11.8 Two-phase pressure drop 391

hNcB = 1.953 × (Tw − Tsat )0.24 × (psat (Tw )− psat (Tsat ))0.75 × S

Qboil = hTP A(Tw − Tsat ) = hTP A(Tw − 50)

[Note that the saturation temperature on the boiling side is 50 °C and that on the condensation side is 100 °C.]

Please note that in this problem too, Tw is unknown, and consequently, “tedious” iterations to solve the problem are unavoidable. Discerning readers will note that problems on boiling and condensation are eminently solvable with a computer code or tools like MATLAB or SCILAB.

Initial guess:

Let Tw = 100 + 50 = 75 C 2

Qcondens = 146967 A J/s =146967 × π Dz

=146967 × π × 0.025 × 0.6

=6925.45W

At inlet, x = 0 (no two-phase) and hence hNcB = 0. Also, at inlet, F = 1 as

1 = 1.6726

Xtt

F = 3.748

hc = 9338.63W/m2K

ReTP = 93788.79

S = 0.375

hNcB = 1.953 × (75 − 50)0.24 × (38.59 × 103 − 12.33 × 103 )0.75 × 0.375 = 3271.2W/m2K

392 CHAPTER 11  Boiling and condensation

hTP,out = hc,out + hNcB,out

=9338.63 + 3271.2 =12609.83W/m2 K

hTP = 2521.53 + 12609.83

2

= 7565.68 W/m2 K

Qboil = hTP π D z (Tw − Tsat )

=7565.68 × π × 0.025 × 0.6 × (75 − 50)

=8912.84 W

Q condens = 6925.45W

Since there is a large difference between Qboil and Qcondens, the assumed Tw is incorrect.

Let Tw = 70 °C. Repeating the above calculations for Tw = 70 °C,

Qboil = 7021.49 W

Qcondens = 7940.24 W

Now Qcondens > Qboil, but the difference is smaller.

Let Tw = 71 °C. Repeating the above calculations for Tw = 71 °C,

Qboil = 7483.4 W

Qcondens = 7740.8W

Qboil and Qcondens are quite close now.

Let Tw = 71.5 °C. Repeating the above calculations for Tw = 71.5 °C,

Qboil = 7565.67 W

Qcondens = 7640.57W

Now the difference between Qboil and Qcondens is very small, and it is approximately 1%. The corresponding average heat transfer coefficients on the boiling side and the

condensation side are 7467.6 W/m2 K and 5689 W/m2 K, respectively. Therefore,

Note

The calculation can be repeated taking liquid properties (other than hfg) in the expression for laminar film heat transfer coefficient corresponding to (Tw + Tsat)/2. The calculation process needs to be repeated until the obtained Tw matches the value used

11.8 Two-phase pressure drop 393

for evaluating the liquid properties. The final Tw is expected to be slightly different from the one obtained above (71.5 °C) and so are the values in (a), (b), (d), and (e).

As the quality continuously changes (and hence the heat transfer coefficient) along the flow direction, accurate design of heat exchangers involving flow boiling or flow condensation needs numerical methods. This is because the local heat transfer coefficient is a complex function of quality and moreover the local heat transfer coefficient also depends on the wall superheat (for boiling) or wall subcooling (for condensation), which is unknown. Hence, for accuracy, it needs to be solved numerically (and by iteration) taking small control volumes and using the energy balance and the heat convected.

Problems

11.1Refer to Fig. 11.4. Derive the expressions for the minimum and maximum sizes of the cavity for nucleation from a heated surface, with the liquid pool at the saturation temperature.

11.2A hot copper sphere with a diameter of 1.5 cm and a temperature of 115 °C is dipped into a vessel containing water at atmospheric pressure and temperature maintained at 100 °C. Boiling commences in the nucleate boiling regime. Estimate the time required for the sphere to reach 105 °C. The density and specific heat of copper are 8933 kg/m3 and 475 J/kgK, respectively. Note the following assumptions:

(i)Sphere temperature is uniform (internal conduction resistance can be neglected).

(ii)Use the Rohsenow correlation (over the entire range: 5 to 15 K superheat).

Csf value in the Rohsenow correlation is 0.01. Fluid properties correspond approximately to Tsat.

11.3Plot the variation of critical heat flux for pool boiling of water on a flat surface as a function of system pressure, considering the variation of saturated fluid properties with pressure.

11.4Consider the copper sphere in Problem 11.2. During a quenching operation, the sphere initially at 300 °C is suddenly dipped into a water bath containing saturated water at atmospheric pressure. Since the temperature difference is large, boiling begins in the film boiling regime. Determine (a) the initial heat transfer coefficient, (b) the initial rate of change of sphere temperature, and (c) the initial rate of heat transfer. Neglect the effect of radiation.

11.5Water boils on a flat polished stainless steel surface (Csf = 0.013) at atmospheric pressure. (a) Estimate the heat flux from the surface to water if the surface temperature is 400 °C. If it is in the film boiling regime, the emissivities of the wall and liquid surfaces may be assumed to be equal to 1. (b) For the same heat flux, determine the wall temperature if it were in the nucleate boiling region.

11.6A furnace wall riser, 19 m long, 76 mm outer diameter, and 6 mm thick receives saturated water at 86 bar and 1.5 m/s velocity. The average wall heat flux based on the outer diameter is 95 kW/m2. Determine

394CHAPTER 11  Boiling and condensation

(a)the local heat transfer coefficient and the inner wall temperature at the riser exit,

(b)the average heat transfer coefficient, and

(c)the pressure drop across the riser.

11.7Consider Example 11.3. Assume an inlet subcooling of 40 K and a singlephase heat transfer coefficient of 5 kW/m2K. Determine

(a)the lower limiting value of the critical heat flux,

(b)the upper limiting value of the critical heat flux,

(c)the critical heat flux,

(d)the exit quality when the critical heat flux occurs, and

(e)the percentage change in the critical heat flux if the mass flux is doubled (keeping the other parameters constant).

11.8In a cogeneration plant, steam is used for maintaining a desired drying temperature. Saturated steam at 100 bar and 0.48 kg/s enters a 5 m long vertical tube of 15 cm diameter and flows downward. Steam transfers heat (by convection and radiation) to the surroundings maintained at 50 °C. The overall heat transfer coefficient between the steam and the surroundings is 104 W/m2 K. Determine the exit quality, and comment on the contributions of the frictional, acceleration (or deceleration), and gravitational head pressure drops to the total pressure drop, using the homogeneous model.

11.9Saturated steam at atmospheric pressure condenses on a vertical flat plate of 1.25 m long, which is maintained at 60 °C by the flow of cool water on the other side of the plate. Determine the flow regime (wave-free laminar, wavy laminar, or turbulent) at the bottom of the plate, and the rate of steam condensation per unit width of the plate.

11.10A condenser design specifies a square (inline) array of 225 horizontal tubes, each of 16 mm outer diameter and 1 m long, to condense saturated steam at 0.123 bar on the outer surface of the tubes, with a tube wall temperature of 45 °C maintained by the flow of cooling water through the tubes. Determine

(a)the heat transfer coefficient for the top-most tube,

(b)the heat transfer coefficient for the bottom-most tube,

(c)the average heat transfer coefficient for the array of tubes, and

(d)the total condensation rate.

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Bromley, L.A., 1950. Heat transfer in stable film boiling. Chem. Engng. Prog. 46, 221–227. Chen, J.C., 1963. A correlation for boiling heat transfer t o saturated fluids in convective flow.

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