown power in the reactor may be determined by using Equation 11.36



Be sure to state and justify any additional assumptions made

Parameter Core Reflector D, cm 1.3 0.81 Σc, cm-1 0.021 0.018 Υ 2.5 --Σf, cm-1 0.016 --αf, $/K -0.0003 --λ = 0.0764 sec-1 Λ = 0.00006 sec Tstart = 300 K rc = 50 cm

β = 0.0065 haxial = 135 cm Ttarget = 3000 K w = 30 cm

-

Derive the one group critical equation for this reactor system

-

Plot the neutron flux as a function of radial position when the control drums are positioned with the absorber plates completely away from the core and the reactor has just reached full power

-

Determine the reactivity worth of the control drums in $

-

Determine the highest possible temperature attainable by the reactor

-

Plot the reactor temperature as a function of time during startup

-

Determine the length of time required at startup for the reactor to reach its target temperature

-

Plot the control drum position (e.g., effective reflector thickness) as a function of core reactivity in $ and determine the reflector thickness when the reactor is at its target temperature

-

At what control drum position (e.g., effective reflector thickness) does the reactor begin to heat up

-

Determine the length of time required at shutdown for the reactor power to drop to 1 MW

-

Plot the reactor power as a function of time during shutdown using a LogLog scale

Solution The first step in solving this problem is to write down the governing neutron diffusion equations for the core and reflector regions of the reactor. Beginning with the core, the governing neutron diffusion equation is found to be:

Since the core is bare in the axial direction, the axial buckling in Equation 1 may be determined from Table 8.2 yielding:

(

⏟

, (

) (2)

( ⏟ [

* ]

By solving Equation 2, it is possible to obtain a relationship of the form:

Assuming that the core has boundary conditions wherein the first derivative of the flux goes to zero at r = 0, Equation 3 for the neutron flux becomes:

| (4)

Using Equation 4 the neutron current in the core may be determined to be:

For the reflector region of the reactor where no fissions are occurring, the governing neutron diffusion equation is found to be:

.

/

In cylindrical geometry, Equation 6 becomes:

(

⏟

,

(

*

Again, since the core is bare in the axial direction, the axial buckling in Equation 7 may be determined from Table 8.2 yielding:

0 ⏟

(

* 1

Solving Equation 8 then yields an equation of the form:

(9)

Assuming that the neutron flux goes to zero at the reflector boundary (e.g., at r = rc+w), Equation 9 can be used to relate C4 in terms of C3 such that:

[

]

[

[

]

]

[

]

Using the results from Equation 10 in Equation 9 it is found that:

2 [

]

[

]

[

]

[

]3

Using neutron flux from Equation 11 the neutron current in the reflector may be determined to be:

[

2

] ]

[

3

To solve for the remaining undetermined coefficients, use will be made of the fact that the neutron flux and neutron current must be continuous at the core/reflector interface. Therefore, equating the expression for the neutron flux in the reflector from Equation 4 and the neutron flux in the core from Equation 11 at the core/reflector interface it is found that:

2

[ [

] ]

3

Likewise, equating the expression for the neutron current in the reflector from Equation 5 and the neutron current in the core from Equation 12 at the core/reflector interface yields:

[

2

] ]

[

3

By dividing Equation 13 into Equation 14, it is possible to eliminate all the unknown coefficients so as to yield an equation of the form:

[

{

]

[ [

{

}

] ]

[

}

]

Since Equation 15 uniquely relates the reactor material properties to the reactor geometry it is the criticality equation for the reactor. Inserting numerical values into the expressions for αc and αr yields:

√

(

)

√ (16)

√

.

8

(

)

Inserting the expressions for the buckling from Equations 16 into the criticality equation as expressed by Equation 15 and plotting the results yields the graph illustrated in Figure 1. Note that the largest root in the criticality equation yields the correct eigenvalue for keff which in this case is 1.00595.

