19 minute read

Chapter 13

Flow Boiling

PROBLEM 13.1 QUESTION

Nucleation in Pool and Flow Boiling (Section 13.2)

A platinum heat surface has conical cavities of uniform size, R, of 10 microns.

1. If the surface is used to heat water at 1 atm in pool boiling, what is the value of the wall superheat and surface heat flux required to initiate nucleation?

2. If the same surface is now used to heat water at 1 atm in forced circulation, what is the value of the wall superheat required to initiate nucleate boiling? What is the surface heat flux required to initiate nucleation?

Answers:

1. ∆Tsat = 3.3 K

Rohsenow method: q″ = 4.68 kW/m2 using Csf = 0.013

Stephan and Abdelsalam method: q″ = 2.85 kW/m2

Cooper method: q″ = 2.67 kW/m2

2. ∆Tsat = 6.5 K q″ = 221 kW/m2

PROBLEM 13.1 SOLUTION

Nucleation in Pool and Flow Boiling (Section 13.2)

All the cavities on the heat surface are assumed to be conical and of uniform size.

���� = 10μm

The following properties of saturated water at 1 atm pressure (101.325 kPa) may be found using tables or computer programs.

– saturation temperature, Tsat = 373 K

– surface tension, σ = 0.05892 N/m

– density, ρf = 958.4 kg/m3 and ρg = 0.5977 kg/m3

– v���� = 1 �������� = 1 0 5977 =1.673 m3 /kg

– specific enthalpy, h f = 419 kJ/kg and h g = 2676 kJ/kg

– h fg = h g = h f = 2257 kJ/kg

– specific heat, cPf = 4.216 kJ/kg K

– viscosity, μf = 281.8 × 10−6 Pa s conductivity, kf = 0.6791 W/m K

1. If the surface is used to heat water at 1 atmosphere in pool boiling, what is the value of the wall superheat and surface heat flux required to initiate nucleation? The wall superheat required to initiate nucleation in pool boiling may be calculated rearranging Equation 12.7.

Several correlations are available for calculating the surface heat flux required to initiate nucleation. Three different correlations have been used in this solution: Rohsenow, Stephan and Abdelsalam, and Cooper.

Rohsenow method Equation 12.18c may be used applying a coefficient Csf = 0.013, which is suitable for a platinum heat surface in water.

Stephan and Abdelsalam method Equation 12.22a may be used:

Where dd is given by Equation 12.22b, assuming a contact angle of θ = 25°:

And the thermal diffusivity of the liquid is:

The main equation can now be solved:

Where M = 18 is the molar mass of water, while the heat transfer coefficient is given by the general relation:

Substituting into the previous equation yields:

2. If the same surface is now used to heat water at 1 atm in forced circulation, what is the value of the wall superheat required to initiate nucleate boiling? What is the surface heat flux required to initiate nucleation? The condition required for bubble nucleation in flow boiling is given by Equation 13.2:

We may solve this equation to obtain the desired heat flux value.

In forced convection, the minimum wall superheat required to initiate nucleation can be expressed by Equation 13.5.

PROBLEM 13.2 QUESTION

Factors Affecting Incipient Superheat in a Flowing System (Section 13.2)

1. Saturated liquid water at atmospheric pressure flows inside a 20 mm diameter tube. The mass velocity is adjusted to produce a single phase heat transfer coefficient equal to 10 kW/m2K, What is the incipient boiling heat flux? What is the corresponding wall superheat?

2. Provide answers to the same questions for saturated liquid water at 290°C, flowing through a tube of the same diameter, and with a mass velocity adjusted to produce the same singlephase heat transfer coefficient.

3. Provide answers to the same questions if the flow rate in the 290°C case is doubled.

Answers:

1. q″ = 1.92×104 W/m2

TW – Tsat =1.92 °C

2. q″ = 230 W/m2

TW – Tsat = 0.023 °C

3. q" = 698.5 W/m2

TW – Tsat = 0.04 °C

PROBLEM 13.2 SOLUTION

Factors Affecting Incipient Superheat in a Flowing System (Section 13.2)

The tube diameter is:

���� = 0 0220 m

The following properties of saturated water at 1 atm pressure (101.325 kPa) may be found using tables or computer programs.

