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2. Problem 1 (35%) – Steady-state natural circulation in a steam generation system
Saturated steam at 3 MPa (properties in Table 1) is used in a certain factory. The steam is
generated by the system shown in Figure 1, which consists of a natural gas-fired heater, a
riser of height L, a steam separator of form loss K, and a downcomer. The makeup flow
can be assumed to be saturated water at 3 MPa. The riser and the steam separator have the
same flow area, A.
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3. i) Using the conservation equations and their constitutive relations, find a single equation
from which the mass flow rate in the loop, m& , could be found as a function of the heat
rate, Q& , and the parameters A, L and K, i.e., f( m& , Q& ,A,L,K)=0. (20%)
ii) Find m& for the two limit cases Q& =0 and Q& = m& hfg. Do you think the m& vs
Q& curve (with fixed A, L and K) could have a maximum between these two limits?
Explain your answer qualitatively. (10%)
iii) For a given Q& , how does m& change if K increases or L increases or A increases?
(5%)
Assumptions:
- Steady state
- Steam separator efficiency is one
- Use HEM for the void fraction in the riser
- Neglect all acceleration and friction terms in the loop momentum equation
- Use the HEM multiplier for the form loss in the separator,
Table 1. Properties of saturated water at 3 MPa.
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4. Parameter Value
Tsat 234°C (507 K)
ρf 822 kg/m3
ρg 15 kg/m3
hf 1,008 kJ/kg
hg 2,803 kJ/kg
Cp,f 4.7 kJ/(kg°C)
Cp,g 3.6 kJ/(kg°C)
μf 1.1×10-4 Pa⋅s
μg 1.7×10-5 Pa⋅s
kf 0.638 W/(m°C)
kg 0.047 W/(m°C)
σ 0.030 N/m
Problem 2 (55%) – Water boiling during a loss-of-flow transient in a home heating
system
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5. A large condo building uses a water forced-convection heating system. The heater consists
of hundreds of round channels of diameter D=2.54 cm and length L=1 m in which water is
heated by an axially uniform heat flux, q″=200 kW/m2 (see Figure 2). The system
operates at 1 MPa and the water temperature at the inlet of the heater channel is Tin=90°C
(hin=365.6 kJ/kg). Under normal operating conditions the mass flux is Go=1000 kg/m2 s
and no boiling occurs in the channel. A pump malfunction occurs at t=0, so that the mass
flux in the heater channel starts to decay exponentially, i.e., G(t) = Goe−t /τ , where τ =10
s. Assume that the heat flux, pressure and inlet temperature remain constant throughout the
transient.
Figure 2. A heater channel
Table 2. Properties of saturated water at 1 MPa.
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6. Parameter Value
Tsat 180°C (453 K)
ρf 887 kg/m3
ρg 5.1 kg/m3
Hf 763 kJ/kg
hg 2,778 kJ/kg
Cp, f 4.4 kJ/(kg°C)
Cp,g 2.6 kJ/(kg°C)
μf 1.5×10-4 Pa⋅s
μg 1.4×10-5 Pa⋅s
kf 0.677 W/(m°C)
kg 0.034 W/(m°C)
σ 0.042 N/m
R* 462 J/kg⋅K
i) Using a simplified version of the energy conservation equation,
calculate the fluid enthalpy and equilibrium quality as functions of z and t. (5%)
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7. ii) At what time does the bulk temperature reach saturation? Assume the specific heat
does not change with temperature. (5%)
iii) At what time does nucleate boiling start? Use the Davis and Anderson model for
ONB and assume that the single-phase forced convection heat transfer coefficient, H,
is proportional to the mass flux, i.e., H = Ho where Ho=9.3 kW/m2 K. (10%)
iv) At what time does a significant amount of vapor first appear in the channel? (10%)
v) Qualitatively sketch the MDNBR vs. time. (5%)
vi) Qualitatively sketch the bulk and wall temperatures vs. time at the channel outlet.
(10%)
vii) Estimate the time at which two-phase density-wave oscillations appear in the
channel. Use the stability map of Figure 3 below. (10%)
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8. Figure 3. Stability map for the heater channel.
Problem 3 (10%) – Short questions on bubble nucleation
i) A steam bubble grows at a cavity with the geometry shown in Figure 4. What can
you say about the steam temperature in this situation? (5%)
Figure 4. Steam bubble growing within a wall cavity.
ii) To obtain bubble nucleation at a cavity of radius 1 μm on a copper surface, a certain
fluid (of contact angle 135° with copper) requires a 2°C superheat. What would the
required superheat be for bubble nucleation at a cavity of radius 3 μm on steel, if the
fluid contact angle with steel were 45°? (5%)
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9. Problem 1 (35%) – Steady-state natural circulation in a steam generation system
i) The flow in the loop is due to natural circulation, driven by the density difference
between the two-phase riser and the single-phase downcomer. The momentum equation
for the loop is:
where the friction and acceleration terms have been neglected, as per the problem
assumptions. The fluid in the downcomer is saturated water therefore its density is
ρdown=ρf, while the density in the riser is:
ρriser = αρ g + (1−α)ρ f (2)
where α is the void fraction. If HEM is used:
per the problem assumption. The flow quality x can be found from the energy balance
for the heater:
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10. where it was assumed that the equilibrium quality is equal to the flow quality, a very
good assumption since the riser is a saturated mixture of steam and water. Eliminating
x in Eqs. (3) and (4) by means of Eq. (5), and substituting Eqs. (2) and (5) into Eq. (1),
one gets the answer:
which could be solved to find
ii) If Q& =0 (no steam), one has x=0, α=0, ρriser=ρf, and therefore m& =0.