Figure 1

With keff determined, the core buckling may now be calculated from Equation 16 yielding:

√

The neutron flux in the reflector from Equation 11 may be determined by using Equation 13 to eliminate the arbitrary constant "C3" in the expression such that:

[

]

[

] [ [

] ]]

{ [

]

[

]

[

]

[

]}

The arbitrary constant "C1" is determined from the core power level. By dividing the total core power by the volume of the core the average core power density may be determined, yielding:

The average core flux may be determined by using the core power density calculated in Equation 19 and by recalling that:

The average core neutron flux may be also be determined by integrating the pointwise neutron flux from Equation 4 over the core volume such that:

∫

∫

[

]

Equating the relationships from Equations 20 and 21 and solving for the arbitrary constant "C1" and using the value for the core buckling obtained from Equation 17, the arbitrary constant "C1" may be calculated to be:

Using the expressions for the neutron flux in the reflector and core from Equations 4 and 18, respectively, and the numerical buckling values calculated earlier then yields:

2 [ =

] {

[ [

] ]

[

]3 }

The flux expressions from Equation 23 in the core and reflector regions of the reactor are plotted in Figure 3.

Figure 2

The maximum reactivity in the reactor in $ may be determined from keff such that:

The minimum reactivity in the reactor occurs when the neutron absorber plates on the control drums are positioned such that they are nearest to the core. In this configuration the reflector has minimal effect on the core reactivity and the reactor keff may be approximated by the use of the bare core relationships expressed in Table 8.2. It is also possible to use the core flux relationship from Equation 0. Since core is bare, the neutron flux must go to zero at the core boundary, therefore:

√

(

* [(

[(

)

(

)

(

) ]

) ]

The minimum reactivity in the reactor in $ may now be determined from Equation 25 such that:

The total control drum reactivity worth in $ may now be determined by subtracting Equation 26 from Equation 24 yielding:

The highest temperature attainable by the reactor will correspond to that temperature which is reached when the core reactivity is maximum, that is, when the control drums are positioned with their absorber plates farthermost from the core, therefore:

⁄ The relationship which illustrates the manner in which the core temperature varies as a function of time is described by Equation 11.73 and is reproduced below. In this particular case, since the maximum reactivity possible was inserted by the control drums, the relationship given by Equation 29 will yield the shortest possible time to reach the desired target temperature. Note that the fuel temperature coefficient of reactivity must be converted from $ units before being used in the relation. After inserting the appropriate parameters from the problem statement and the maximum attainable temperature from Equation 28, it is possible to now obtain an equation of the form:

6( *

(

)

(

*

(

)

7

6(

*

0(

(

*

(

*

*

7

1

Plotting the results of Equation 29 then yields a description of the time– temperature history of the startup:

Figure 3

The criticality relationship from Equation 15 may also be used to determine the relationship between the effective reflector thickness and the keff of the reactor. Plotting this relationship as a function of reflector thickness and reactivity in $ then gives an estimated control worth profile.

Figure 4

Note in Figure 4 that the reactor is subcritical until the control drums reach a position wherein the equivalent reflector thickness is equal to 5.52 cm. As a consequence no power will be generated in the reactor until the control drums reach this position. After the control drums reach this critical position, the reactor becomes supercritical and the power will start to rise causing the core to begin heating up.

After the rocket engine completes its thrust maneuver, the reactor must be shut down and brought into a cold subcritical configuration. This is accomplished by rotating the control drums from their full power position to a position wherein the neutron absorber plates are positioned nearest the core.

The positive reactivity required to achieve the desired hot full power condition may be determined from:

Using the positive reactivity required to achieve full power and the reactivity associated with the control drums positioned with their absorber plates closest to the core where the core reactivity is minimum, it is found that the negative reactivity available for shutdown is:

Using the relationship for reactor power as a function of shutdown reactivity as expressed by Equation 11.36 and the shutdown reactivity from Equation 31, it is found that:

2

{

[

.

]3

6

(32) ⌉-

⌈

{

⌈

⌉}

Plotting this shutdown power decay profile as a function of time illustrates the fairly lengthy time required to cool down the reactor after powered operation ceases. This is primarily due to fission product decay rather than delayed neutrons causing additional fissions to occur.