– saturation temperature, Tsat = 373 K

– surface tension, σ = 0.05892 N/m

– specific volume, vg = 1.673 m3/kg

– specific enthalpy, hf = 419 kJ/kg and hg = 2676 kJ/kg

– hfg = hg − hf = 2256 kJ/kg

– viscosity, μf = 0.0002818 Pa s

– conductivity, kf = 0.6791 W/m K

The mass velocity is adjusted to produce a single-phase heat transfer coefficient equal to:

1. What is the incipient boiling heat flux? What the corresponding wall superheat? In forced convection, the minimum wall superheat required to initiate nucleation is given by Equation 13.5.

At the point of boiling inception, the condition of minimum wall superheat has to satisfy also the single-phase heat transfer equation:

Notice that per the problem statement at boiling inception the coolant bulk temperature is

. Combining those two equations and solving:

2. Provide answers to the same questions for saturated liquid water at 290°C, flowing through a tube of the same diameter, and with a mass velocity adjusted to produce the same single-phase heat transfer coefficient. The following properties of saturated water at 290°C (563.1 K) temperature may be found using tables or computer programs.

Solving the equations already seen in Part 1, and keeping the same heat transfer coefficient:

3. Provide answers to the same questions if the flow rate in the 290°C case is doubled. The single-phase heat transfer coefficient may be expressed using the Dittus-Boelter/Mc Adams correlation, Equation 10.91, as:

Since the mass flow rate is doubled, G is doubled with respect to part 2 of this problem. All the terms in the equation above are constant except h and G. The new heat transfer coefficient may be calculated as:

Updating the equations already seen in part 2 with the new heat transfer coefficient and solving:

PROBLEM 13.3 QUESTION

Heat Transfer Problems for a BWR Channel (Section 13.3)

Consider a heated tube operating at BWR pressure conditions with a cosine heat flux distribution. Relevant conditions are as follows:

Channel geometry

D = 11.20 mm

L = 3.588 m

Operating conditions

P = 7.14 MPa

Tin = 278.3 C

G = 1625 kg/m2s q′max= 47.24 kW/m

1. Find the axial position where the equilibrium quality, xe, is zero.

2. What is the axial extent of the channel where the actual quality is zero? That is, this requires finding the axial location of boiling incipience, commonly called ONB, onset of nuclear boiling. (It is sufficient to provide a final equation with all parameters expressed numerically to determine this answer without solving for the final result).

3. Find the axial location of maximum wall temperature assuming the heat transfer coefficient given by the Thom et al. correlation for nuclear boiling heat transfer (Equation 13.23b). Hint! Example 14.5 treats a similar question.

4. Find the axial location of maximum wall temperature assuming the heat transfer coefficient is not constant but varies as is calculated by relevant correlations. Here, you are not asked for the exact location, but whether the location is upstream or downstream from the value from Part 3.

Answers:

1. Z OSB = 0.61 m (from the bottom of the channel)

2. Z ONB = 0.21 m

3. 1.79 m

4. Upstream

PROBLEM 13.3 SOLUTION

Heat Transfer Problems for a BWR Channel (Section 13.3)

The following parameters are known:

– tube diameter, D = 11.2 ×10−3 m

– tube length, L = 3.588 m

– inlet temperature, Tin = 278.3 °C

– pressure, P = 7.14 MPa

– mass flux, G = 1625 kg/m2s

– peak linear heat generation rate, q′ max = 47.24 kW/m

The following parameters may be taken from tables or software with water properties.

For a cosine heat flux distribution, the axial peaking factor is equal to π/2. The total channel power is hence: this location from the channel inlet instead of the midplane: liquid only Reynolds number

1. Find the axial position where the equilibrium quality, xe, is zero. Applying the conservation of energy to the heated tube with a cosine heat flux distribution, Equation 14.33b is obtained, which yields the equilibrium quality as a function of z. The equilibrium quality is set to be equal to zero to obtain zOSB.

2. What is the axial extent of the channel where the actual quality is zero? That is, this requires finding the axial location of boiling incipience commonly called ONB, onset of nucleate boiling. (It is sufficient to provide a final equation with all parameters expressed numerically to determine this answer without solving for the final result). The location of axial boiling incipience may be determined as the location where the single-phase heat transfer curve intersects the nucleate boiling curve described by the Thom correlation.

