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hw09_11_solns - NEEP 411: HW#9 Solutions Two—Phase Flow...

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Unformatted text preview: NEEP 411: HW#9 Solutions Two—Phase Flow Fundamentals Due: Friday, December 2nd, 2011 1. 121, El—Wakil Problem Statement: Water enters a 6—ft-10ng uniformly heated one inch tube at 460 OF and 600 psia. The entrance velocity is 2.5 ft/s. Neglecting pressure losses in the channel, determine the maximum heat input, in Btu/hr-ft of channel length, if the exit void fraction is not to exceed 60 percent. Solution: We’ll need an average slip ratio so that we can ﬁnd a corresponding exit quality, x3. We have enough information to determine the inlet condition and mass ﬂowrate, so we can infer the total heat added and then ﬁnd q; = Q/ L . Property values below were obtained with the latest version of EES. At 600 psia, T sat = 4862 OF. Relevant saturation properties at 600 psia are: 12]. : 0.02 ft3/lbm ; vg : 0.77 ft3/lbm 17f : 471.7 Btu/lbm ; hfg 2 732.2 Btu/lbm Also, the inlet conditions are: Vin : 0.0196 ft3/1bm (,0in = 51.01bm/ft3); 11m = 441.6 Btu/lbm Slip ratio: Figure 12—8 provides S E 2 at conditions similar to those of this problem (p = 600 psia, Vin ~ 2 ft/s, x ~ 0.05). Using this s1ip ratio and Equations 12—11 and 12—10: V y/z—szo'—02--2=0.0519 vg. 0.77 1 1 r: : —0.0723 *6 1[1—0581 1'04 1 a w 'o.60.0519 6 Mass ﬂow: 771 : pin V. A : (51.01bm/ft3)(2.5 ft/s)§(1 in)2 (1 ft/12 in)2 (3 600 s/hr) = 2503 lbm/hr In Total power: Q : 777(17/ + xehﬂg — h. )2 25031bm/hr(47l.7 + 0.0723(7322) ~ 441.6) Btu/lbm [[1 Q : 2.079 ><105 Btu/hr Linear heat rate: q; = Q/L = 2.079 ><105 Btu/hr/(6 ft) = 3.46 x 104 Btu/hr-ft 2. 12—2, El—Wakil Problem Statement: A boiling—water reactor operates at 600 psig. In one of its channels the inlet velocity is 1.4 ft/s, the heat generated is 5 x 106 Btu/hr, and the total coolant ﬂow is 76,000 1bm/hr. The incoming water is 11 OF subcooled. Estimate the exit void fraction in that channel. Solution: Here we have enough information in the problem statement to ﬁnd xe. It is then just a matter of finding a suitable slip ratio S to calculate as. It is somewhat unusual to be given the system pressure as a gage pressure (614.7 psia), but it appears to be given so that Figure 12.7 can be used to ﬁnd S. Property values below were obtained with the latest version of EES. At 614.7 psia, T sat = 488.9 OF. Relevant saturation properties at 614.7 psia are: 12]. 2 0.0202 ft3/lbm ; vg 2 0.751ft3/1bm hf 2 474.8 Btu/lbm ; hfg 2 728.9 Btu/lbm Also, the inlet conditions are: vin 2 0.01995 ft3/1bm (,0in 2 50.11bm/ft3); km 2 462 Btu/1bm We can then rearrange our energy balance over the core, 2 n'tUtf + x911 jg — hm ) , to ﬁnd the quality: _ Q/m - (hf — hm) w (5 ><106 Btu/hr/76,0001bm/hr)— (474.8 — 462)Btu/lb1n 728.9 Btu/lbm X C’ hfg x8 2 0.0727 From Figure 12.7 at Vin = 1.4 ft/s and Xe = 0.073, S~ 2.75. Then 0.0202 0.751 Vf w2~—S2 -2.75=0.074 VI; and from Equation 12.9: 0‘8 Z 11' Z 1 0 01727 20514 1+ “‘6 w 1+————'—-20.074 x 0.0727 6 are: 0.514 3. 12-6, El-Wakil Problem Statement: A 6-ft—high BWR channel operates at an average pressure of 700 psia. Water enters the channel 23.1 0F subcooled and leaves with a 6 percent steam quality. Calculate the nonboiling and boiling heights if heat is added along the channel (a) uniformly and (b) sinusoidally. Neglect the extrapolation lengths. Solution: In both cases, we may write: Q3 hf — hm Q, hf + xehfg — h. [11 Again, from EES, relevant saturation properties at 700 psia are: hf: 491.6 Btu/lbin and kg = 710.3 Btu/lbin. With 23.1 0F subcooling, hm = 464.5 Btu/lbin. Q. a 491.6 — 464.5 Btu/1bm _ 27.1 Q, 4916 + 0.06(710.3) — 464.5 Btu/lbm 69.7 (a) For uniform heat addition, : 0.389 H_Q[ H 0 Q“ 20.389 HO = 2.33 ft; HB = 3.67 ft (b) For sinusoidal heat addition, 1 H —[1 — cos(7rHU /H)] = Q" : 0.389 ; solving, 0 = 0.429 2 Q, H HO = 2.57 ft; HB = 3.43 ft 4. 12—7, El—Wakil Problem Statement: A 6-ft—high boiling water reactor channel operates at 700 psia average, 23.1 0F subcooling and 6 percent exit quality (as in the previous problem, 126). The voids in the upper part of the channel, however, cause a strong neutron-ﬂux depression there, so that the axial ﬂux distribution is represented by: (p(z) = C exp(—7zz / H) sin(7zz / H) where C is a constant, 2 : 0 indicates the channel entrance, and H is the channel height. Find (a) the height 2 at which the ﬂux is a maximum, (b) the value of C in terms of the maximum ﬂux, and (c) the boiling and nonboiling heights. Solution: (a) Find the axial position of maximum ﬂux: 5‘5? 2 C —£exp(—Jzz/H)sin(7zz/H) + leos(7zz/H)exp(—7zz/H) = 0 dz H H h The term in simpliﬁes to tan(7zz/H) = l or th/HZ 7t/4 so 2 = H/4 = 1.5 ft. (b) Find the value of C in terms of the maximum ﬂux: goo : Cexp(~7r/4) sin(7r/4) so C: 3.102%. (0) Solving for the boiling and nonboiling heights: .,u I Q. H J: exp(_”Z/H) 5111(7lZ/HMZ _ — (H/27r) exp(—7zz/H)[sin(7zz/H) + cos(7zz/H)] |f" _ H Q, fexp(_7z2 / H) Smog / HMZ T ~ (H / 27x) exp(—7zz/H)[sin(7zz / H) + cos(7zz/H)] lo H 1— exp(—7TH0 /H)[sin(7rH0 /H) + cos(7rH0 /H)] T 1+ exp(—7z) : 0.389 (from previous problem) This can be solved numerically using Maple or Matlab or whatever additional solver suits you; the result is HO/H = 0.275. H0 = 1.65 ft; HB = 4.35 ft ...
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hw09_11_solns - NEEP 411: HW#9 Solutions Two—Phase Flow...

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