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382 CHAPTER 11  Boiling and condensation

Table 11.4  Properties of saturated steam (Meyer et al., 1993; Harvey et al., 2000).

model overestimating the acceleration pressure drop, as it neglects the slip between the phases and assumes equal acceleration for both liquid and vapor phases. It may be noted that the homogenous model gives accurate results for dispersed flow – bubbly flow with small bubbles dispersed in liquid and spray flow or drop flow with small droplets dispersed in vapor, as the vapor velocity is approximately equal to the liquid velocity. The homogenous model also gives accurate results for higher pressures where vapor properties approach liquid properties. For accurate estimates of pressure drop over a wide range of operating conditions and vapor qualities, the correlations based on the separated flow model are recommended (Collier and Thome, 1994; Stephan, 1992; Jayaramu et al., 2019).

384 CHAPTER 11  Boiling and condensation

Therefore,

= 3.988 ×106 qcrit 0.384 + 0.9

= 3.104 ×106 W/m2

From energy balance, the exit quality can be determined as follows:

It is instructive to note the order of magnitude of critical heat flux for flow boiling. As a rule of thumb it goes as MW/m2, for water.

Example 11.4: For the problem in Example 11.3, with heat flux equal to the critical heat flux, determine the local heat transfer coefficients and the corresponding wall temperatures at two locations: z = 0.4 m and 0.8 m from the inlet. Use the Chen correlation.

Solution:

Saturated water properties corresponding to 10 bar are:

k f = 0.677 W/mK, Prf = 0.967, σ = 0.04226 N/m, cpf = 4.403 × 103 J/kg K,

ρg = 5.16 kg/m3 , ρf = 887 kg/m3 , hfg = 2014.9 × 103 J/kg, µf = 148.5 × 10−6 Ns/m2 , µg = 15.02 × 10−6 Ns/m2

Using Eq. (11.34),

386 CHAPTER 11  Boiling and condensation

Tw = Tsat (10 bar) + 50 C

=180 + 50 C

=230 C

psat (Tw ) = 27.98 bar = 27.98 × 105 N/m2

psat (Tw ) − psat (Tsat ) = (27.98 − 10) × 105 = 17.98 × 105 N/m2

q = hTP (Tw − Tsat )

=24555 + 0.0963 × (Tw − Tsat )0.24 (psat (Tw ) − psat (Tsat ))0.75 (Tw − Tsat )

=24555 + 0.0963 × 500.24 × (17.98 × 105 )0.75 × 50 =1.832 × 106 < 3.104 × 106 W/m2

Let (Tw − Tsat ) = 65 C

Tw = Tsat (10 bar) + 65 C =180 + 65 C

= 245 C

psat (Tw ) = psat (245 C) = 36.62 × 105 N/m2

psat (Tw ) − psat (Tsat ) = (36.62 − 10) × 105 = 26.62 × 105 N/m2

q = hTP (Tw − Tsat )

q= 24555 + 0.0963 × 650.24 × (26.62 × 105 )0.75 × 65 = 2.719 × 106 < 3.104 × 106 W/m2

Let (Tw − Tsat ) = 70 C

Tw = 180 + 70 C = 250 C psat (Tw ) = 39.776 × 105 N/m2

psat (Tw ) − psat (Tsat ) = 29.776 × 105 N/m2

q = hTP (Tw − Tsat )

= 3.058 × 106 ≈ 3.104 × 106 W/m2

Therefore, at z = 0.4 m,

11.8 Two-phase pressure drop 387

=2.35 × (7.07 + 0.213)0.736

=10.13

hc = 0.023 Re0f .8 Prf0.4 kDf F

=36349W/m2K

ReTP = Ref × F1.25

=9.244 × 105

Using S = 0.0398 in the expression for hNcB,

hNcB = 0.0561× (Tw − Tsat )0.24 (psat (Tw )− psat (Tsat ))0.75

hTP = hc + hNcB

q = hTP (Tw − Tsat ) = 36349 + 0.0561× (Tw − Tsat )0.24 (psat (Tw )− psat (Tsat ))0.75 (Tw − Tsat )

Let(Tw − Tsat ) = 65 C

Tw = Tsat + 65 C = 180 + 65 C = 245 C

psat (Tw )− psat (Tsat ) = 26.62 × 105 N/m2

Substituting the above values in the expression for q,

q = 3.017 ×106 ≈ 3.104 ×106 W/m2

Therefore, at z = 0.8 m,

hTP = 3.017 ×106 = 46415W/m2K 65

Tw = 245 C

Note that this sample problem does not ask for the calculation of the heat transfer coefficient at z = 0.9 m (tube exit). It is intentional, as the heat flux is equal to the critical heat flux, which indicates dryout at the exit. The Bowring CHF correlation presented in the chapter (or for that matter any other CHF correlation) estimates the CHF that corresponds to the onset of dryout (or DNB) at the tube exit. The Chen correlation should not be used to calculate the two-phase heat transfer coefficient for dryout and post-dryout (spray flow or drop flow) heat transfer, though the correlation gives some values of the heat transfer coefficient. The heat transfer coefficient calculated at the tube exit using the Chen correlation can at best be taken as the value prior to the onset of dryout at the tube exit.

388CHAPTER 11  Boiling and condensation

Example 11.5: For the problem in Example 11.3, with heat flux equal to the critical heat flux, evaluate the pressure drop across the tube using the homogeneous model.

Solution:

Fluid properties corresponding to 10 bar are:

f = 148.5 × 10−6 Ns/m2 , g = 15.02 × 10−6 Ns/m2 Specific volumes v f = 1.127 × 10−3 m3 /kg, vg = 0.1937m3 /kg

v fg = vg − v f = 0.1926 m3 /kg

The tube exit quality, xe = 0.4436

Fanning friction factor at the tube inlet:

Using the Blasius equation,

Fanning friction factor at the tube exit:

Two-phase viscosity can be calculated from the relation

Using the Blasius equation,

= 0.079 =

fTP ReTP1/4 0.0031

The average value of fTP over the tube length is given by

The total pressure drop can be determined using Eq. (11.91). The frictional pressure drop from the homogeneous model is given by

= 3054 N/m2