The flow is turbulent. The Dittus-Boelter/McAdams relation may be applied (Equation 10.91) to obtain the heat transfer coefficient hDB:

Solving numerically:

Measuring this location from the channel inlet instead of the midplane: ����������������

3. Find the axial location of maximum wall temperature assuming the heat transfer coefficient given by the Thom et al. correlation for nuclear boiling heat transfer (Equation 13.28b). Equation 13.28b provides the heat flux as:

The parameters in bold are constant in our model. Therefore, observing the Thom et al. correlation we can notice that the maximum wall temperature occurs at the location of maximum heat flux, which is the core midplane:

Measuring this location from the channel inlet instead of the midplane:

4. Find the axial location of maximum wall temperature assuming the heat transfer coefficient is not constant but varies as is calculated by relevant correlations. Here, you are not asked for the exact location, but whether the location is upstream or downstream from the value from Part 3. The solution to this question may be given qualitatively. The general heat transfer equation is:

In the location of interest the water bulk has an enthalpy comprised between hf and hg. Therefore, its temperature is fixed as Tsat. The wall temperature is hence:

Observing this equation we notice that the location of maximum wall temperature should have a high heat flux q" and a low heat transfer coefficient ℏ. The heat flux has a cosine axial shape and is maximum at the channel midplane, while the heat transfer coefficient increases as z increases. This occurs because the boiling process and the turbulence caused by boiling enhance the heat transfer. Therefore, if the location of maximum wall temperature is not at z = 0, it must be necessarily where the heat transfer coefficient is lower: upstream from that point.

PROBLEM 13.4 QUESTION

Thermal Parameters in a Heated Channel in Two-Phase Flow (Sections 13.3 and 13.4)

Consider a 3 m long water channel of circular cross-sectional area 1.5 × 10−4 m2 operating at the following conditions:

���� = 0.29 kg s⁄

���� = 7 2 MPa ℎ�������� = saturated liquid

���� ″ = axially uniform

���������������� = 0.15

Compute the following:

1. Fluid temperature

2. Wall temperature using Jens and Lottes correlation

3. Determine CPR using the Groeneveld lookup table

Answers:

1. 287.7 °C

2. 294.3 °C

3. CPR= 2.75

PROBLEM 13.4 SOLUTION

Thermal Parameters in a Heated Channel in Two-Phase Flow

(Sections 13.3 and 13.4)

The following parameters are known:

– flow area, A = 1.5×10−4 m2

– mass flow rate, ṁ = 0.29 kg/s

– channel length, L = 3m

– pressure, P = 7.2 MPa

– outlet equilibrium quality, xe,out = 0.15

The following parameters may be taken from water properties:

The following parameters are calculated:

The heat flux, which is axially constant, is given by solving the following energy balance (Equation 13.44b):

1. Compute the fluid temperature The fluid is at saturation, so it has the same temperature throughout the channel, which is:

2. Compute the wall temperature using Jens and Lottes correlation A numerical solution of the Jens and Lottes correlation, Equation 13.23a, yields the wall temperature:

3. Determine MCPR using the Groeneveld Lookup Table First of all, we determine the diameter correction factor as given in Table 13.5:

The Groeneveld CHF lookup table provides the critical heat flux requiring three input parameters: quality, mass flux and pressure. In our case, mass flux and pressure are known and constant, while the quality varies with the channel power.

The critical heat flux decreases with increasing quality. In this problem we are dealing with a constant heat flux distribution, so the location where we expect dryout to occur is the channel outlet.

The Critical Power Ratio, which is the ratio between the channel power that leads to critical condition and the operating power, is calculated iteratively, increasing the power until the heat flux matches the CHF. The results of these iterations can be seen in Table SM-13.1, where the nominal power is multiplied by a tentative coefficient called “power ratio”.

The exit quality is calculated using Equation 13.44b (see above).

The uncorrected CHF is determined using the lookup table at the given exit quality, and then multiplied by K1 to obtain the corrected CHF.

Note: See www.CRCPRESS.com/PRODUCT/ISBN/9781433980887 for the interpolation procedure of the Groeneveld lookup table.

The Critical Power Ratio is hence equal to 2.75.

PROBLEM 13.5 QUESTION

CHF Calculation with the Bowring Correlation (Section 13.4)

Using the data of Example 13.3, calculate the minimum critical heat flux ratio using Bowring correlation. Consider an axially-uniform linear heat generation rate: q′ = 35.86 kW/m.