Figure 5 Note in the plot described by Figure 5 that the first drop in power at about 10-5 sec to about 200 MW is due to the prompt jump effect. The second drop in power at about 1 sec to about 10 MW is due to the decay of the delayed neutron precursor nuclides. After this point, the power drop off is due almost entirely to the decay of the fission product inventory which built up during the time at which the engine operated at full power. This power drop is much more gradual than one might wish and that even after about a half hour after shutdown, the reactor is still producing almost a megawatt of power due to this decay of these fission products. This power is significant and likely will require some type of cooling for perhaps several hours to prevent the core from overheating.

Chapter 12 1. Assuming a constant and along with the thermal neutron absorption cross section having a dependence, show that the nuclear temperature coefficient of reactivity in a thermal reactor is:

√

√

Solution Recalling that the definition of

is equal to:

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

Writing the definition of reactivity in terms of

from Equation 1 then yields:

Recalling that the definition of

is equal to:

Incorporating the definition of

from Equation 3 into Equation 2 then yields:

Recalling the definition of a

absorber from Equation 5.7:

√

√

where:

or Boltzmann's constant

or

Substituting the temperature-dependent absorption cross section from Equation 5 into the reactivity expression from Equation 4 then yields:

√

Finally, taking the derivative of the reactivity in Equation 6 with respect to temperature then yields an expression for the temperature coefficient of reactivity of:

√

Chapter 13

√

1. For deep space robotic missions, thermoelectric generators are sometimes used to power the spacecraft systems. The power to run these generators often comes from General Purpose Heat Source (GPHS) units which produce heat from the radioactive decay of The used in these units is produced by the neutron bombardment of in special production reactors designed for this purpose. does not occur naturally, but is available in quantity from reprocessed spent nuclear reactor fuel. The production cycle consists of:

Assuming that only is initially present, determine an expression which yields the time-dependent concentration of . Also determine an expression which yields the time required to reach a maximum concentration.

Solution Since the decay of is fairly rapid at only 2.1 days, it will not introduce much error to assume that is produced directly from . With this assumption, the transmutation chain for the production of may be given by:

Integrating Equation 1 for the time-dependent behavior of

∫

yields:

∫

Substituting Equation 3 into Equation 2 yields a differential equation for the time-dependent behavior of such that: ⏟

⏟

The differential equation for the time dependence of as expressed by Equation 4 may be solved through the use of an integrating factor ( ) wherein: ∫

∫

Using the integrating factor found in Equation 5, the solution to the differential equation for the production of as expressed by Equation 4 is then: ∫

∫ (6) ∫

Rearranging Equation 6 and integrating then yields:

(

)

Assuming that initially there is no present, Equation 7 may be used to determine a value for the arbitrary constant “ ” such that at :

Substituting Equation 8 into Equation 7 then yields for the time-dependent concentration of :

(9)

By taking the derivative of the time-dependent concentration of from Equation 7 and setting the result equal to zero, an expression can be found which will yield the time at which the maximum concentration of occurs: (

) (10)

Solving Equation 10 for the time then yields the time at which the maximum concentration of occurs: (

)

.

/

2. Assume that a reactor has been operating at steady state for a long period of time and the reactor "scrams". Derive a relationship that will yield the time at which the maximum xenon concentration occurs.

Solution The time dependent concentration of xenon during a reactor shutdown assuming that the xenon was initially at steady state conditions can be calculated using Equation 13.24 such that: [

]

By taking the time derivative of the xenon concentration from Equation 1 and setting the result equal to zero, it is possible to derive an expression which when solved for the time will yield an expression for the time at which the maximum xenon concentration occurs of the form:

⏟

⏟ (2) ⏟

Rearranging Equation 2 then yields: (

)

By solving for the time in Equation 3 and substituting in the definitions for , , and from Equation 2, it is possible to derive an expression for the time at which the maximum xenon occurs of the form: 2

Chapter 14

0

.

/.

/13

1. To save shielding weight on a Mars mission, it is desired to take advantage of the shielding effects provided by the liquid hydrogen fuel tanks with regard to protecting the crew from the gamma radiation emanating from the propulsion system's nuclear reactor. The vehicle configuration is illustrated in the figure below.