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11. An increase in heat rate, Q& , increases the density difference between the riser and the
downcomer, which would tend to increase the flow. However, an increase in Q& also
increases the quality and thus the two-phase form loss multiplier, which of course would
tend to reduce the flow. Because there are two conflicting effects, a maximum in the m&
vs Q& curve is possible. bThis curve is shown for some representative values of A, K and
L in Figure 1, and it does in fact have a maximum.
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12. iii) For a given
- decreases with increasing K because the resistance to the flow is higher
- increases with increasing L because the gravity head driving the flow is higher
-increases with A because a larger flow area reduces the velocity and thus
reduces the form pressure loss in the separator.
Problem 2 (55%) – Water boiling during a loss-of-flow transient in a home heating
system
i) The energy equation can be readily integrated to give:
where Ph=πD=7.98 cm and A=π/4⋅D2 =5.1 cm2 . Then the equilibrium quality, xe, is:
ii) Before reaching saturation h-hin can be expressed as Cp,f(Tb-Tin), where it is was
assumed that the specific heat is independent of temperature, as per the hint. Thus, from
Eq. (8) one gets:
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13. Obviously, saturation is first reached at the channel outlet, so setting Tb=Tsat and z=L
in Eq. (10) and solving for t, one gets the time at which saturation first occurs in the
channel:
An identical result would have been obtained by setting h=hf in Eq. (8) or xe=0 in Eq.
(9).
iii) The Davis and Anderson model for the Onset of Nucleate Boiling (ONB) gives a
relation between the heat flux and the wall superheat, Tw-Tsat, at ONB, as follows:
where P=1 MPa is the system pressure. To find the time at which the wall temperature
reaches 182.2°C, we can use Newton’s law of cooling:
q"= H (Tw −Tb ) (13)
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14. where H = is the heat transfer coefficient, as per the problem statement.
Substituting Go Eq. (10) into Eq. (13), setting Tw =Tw,ONB, recognizing that at any
given time the maximum wall temperature is at z=L, and solving for t, one gets the
time at which ONB first occurs in the channel:
iv) The Onset of Significant Void (OSV) will first occur at z=L, and can be predicted
with the Saha and Zuber correlation:
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15. where Pe≡(GDCp,f)/kf. Since OSV will occur after ONB, and Pe≈5×104 at ONB, we
can conclude that Pe
v) The DNBR is defined as q′ DNB ′ / q ′′ at any location in the channel. Since q′ DNB ′
decreases with increasing xe, the minimum DNBR (MDNBR) is at the channel outlet at
any given time. The MDNBR vs time is sketched qualitatively in Figure 2 below. Note
that the MDNBR decreases rapidly with time because of the combined effect of the mass
flux exponential decay ( G(t) = Goe−t /τ ) and xe exponential growth (Eq. 9). Therefore,
DNB will occur (MDNBR=1) soon after ONB. This can be avoided if the normal mass
flux is re-established or the heat flux is significantly reduced.
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16. vi) The bulk temperature increases exponentially per Eq. (10) until it reaches Tsat;
then it stays at Tsat until xe=1. The wall temperature is found from Newton’s law of
cooling as
Tw = Tb + q"/ H (17)
where H is the heat transfer coefficient at time t. For ttONB H increases as the heat
transfer regime becomes partial and then fullydeveloped subcooled nucleate boiling.
However, at t=tDNB H drops dramatically because the transition to film boiling occurs.
Failure (burnout) of the heater channel is expected soon after this transition. The
qualitative time history of the bulk and wall temperatures at the channel outlet is shown
in Figure 3. Note that without a quantitative calculation of qDNB ′′ vs. time, it is not
possible to determine a priori whether tDNB>tsat or vice versa.
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17. Figure 3. Time history of the bulk and wall temperatures at the channel outlet
(not to scale)
vii) To determine the onset of dynamic instability, one first has to calculate the
subcooling number, Nsub:
and the phase change number, Npch:
At normal operating conditions the values for the heater channel are Nsub≈34 and
Npch≈3, which identify a stable point on the stability map. However, for t>0 the phase
change number increases because the mass flux decreases, while Nsub remains constant
because the inlet enthalpy and pressure are fixed throughout the transient. Therefore, the
channel “trajectory” on the stability map is a straight horizontal line (see Figure 4 below).
The Npch value at which instability occurs is 38, found by intersecting the trajectory with
the stability line, Nsub=Npch - 4. Then, solving Eq. (19) for G, one gets Gunst≈70.5
kg/m2 s. The time at which G=Gunst is:
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18. Figure 4. Trajectory of the channel on the stability map.
Problem 3 (10%) – Miscellaneous short questions
i) Since the steam/liquid interface is flat (i.e., the radius of curvature is infinite), the steam
pressure is equal to the liquid pressure. This can happen only if the steam is at the
saturation temperature corresponding to the liquid pressure, i.e., 100°C assuming the
liquid is at 1 atm.
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19. ii) The critical (or maximum) superheat, ΔTsat,cr, is inversely proportional to the
minimum radius of curvature of the bubble, as it grows at the cavity mouth:
where K is the proportionality constant (K=2σT 2 R* sat /(hfgPℓ)), which depends on
fluid and pressure, and rmin depends on the cavity radius, rc, and the contact angle, θ,
as follows:
Using Eqs. (21) and (22) for ΔTsat,cr=2°C, rc=1 μm and θ=135°, one finds K≈2.828
μm°C. Thus, for rc=3 μm and θ=45°, ΔTsat,cr≈0.94°C.
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