Answer:

MCHFR = 2.13

PROBLEM 13.5

Solution

CHF Calculation with the Bowring Correlation (Section 13.4)

The following parameters are known:

– tube diameter, D = 10.0 × 10−3 m

– pressure, P = 6.89 MPa

– tube length, L = 3.66 m

– inlet temperature, Tin = 204 °C

– mass flux, G = 2000 kg/m2s

– linear heat generation rate, q′ = 35.86 kW/m

The following specific enthalpy values may be taken from tables or software with water properties:

Calculate the minimum critical heat flux ratio using Bowring correlation. The Bowring correlation is given by Equation 13.49a

The parameters A, B and C are calculated as follows, using the equations from 13.49b to 13.49h.

For the particular case of Pr = 1, it may be easily verified (Equations 13.49g and 13.49h) that the coefficients are all equal to unity.

The critical heat flux decreases when quality increases. In this problem we are dealing with a constant heat flux distribution, so the location where we expect the minimum CHFR is the channel outlet.

PROBLEM 13.6 QUESTION

Boiling Crisis on the Vessel Outer Surface during a Severe Accident (Section 13.4)

Consider water boiling on the outer surface of the vessel where water flows in a hemispherical gap between the surface of the vessel and the vessel insulation (Figure 13.31). The gap thickness is 20 cm. The system is at atmospheric pressure.

1. The water inlet temperature is 80 °C and the flow rate in the gap is 300 kg/s. The heat flux on the outer surface of the vessel is a uniform 350 kW/m2 in the hemispherical region (0

90°) and zero in the beltline region (θ > 90°). If a boiling crisis occurred in this system, what type of boiling crisis would it be (DNB or Dryout)?

2. At what angle θ within the channel would you expect the boiling crisis to occur first and why?

Answers:

1. DNB

2. θ = 90°

PROBLEM 13.6 SOLUTION

Boiling Crisis on the Vessel Outer Surface during a Severe Accident (Section 13.4)

The system shown in Figure 13.31 is at atmospheric pressure.

Some of the known parameters of the problem are written in the figure. The heat flux on the outer surface of the vessel, the mass flow rate and the pressure are, respectively: ����

The heated surface can be calculated as follows, considering a hemisphere of radius r = 5.2 m:

The total heat produced by the hemispherical surface is:

1. If a boiling crisis occurred in this system, what type of boiling crisis would it be (DNB or Dryout)? Let us determine the water enthalpy at the inlet and at the outlet of the control volume. The inlet enthalpy can be determined from water properties as:

The outlet enthalpy is given by a simple energy balance:

The DNB is a boiling crisis observed at low-quality or even subcooled conditions, while film Dryout is observed at moderately high quality. Since our system has a maximum equilibrium quality of just -1.6% (and an outle temperature of 91.8ºC), the boiling crisis type which is expected to occur is DNB.

To find the coolant temperature of 91.8 degrees C where cP,f = 4.2 kJ/ºC, solve the following equation

2. At what angle θ within the channel would you expect the boiling crisis to occur first and why? Let us analyze the trends of the coolant and heat transfer parameters in the control volume.

– Pressure: Decreases from inlet to outlet due to friction, gravity and acceleration.

– Equilibrium quality: increases from inlet to outlet.

– Mass flux: Decreases from inlet to outlet, as the flow area increases.

– Heat flux: Constant.

Therefore, at the exit of the control volume the coolant has the lowest pressure and mass flux and the highest equilibrium quality. We may notice observing the Groeneveld LUTs (or using other critical condition prediction methods) that at, a pressure around 1 atm and equilibrium quality lower than 0.1, the critical heat flux:

– Decreases when equilibrium quality increases.

– Decreases when mass flux decreases.

– Decreases when pressure decreases.

According to the trend of those three parameters, the lowest critical heat flux occurs at the outlet. Since the heat flux is constant on the whole surface, the location with the lowest CHFR is the outlet. Therefore, we expect that DNB could occur first at an angle:

PROBLEM 13.7 QUESTION

Calculation of CPR for a BWR Hot Channel (Section 13.4)

Consider a BWR channel operating at 100% power at the conditions noted below. Using the Hench–Gillis correlation (Equation 13.57a) determine the CPR at 100% power.