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During engine operation, the liquid hydrogen (saturated at 20 K) in the tank is depleted such that the hydrogen level in the tank decreases at a rate of 2.5 cm/sec. The amount of power produced by the NTR associated with the 6 MeV gamma rays is 2 kW. Assuming that the reactor can be treated as a point radiation source, determine: -

The whole body dose (in Sv) due to 6 MeV prompt fission gamma rays received by an astronaut located at the center of the habitat after the engine firing is complete.

Formatted: Indent: Left: 0.25", Bulleted + Level: 1 + Aligned at: 0.5" + Indent at: 0.75"

-

A plot of the 6 MeV prompt fission gamma ray dose (in Sv) received by the astronaut as a function of time.

Formatted: Indent: Left: 0.25", Bulleted + Level: 1 + Aligned at: 0.5" + Indent at: 0.75"

-

The dose (in Sv) due to 6 MeV prompt fission gamma rays which the astronaut would have received assuming that the hydrogen provided no shielding.

Formatted: Indent: Left: 0.25", Bulleted + Level: 1 + Aligned at: 0.5" + Indent at: 0.75"

Assume that the astronaut has a mass of 75 kg and an effective cross-sectional area of 8000 cm2. Also assume the hydrogen in the fuel tank is fully depleted at the end of engine thrusting and that there is no radiation shielding other than the hydrogen in the tank. If any other assumptions are required to solve the problem, they should be clearly stated and justified. Comment on the effectiveness of using the hydrogen tank as a means of reducing the radiation dose to the crew.

Solution Because the nuclear engine is small and far away as compared to the total size of the vehicle, the reactor can be treated as a point source for the purposes of the present analysis. The 6 MeV gamma ray energy flux to the astronaut, therefore, can be modeled using the differential form of the gamma attenuation expression from Equation 14.12 such that: [ where:

]

energy flux to the astronaut due to 6 MeV prompt fission gamma rays

instantaneous level of hydrogen in the propellant tank amount of power produced by the NTR associated with the 6 MeV gamma rays (distance between the NTR and the astronaut) From the problem statement, the propellant level in the tank as a function of time may be determined to be:

The time needed to drain the tank may be computed from Equation 2 yielding:

where:

time required to drain the propellant tank

Using Equations 1 and 2, the differential 6 MeV gamma ray energy flux to the astronaut may be written as: [

]

By integrating Equation 4 over the time needed to drain the propellant tank, an expression for the total 6 MeV gamma ray energy received by the astronaut during the engine firing may be determined such that: ∫[

] (5) ⏟

[

(

) (

)

From Figure 14.5 the specific mass attenuation coefficient for liquid hydrogen at 6 MeV is about 0.045 cm2/gm and the density of liquid hydrogen is about 0.07 gm/cm3; therefore, the mass attenuation coefficient for liquid hydrogen may be determined to be:

From Table 14.1 the coefficients for Taylor gamma ray buildup formulation for liquid hydrogen may be determined to be:

]

Substituting the numerical values from Equations 6 and 7 and from the problem statement into Equation 5 then yields for the 6 MeV gamma ray energy flux to the astronaut:

[

(

(

)

)

]

Since the astronaut has a cross-sectional area of 8000 cm2 and a mass of 75,000 gm, the dose he receives assuming an RBE of one (Table 14.2) and the gamma radiation from Equation 8 evaluates to:

(9)

The dose to the astronaut as a function of time may be determined by plotting Equation 9 using the time-based functional relationship from Equation 8. This functional relationship is illustrated in Figure 1.

Figure 1 Note that most of the radiation dose to the astronaut comes very near the end of the engine burn when the hydrogen is nearly depleted and can no longer provide shielding to the astronaut.

To determine the dose that the astronaut would have received had there been no liquid hydrogen to provide shielding, it is noted that the only attenuation of the gamma radiation comes as a result of the geometric spreading of the gamma radiation leaving the engine and not through any capture processes in the hydrogen, thus Equation 5 is modified such that: ∫[

] (10) ∫

Substituting numerical values from the problem statement into Equation 10 then yields:

(11)

Using Equation 9 to convert the unshielded gamma radiation from Equation 11 to a biologically equivalent dose to the astronaut then yields:

Clearly, the hydrogen provides significant shielding to the astronaut by reducing the biological dose by over an order of magnitude. Also, be aware that the astronaut dose calculated using Equations 9 and 12 assumes that all the gamma rays emanating from the reactor have an energy of 6 MeV. To get a more accurate dose, Equation 8 would have to be integrated over the actual reactor gamma ray energy spectrum using energy-dependent hydrogen attenuation and buildup factors so as to yield a spectrum weighted astronaut dose.

2. A nuclear rocket engine has a tungsten shield attached to the side of the reactor protecting a piece of radiation-sensitive equipment that is attached to the opposite side of the shield. In order to provide adequate gamma radiation protection to the equipment, it has been determined that the shield must be designed so as to reduce the radiation level coming from the reactor by a factor of 10000.

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

Assume that the reactor is composed of solid uranium: 

What shield thickness "t" is required to protect the equipment

Formatted: Bulleted + Level: 1 + Aligned at: 0.25" + Indent at: 0.5"



Plot the relative gamma ray intensity for each energy level as a function of shield thickness

Formatted: Bulleted + Level: 1 + Aligned at: 0.25" + Indent at: 0.5"



Plot the total relative gamma ray intensity as a function of shield thickness

Formatted: Bulleted + Level: 1 + Aligned at: 0.25" + Indent at: 0.5"



Plot the relative gamma ray intensity at the outside of the shield as a function of energy level. Discuss

Formatted: Bulleted + Level: 1 + Aligned at: 0.25" + Indent at: 0.5"

Solution Since the equipment package in question is small compared to the much larger adjacent reactor and the shield combination, the equations appropriate for determining the shielding required to attenuate the radiation emanating from a point gamma ray source will not apply. A more suitable analysis, therefore, will be to model the gamma ray attenuation as being through a semiinfinite slab shield adjacent to an infinitely large fission-induced gamma ray source.

The first step in the analysis will be to determine the gamma ray source term for each energy interval at the shield surface facing the reactor. From Table 14.1, it can be seen that there is information available to calculate the radiation intensity at seven different energy intervals, Ei (e.g., centered at 1, 2, 3, 4, 6, 8, and 10 MeV). From Equation 14.20, note that the gamma intensity at the surface of the shield closest to the reactor may be determined from core power density such that:

To determine the number of prompt gamma rays produced over the different energy intervals, the prompt gamma ray energy distribution from Figure 14.3 is integrated over an appropriate energy interval centered on the energy of interest such that:

∫

Carrying out these calculations for the number of gamma rays produced over the various energy intervals then yields:

∫

∫

∫

∫

∫

∫

∫

∫

(3)

Note that in the calculation of , the energy interval is not exactly centered on 6 MeV due to the way the energy ranges are structured. Also note that in the calculation of , there is a small extrapolation of the data from that given in Figure 14.3. In both these cases the errors introduced should be minimal since the irregularities involved are quite small. For energies below 1 MeV, the gamma ray intensity falls off quite rapidly within the shield, so these gamma rays can be safely neglected. In addition, while gamma rays for energies above 10 MeV are quite

penetrating, relatively few gamma rays of such high energies are produced in fission so these gamma rays can be also safely neglected.

The gamma ray mass attenuation factors may be obtained by multiplying the values obtained from Figure 14.5 by the density of the uranium fuel such that:

With the information now available, it is possible to calculate the gamma source present at the surface of the shield facing the reactor by using Equation 1. Since only relative gamma attenuation values are ultimately desired, the fission gamma volume source term may be arbitrarily specified. For the calculations below it will be assumed that the gamma volume source term is equal to 1 γ/cm3 yielding surface source terms of:

Using the gamma source terms from Equations 5, the attenuation of the gamma rays as a function of position within a shield having an extended source term may be calculated from Equation 14.21 which is reproduced below.

{

[

]

[

]}

The values for the Taylor coefficients for tungsten in Equation 6 may be determined from Table 14.1 and the gamma ray mass attenuation factors for tungsten may be obtained by multiplying the values obtained from Figure 14.5 by the tungsten density yielding:

At this point, all the information necessary to calculate the gamma attenuation as a function of position for each energy group within the shield has been determined, and upon using Equation 6, it is found that:

{

[

]

[

]}

{

[

]

[

]}

{

[

]

[

]}

(8) {

[

{

]

[

{

]

[

{

[

[

]

[

]}

[

]

]}

]}

[

]}

Plotting the gamma fluxes by energy group as a function of position using Equation 8 then yields:

Figure 1

By summing the groupwise gamma fluxes from Equation 8, the total gamma flux from the reactor may be determined such that:

∑

Plotting the normalized total gamma flux as a function of position within the shield then yields:

Figure 2 Note from the plot in Figure 2 that it takes about 9.2 cm of tungsten to drop the gamma radiation flux by the required factor of 10000. Plotting the gamma radiation flux by energy group at the reactor face and at the exit face yields the plot in Figure 3. From the plot, it can be seen that the lower energy gamma rays attenuate faster than the higher energy gamma rays. This effect leads to a shift in the gamma ray energy spectrum toward the higher energies as the radiation traverses the shield.

Figure 3

Chapter 15 1. Derive an expression that for a given pressure vessel internal volume yields the ratio which minimizes the pressure vessel mass. Assume that " " is the length of the cylindrical portion of the pressure vessel and " " is the diameter of the cylindrical portion of the pressure vessel.

Solution The mass of a cylindrical pressure vessel with hemispherical heads may be approximated by Equation 15.3 such that:

where:

Formatted: Indent: Left: 0", Numbered + Level: 1 + Numbering Style: 1, 2, 3, … + Start at: 1 + Alignment: Left + Aligned at: 0.25" + Indent at: 0.5"

The volume of the pressure vessel can be determined from: (

*

Solving Equation 2 for the radius of the pressure vessel yields:

(

)

Substituting the radius of the pressure vessel from Equation 3 into the expression for the mass of the pressure vessel from Equation 1 yields an equation of the form:

4

where:

5

dimensionless mass

Plotting the dimensionless mass from Equation 4 as a function of the ratio in Figure 1 indicates that the minimum pressure vessel mass occurs at an ratio of zero. The minimum mass of the pressure vessel is thus a sphere.

Figure 1 Substituting an ratio of zero into Equation 4 for the mass of a pressure vessel yields a minimum pressure vessel mass for a given volume of:

Chapter 17 The mean failure rate percentage of a nuclear rocket engine component in a limited number of tests has been estimated to be 4%. How many tests would be required to have 90% confidence that the actual failure rate of the component is between 3% and 5%?

Solution To establish the confidence limits on the engine component in question, use will be made of the cumulative distribution function for the binomial distribution as expressed by Equation 17.7 which is repeated below.

∑

Since the failure rate has been estimated to be 4%, the number of failures which would be expected in "n" trials would be:

Using Equation 1 to calculate the upper confidence bounds of the component yields an expression of the form:

∑

∑

Since Equation 3 cannot be solved explicitly for the number of trial tests as a function of the confidence limit, the expression is plotted so as to more easily analyze the results. As can be seen from Figure 1, at least 1350 trials are required to achieve the desired 90% upper confidence limit.

Figure 1

In like manner, Equation 1 may also be used to determine the lower confidence bound. In this case, the expression becomes:

∑

∑

Plotting Equation 4 as a function of the number of test trials versus the confidence limit yields the graph shown in Figure 2. Note that in this case, only about 950 trials are necessary to establish a lower confidence limit of 90%. Nevertheless, since 1350 trials are needed to meet the upper confidence limit criteria, the lower confidence limit will be somewhat higher than 90%.

As observed in the plot, the confidence limit which will actually be achieved in the test program using 1350 trials will be around 95.3%.

Figure 2