= 1.17

PROBLEM 13.7 SOLUTION

Calculation of CPR for a BWR Hot Channel (Section 13.4)

The following parameters are given:

– reference linear heat generation rate, q′ref = 104.75 kW/m

– axial heat distribution coefficient, α = 1.96

– inlet temperature, Tin = 278.3 °C

– pressure, P = 7.14 MPa

– mass flux, G = 1569.5 kg/m2s

– heated length, L = 3.588 m

– rod diameter, D = 0.0112 m

– subchannel flow area, Afch = 1.42 × 10−4 m2

– bundle flow area, Af = 9.718 × 10−3 m2

The linear heat distribution is given by the following equation:

The following parameters may be taken from tables or software with water properties:

The mass flow rate in the subchannel of interest is:

Calculation of zOSB The equilibrium quality distribution is given by Equation 14.33a as: quality profile Applying the same equation, the equilibrium quality profile is:

Setting the equilibrium quality equal to zero and integrating numerically, the OSB location zOSB can be found.

Hench-Gillis correlation The Hench-Gillis correlation is given by Equation 13.57a:

To use the Hench Gillis correlation, we have to convert the mass flux into British units.

Substituting all those values in Equation 13.57a:

The boiling length LB is defined as the distance from the OSB location:

Determination of the CPR The critical power ratio can be determined graphically, by plotting the distributions of the Hench-Gillis critical quality x c ( z ) and of the equilibrium quality x e ( z ) while increasing the channel power. We will plot equilibrium quality and critical quality as a function of the boiling length L B , which is defined as the distance from the OSB location:

Notice that the OSB location is equal to -1.26 m, as calculated above, only for nominal power. When the power increases, the OSB location must be re-calculated.

As the following figure shows, the CPR is 1.17.

PROBLEM 13.8 QUESTION

Reduction in a PHWR CHF Margin upon a Local Channel Power Increase due to Increased Two Phase Pressure Drop and Corresponding Reduction in Channel Mass Flux

Consider a horizontal flow channel of typical PHWR geometry operating at typical PHWR full power conditions of 2700 ���� ��������ℎ with 480 channels. The relevant geometric and steady state channel operating conditions are given in Table 13.7 below.

In this example we will postulate a local increase in power and assess the reduction in CHF margin. Since there are more than hundred channels connected to the same inlet and outlet manifolds/headers (boundary conditions), a heat up in one or a few channels would not affect the boundary conditions (i.e. the inlet and outlet pressure conditions) noticeably. Local heat up / power increase could be due to movement of local reactivity devices, or (abnormal) online fueling of a channel. (Normally the on line fueling causes about 10% increase for central channels and as much as 15% for peripheral channels).

Using the Groeneveld 2006 CHF lookup table [34] determine the following quantities: a) The margin to CHF at steady state full power channel conditions of 6 MW b) The changed margin to CHF upon an increase in channel power by 15%

Assume: a) The heat flux is uniform along the channel b) Treat the friction factor in the subcooled region of the channel as single phase coolant c) Use heavy water properties d) Homogeneous Equilibrium Model (HEM)

Answer: a) Steady State: MCHFR = 5.82 b) Transient (channel power increase by 15%): MCHFR = 3.17

PROBLEM 13.8 SOLUTION

Reduction in a PHWR CHF Margin upon a Local Channel Power Increase due to Increased Two Phase Pressure Drop and Corresponding Reduction in Channel Mass Flux

1. Steady state

In steady state, the mass conservation equation (Equation 5.63) applied to the heated channel reduces to: where “ss” indicates the steady state. In the energy conservation equation (Equation 5.159), the time-dependent terms are zero because it is a steady state problem, the potential energy terms cancel out because the channel is horizontal, we neglect the kinetic terms, we neglect the shear term and the heat source term is zero. After integration we obtain: where ����̇ �������� is given in Table 13.7, ℎ��������,�������� can be calculated from ������������,�������� and ������������,�������� , and ℎ��������,������������ can be expressed as a function of outlet quality ������������,������������ and saturation enthalpies at the outlet conditions, using heavy water properties:

The inlet quality can be calculated from the saturation enthalpies at the inlet conditions. At the inlet the fluid is sub-cooled since at the inlet pressure 11 ������������ the saturation temperature is 317 06 ℃

Since ℎ����,��������,�������� = 2511.8 kJ kg and ℎ����,��������,�������� = 1389.95 kJ kg , then:

To calculate the coordinate at which the fluid transitions to 2-phase conditions, we start by guessing a value of ������������ , then we calculate the corresponding ℎ��������,������������ and ������������,������������ from Equation (2) and Equation (3)

Denoting ���� the channel length and ���� the coordinate at which the fluid transitions from 1-phase to 2-phase conditions, and assuming that the fluid’s quality grows linearly in the channel (which is reasonable, as the heat flux is constant), then:

Hydraulic loss coefficients provided in Table 13.7 are derived as follows:

���������������� – There are 12 fuel bundles, each with two separate minor losses. One at mid-length where the spacers are located, and one at bundle beginning/end where the fuel elements are attached to the bundle end plates. The ���������������� provided in Table 13.7 is the sum of all minor losses in the 6 ���� long channel.

������������ and ���������������� Provided to capture all hydraulic losses (friction and minor losses) in the inlet feeder pipe (between inlet header and beginning of heated section), and ���������������� in the outlet feeder (between the exit of heated channel and the outlet header).

�������������������� – The fuel channel skin friction loss is given as �������������������� in Table 13.7.

In the heated channel, to calculate the pressure drop due to friction and minor losses, minor losses (���������������� ) in the heated length of the channel can be treated continuously for simplicity. Then, treating the single and two-phase regions separately: where we used the liquid saturation density at an intermediate pressure (10.5 MPa):

For the HEM model, the 2-phase flow multiplier is expressed as:

We make the following simplifying assumptions:

,

2 is an average quality value in the 2-phase zone, and �������� and �������� are calculated at the outlet conditions.

In Equation 8 we assume that ���������������� =

Finally, Equation 5 becomes:

The acceleration pressure drop term is:

We make the following assumptions:

Finally, Equation (11.59a) becomes:

The only unknowns in Equation (11) and Equation (12) are ������������ and ������������ ,������������ . We can solve the system of equations iteratively, guessing a mass flow rate ������������ , then calculating the corresponding ������������,������������ until these values are consistent. The solution is:

The corresponding steady state mass flux is:

The Groeneveld lookup table [34] provide the values of Critical Heat Flux (CHF), as function of mass flux ���� , fluid pressure ���� and thermodynamic quality ���� . In case of constant heat rate profile, the minimum departure from nucleate boiling ratio is at the channel exit. Thus, we use the tables to find CHF��������,������������ :

The steady state heat flux is:

The Minimum CHF Ratio (MCHFR) is then:

2. Transient

The transient consists of a local heat rate increase of 15%, due potentially to at-power fueling of the fuel channel, which brings the system to a new equilibrium state.

The boundary conditions are pressure and temperature at the inlet plenum and pressure at the outlet plenum; they are assumed not to change with respect to the steady state analysis since increasing the power in a single channel does not change the boundary conditions since there are 90 to 120 channels connected to the same plena.

On the other hand, the fluid quality at the outlet is higher in the new post-transient equilibrium state due to the larger heat rate. Saturation is reached earlier in the channel, so the 2-phase portion of the channel is longer. This results in a decreased post transient mass flow rate �������� because of the increased hydraulic flow resistance within the channel. We follow a procedure analogous to the steady state case, with an initial guess of �������� lower than in the steady state:

Then we use the same equations as in the steady state to calculate:

The new coordinate �������� (e.g., the coordinate where the fluid transitions from 1 to 2-phase conditions) is:

We assume that form, minor and friction loss coefficients do not change in the transient. After a few iterations we find that the post-transient mass flow rate and exit quality are:

The post-transient mass flux is ������������ = 7000 kg s. Using once again the Groeneveld lookup tables, we find that:

The post-transient steady state heat flux is 15% higher than the steady state value:

The post-transient Minimum Departure from Nucleate Boiling Ratio (MDNBR) is then:

A 15% increase of the heat rate results in the MCHFR decreasing by almost half!

3. Transient*: result due to power increase only

If we only accounted for the power increase (e.g. if we neglected the associated reduction in flow rate), the results would be different. We indicate the quantities related to this case with an asterisk (*).

To show this, we assume that but the mass flow rate doesn’t change with respect to the steady state:

At this point, the exit quality ���������������� ∗ is the only unknown in the following equation., where is from Table 13.7.

The mass flux is the same as in the steady state:

The heat flux is the same as in the transient:

Using once again the Groeneveld lookup tables, we find that:

The Minimum CHF Ratio (MCHFR) is then:

This article is from: