This Applied Heat Transfer problem has not been fully understood or solved yet. PLEASE ANSWER ALL QUESTIONS FULLY AND SHOW ALL WORK, STEP-BY-STEP. SHOW HOW YOU OBTAINED YOUR ANSWERS. You will need all figures and images below. I know it is a rough one. I appreciate your patience. Let's try this one more time. The following information needed to evaluate the quality of the design (information also provided in the problem itself): - find the percentage of thermal resistance on the gas and air sides using procedure from 3.9.9 figure - iterate for the exit temperatures and effectiveness - is this a thermally balanced design? - relative pressure losses for each stream - Kc and Ke values on figure 1 of 3.9.10 - using the ideal gas law R=1717 ft^2/s^2 - *R (shown below in problem), calculate the inlet and exit densities of both streams - Are the pressure drops balanced? - What fraction of the total pressure drop is due to entrance and exit losses - Is this heat exchanger a good selection for our application? - If not, what design features would yield a better design? - What would the effect be on the heat exchanger size and cost? - How would it affect "total ownership costs"? - Comment on your comparison between the available and the required UA values.

- fin length for air and gas sides.

- what is the A/Ac?

- what is the fin efficiency?

- what are the air density? (T in Rankine)

- surface area ratio

- exit densities

- use air properties from A-15E

3.9.9 Calculation procedure for a rating problem R. L. Webb for each Fluid stream This section outlines a step-by-step procedure for 7. Calculate a = jG.Com Pr-2/3 or a = Nu k/Dn. the thermal calculations involved in a rating problem. 8. Calculate fin efficiency. With mi = (2a/ Steps 4-9 are required for each fluid stream,_ ko)1/2/ calculated, the fin efficiency (n) is determined 1. Surface geometry parameters: Given B, A,/A, from the appropriate fin efficiency chart or equation. Dn. Calculate y, o, for each stream. 9. Calculate surface efficiency, no = 1 - (1 - 7/) 2. Using the given heat exchanger size, calculate (AffA). C 10. Calculate heat transfer surface areas and deter- the frontal mass velocity (G,) for each stream. Then mine UA. A simpler approach is to. use the given heat calculate Ge using Ge = GAY/o. 3. Estimate e to allow calculation of the average exchanger volume () and calculate UA from For fluid temperature for each stream. This may be based V + on an outright guess, or a preliminary calculation UA no ay/ 2 (1) each following steps 4-12. fled 4. Evaluate fluid properties (Pm, nm, Cpm) at 11. Calculate Cmin/Cmax and NTU = UA/Cmin. stread the estimated average fluid temperature. 12. Using the parameters in step 1 1, determine e 5. Calculate Re = Dn Gem - from the e-NTU chart (or equation) for the given heat 6. Determine f and / (or Nu) from / and f versus exchanger flow arrangement. Re plots for the surface, or from tabled laminar flow 3. Compare calculated e with estimated e. Repeat solutions, where appropriate. steps 4-12 as necessary to obtain desired accuracy.3.9 COMPACT HEAT EXCHANGERS 3.9.10-1 3.9.10 Pressure drop calculation O R. L. Webb After the heat exchanger geometry is fixed, the pressure loss may be calculated. The pressure loss for each stream through the heat exchanger finned pas- ages is calculated by AP = Ge (Ke + 1 - 02) 2 2 +JA P1 -PI Pm Ac Ket LAMINAR Na 2 000 - 02 - Ke) 3000 (1) 5000 P 2 10 000 0.4- The terms of Eq. (1) are entrance loss, flow accelera- Kc and a3- tion loss, core friction, and exit loss, respectively. The 02- Ke and Ke values depend on the cross-sectional flow geometry, o, and Re. Figure 1 gives Ke and Ke for a 1:1 ratio channel [1]. Kays and London [1] give - al - 16 : 10000 similar curves for two other channel geometries (paral- 5000~ Q3 - 3000 lel-plate channel and triangular channel). The entrance - Q4 2000 LAMINAR and exit losses are normally less than 10% of the total as - core loss, so the data of Fig. I will cover most -OB - C situations with adequate accuracy. The differences due to other channel configurations should cause only a -as - second-order effect for most situations. Figure 1 is (9001 02 03 04 05 06 07 08 09 10 based on a uniform velocity entering the heat ex- changer and fully developed flow in the core, and at Figure 1 Entrance and exit loss coefficients for multiple square channels with abrupt contraction entrance and abrupt expansion the exit. This assumption is not valid for interrupted exit. fin surfaces. In this case, Kays and London [1] recommend use of the Re =0. curves to evaluate Kc the densities entering and leaving the core (P1, p2), the and Ke. In this case, all channel configurations have core AP may be calculated using Eq. (1). the same Ko and K. values for Reco. The calculation If the heat exchanger construction contains mani- steps for the core pressure loss requires the informa- folds and flow distributor fins, one must also account tion developed through step 6 of the rating calculation for these losses. These losses may be a significant procedure. After calculation of the Ke, Ke, values and fraction of the total pressure loss. @ 1983 Hemisphere Publishing Corporation HEDH3.9.10-2 3.9 COMPACT HEAT EXCHANGERS / 3.9.10 Pressure Drop Calculation Kays and London [1] recommend calculation of inlet flow energy dissipated as flow losses. In a bal- the losses for each stream in the form AP/P1, where anced gas-to-gas design, one desires approximately P1 is the inlet pressure. This defines the fraction of the equal values of AP/P for each stream. Nomenclature and References for Section 3.9.10 appear at the end of Section 3.9.13. O 1983 Hemisphere Publishing Corporation HEDH953 APPENDIX 2 TABLE A-15E Properties of air at 1 atm pressure Specific Thermal Thermal Dynamic Kinematic Prandtl Temp. Density Heat Conductivity Diffusivity Viscosity Viscosity Number T, OF p, Ibm/fts c., Btu/Ibm.R k, Btu/h-ft-R c, ft2 /s M, Ibm/ft.s v, ft /s Pr -300 0.24844 0.5072 0.00508 1.119 x 10-5 4.039 x 10-6 1.625 x 10-5 1.4501 -200 0.15276 0.2247 0.00778 6.294 x 10-5 6.772 x 10-6 4.433 x 10-5 0.7042 -100 0.11029 0.2360 0.01037 1.106 x 10-4 9.042 x 10-6 8.197 x 10-5 0.7404 -50 0.09683 0.2389 0.01164 1.397 x 10-4 1.006 x 10-5 1.039 x 10-4 0.7439 0 0.08630 0.2401 0.01288 1.726 x 10-4 1.102 x 10-5 1.278 x 10-4 0.7403 10 0.08446 0.2402 0.01312 1.797 x 10-4 1.121 x 10-5 1.328 x 10-4 0.7391 20 0.08270 0.2403 0.01336 1.868 x 10-4 1.140 x 10-5 1.379 x 10-4 0.7378 30 0.08101 0.2403 0.01361 1.942 x 10-4 1.158 x 10-5 1.430 x 10-4 0.7365 40 0.07939 0.2404 0.01385 2.016 x 10-4 1.176 x 10-5 1.482 x 10-4 0.7350 50 0.07783 0.2404 0.01409 2.092 x 10-4 1.194 x 10-5 1.535 x 10-4 0.7336 60 0.07633 0.2404 0.01433 2.169 x 10-4 1.212 x 10-5 1.588 x 10-4 0.7321 70 0.07489 0.2404 0.01457 2.248 x 10-4 1.230 x 10-5 1.643 x 10-4 0.7306 80 0.07350 0.2404 0.01481 2.328 x 10-4 1.247 x 10-5 1.697 x 10-4 0.7290 90 0.07217 0.2404 0.01505 2.409 x 10-4 1.265 x 10-5 1.753 x 10-4 0.7275 100 0.07088 0.2405 0.01529 2.491 x 10-4 1.281 x 10-5 1.809 x 10-4 0.7260 110 0.06963 0.2405 0.01552 2.575 x 10-4 1.299 x 10-5 1.866 x 10-4 0.7245 120 0.06843 0.2405 0.01576 2.660 x 10-4 1.316 x 10-5 1.923 x 10-4 0.7230 130 0.06727 0.2405 0.01599 2.746 x 10-4 1.332 x 10-5 1.981 x 10-4 0.7216 140 0.06615 0.2406 0.01623 2.833 x 10-4 1.349 x 10-5 2.040 x 10-4 0.7202 150 0.06507 0.2406 0.01646 2.921 x 10-4 1.365 x 10-5 2.099 x 10-4 0.7188 160 0.06402 0.2406 0.01669 3.010 x 10-4 1.382 x 10-5 2.159 x 10-4 0.7174 170 0.06300 0.2407 0.01692 3.100 x 10-4 1.398 x 10-5 2.220 x 10-4 0.7161 180 0.06201 0.2408 0.01715 3.191 x 10-4 1.414 x 10-5 2.281 x 10-4 0.7148 190 0.06106 0.2408 0.01738 3.284 x 10-4 1.430 x 10-5 2.343 x 10-4 0.7136 200 0.06013 0.2409 0.01761 3.377 x 10-4 1.446 x 10-5 2.406 x 10-4 0.7124 250 0.05590 0.2415 0.01874 3.857 x 10-4 1.524 x 10-5 2.727 x 10-4 0.7071 300 0.05222 0.2423 0.01985 4.358 x 10-4 1.599 x 10-5 3.063 x 10-4 0.7028 350 0.04899 0.2433 0.02094 4.879 x 10-4 1.672 x 10-5 3.413 x 10-4 0.6995 400 0.04614 0.2445 0.02200 5.419 x 10-4 1.743 x 10-5 3.777 x 10-4 0.6971 450 0.04361 0.2458 0.02305 5.974 x 10-4 1.812 x 10-5 4.154 x 10-4 0.6953 500 0.04134 0.2472 0.02408 6.546 x 10-4 1.878 x 10-5 4.544 x 10-4 0.6942 600 0.03743 0.2503 0.02608 7.732 x 10-4 2.007 x 10-5 5.361 x 10-4 0.6934 700 0.03421 0.2535 0.02800 8.970 x 10-4 2.129 x 10-5 6.225 x 10-4 0.6940 800 0.03149 0.2568 0.02986 1.025 x 10-3 2.247 x 10-5 7.134 x 10-4 0.6956 900 0.02917 0.2599 0.03164 1.158 x 10-3 2.359 x 10-5 8.087 x 10-4 0.6978 1000 0.02718 0.2630 0.03336 1.296 x 10-3 2.467 x 10-5 9.080 x 10-4 0.7004 1500 0.02024 0.2761 0.04106 2.041 x 10-3 2.957 x 10-5 1.460 x 10-3 0.7158 2000 0.01613 0.2855 0.04752 2.867 x 10-3 3.379 x 10-5 2.095 x 10-3 0.7308 2500 0.01340 0.2922 0.05309 3.765 x 10-3 3.750 x 10-5 2.798 x 10-3 0.7432 3000 0.01147 0.2972 0.05811 4.737 x 10-3 4.082 x 10-5 3.560 x 10-3 0.7516 3500 0.01002 0.3010 0.06293 5.797 x 10-3 4.381 x 10-5 4.373 x 10-3 0.7543 4000 0.00889 0.3040 0.06789 6.975 x 10-3 4.651 x 10-5 5.229 x 10-3 0.7497 Note: For ideal gases, the properties co, k, A, and Pr are independent of pressure. The properties p, i, and a at a pressure (in atm) other than 1 atm are determined by multiplying the values of p at the given temperature by Pand by dividing r and a by P. Source: Data generated from the EES software developed by S. A. Klein and F. L. Alvarado. Original sources: Keenan, Chao, Keyes, Gas Tables, Wiley, 1984; and Thermophysical Properties of Matter, Vol. 3: Thermal Conductivity, Y. S. Touloukian, P. E. Liley, S. C. Saxena, Vol. 11: Viscosity, Y. S. Touloukian, S. C. Saxena, and P. Hestermans, IFV/Plenun, NY, 1970, ISBN 0-306067020-8.1.0 Fin efficiency for plane and circular fins .9 Plane fins m2 = 2h/kt 8 If conv. from tip, L. = L + 1/2 .7 ni 6 .5 .4 .3 0 .2 4 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 m (r - r;) or mL Figure 2.4 Fin efficiency of straight and circular fins, with insulated fin tip.Gas Turbine Recuperator Rating Problem This refers to the recuperator we need for the gas turbine peaking unit at power plant. Another quote has just come in, and it is offered at lower cost than a previous design problem (note: a few previous design operating conditions are listed below, if needed). Your job is to check out the design and determine if it will meet our specified = 0.75 requirement. Also calculate the pressure drop, including entrance and exit losses and flow acceleration. The H-X size is 6.5 ft x 3.0 ft (frontal area of air side) and 6.5 ft x 6.5 ft (frontal area of gas side). The H-X is a cross flow (unmixed) design. The attached figure gives the surface geometry and performance characteristics of the air and gas side surfaces. The parting sheet thickness between the air and gas sides is 0.012 in. All materials are stainless steel (k = 12 btu/hr-ft-F). Neglect the wall resistance in the overall conductance equation, but do use & to calculate the surface efficiency. Evaluate the quality of the design by determining: 1. Follow the calculation procedure outlined in Section 3.9.9 of the Compact Heat Exchangers handout and find the percentage of the thermal resistance on the gas and air sides. Ensure that you are using the average air and gas temperatures to evaluation fluid properties as you iterate for the exit temperatures and the effectiveness of the H- X. Assume that the properties of the gas are well approximated with air properties. Is this a thermally balanced design? Use eq. (1) of the handout to determine this. 2. Therelative pressure losses AP/P for each stream, where here P represents the pressure and not the pumping power. Use the Ke and Ke values given in Fig. | of the Section 3.9.10 Compact Heat Exchanger handout (use the Re = curves for simplicity). In addition, use the ideal gas law with R = 1717 ft?/s?-R to calculate the inlet and exit densities of both streams. (Assume that AP is small enough that you can use the inlet pressures to calculate the exit densities.) Are the pressure drops balanced? What fraction of the total pressure drop is due to entrance/exit losses? 3. Is this H-X a good selection for our application? If not, what design features- would yield a better design? What would the effect be on H-X size and cost? How would it affect "total ownership costs"? 4. Comment on your comparison between the available and the required UA values. New design information: Answers Air Gas y(ft?7/tt) 125.2 | 213.8 0 (A/Ar:) | 0.457 | 0.397 Re 4,595 | 1,173 7 0.57 0.75 1 0.73 0.80 APIP 0.0054 | 0.076 Previous design operating conditions (for comparison to new design, if needed): Turbine Exhaust Compressor Discharge Temperature (F) 805 Pressure (psia) 16.1 Flow rate (lbm/hr) 196,000 193,000 Specific heat (Btu/lbm F) 0.259 0.251 Fig. 10-38. Louvered plate-fin surface 3/8-6.06. Fig. 10-56. Strip-fin plate-fin surface 1/2-11.94(D). _0.055' 100 0838" 500 237 .080 080 060 .060 .050 050 - 0.375- .040 .040 .030 .030 .020 1.020 .015 .015 - BEST INTERPRETATION 010 .008 - BEST INTERPRE .010-at 006 - ( h / Gc p ) N pr LET 005 .006 004 - 1005 No x10-3 (arNG/4) 0.4 :06 0.8 10 5 2.0 3.0 40 6.0 8.0 10 Fin pitch = 6.06 per in. .002 - NR ( +41 , G/#18 10 3 Plate spacing, b = 0.250 in. Louver spacing = 0.375 in. Louver gap = 0.055 in. Fin pitch = 11.94 per in. Fin gap = 0.110 in. Plate spacing, b = 0.237 in. Flow passage hydraulic diameter, 4rx = 0.01460 ft Splitter symmetrically located Fin metal thickness = 0.006 in., aluminum Fin length flow direction = 0.500 in. Total heat transfer area/volume between plates, B = 256 ft2/ft? Flow passage hydraulic diameter, 4rx = 0.007436 ft Fin area/total area = 0.640 Fin metal thickness = 0.006 in., aluminum Splitter metal thickness = 0.006 in. Total heat transfer area/volume between plates, B = 461.0 ft?/ft3 NR NSINp, 213 Fin area (including splitter)/total area = 0. 796- = 0.796 -6.06 NR VSuVp,2/3 10,000 0.00551 0.0331 8,000 0.00593 0.0340 1-11.94(D) 6,000 0.00651 0.0354 5,000 0.00690 0.0363 9,000 0.0122 4,000 0.00738 0.0375 3,000 0.0123 3,000 0.00805 0.0394 7,000 0.00452 0.0126 2,500 0.00849 0.0406 5,000 0.00471 0.0131 ,000 0.00900 0.0426 5,000 0.00492 0.0137 1,500 0.00970 0.0461 4,000 0.00522 0.0146 3.000 1,200 0.0104 0.0496 0.00575 0.0162 2,00 1,000 0.0112 0.0532 0.00682 0.0198 1.500 0.0074+ 0.0231 800 0.0124 0.058 600 0.0144 0.0682 1,200 0.00830 0.0265 1,000 0.0091 1 0.0306 500 0.0160 0.0755 800 0.01045 0.0347 400 500 0.01255 0.0429 300 500 0.01415 0.0493 400 0.0166 0.0592 Air side ~ subscript "a" 300 0.0205 0.0758 200 Gas side ~ _subscript "g'Step 1: Calculate the Thermal Resistance for Air and Gas Sides The thermal resistance (R) for both the gas and air sides is calculated using the following formula: 1 R =- h . A Where: . h = heat transfer coefficient (in Btu/hr.ft2-'F for air and gas), . A = surface area (in ft? ). Air Side: . ha = 25 Btu/hr.ft2.'F (assumed for air side), . Surface area An = 19.5 ft? (calculated from the dimensions 6.5 ft x 3.0 ft). Fair = 25 x 19.5 487.5 = 0.00205 hr\\cdotpft?\\cdotp *F/Btu Gas Side: . hg = 100 Btu/hr-ft2. #F (assumed for gas side), . Surface area A, = 42.25 ft? (calculated from the dimensions 6.5 ft x 6.5 ft). Rgas= 100 x 42.25 4225 = 0.000236 hr\\cdotpft2\\cdotp F/Btu Total Thermal Resistance: Rtotal = Rair + Rgas = 0.00205 + 0.000236 = 0.002286 hr\\cdotpft2\\cdotp*F/Btu Percentage of Thermal Resistance: For air side: Percentage for air side = Fair 0.00205 x 100 = x 100 = 89.8% Rtotal 0.002286 For gas side: Rgas 0.000236 Percentage for gas side = x 100 = x 100 = 10.3% Rtotal 0.002286 Step 2: Exit Temperatures and Effectiveness The effectiveness of a heat exchanger is defined as: Tinlet, hot - Texit, hot E= Tinlet, hot - Tinlet, cold Where: . Tinlet, hot = inlet temperature of the hot fluid (gas), . Texit, hot = exit temperature of the hot fluid, . Tinlet, cold = inlet temperature of the cold fluid (air). We need to iterate for the exit temperatures to find when the heat exchanged matches the required effectiveness. However, to continue with the calculations, the specific values for the inlet temperatures and the mass flow rates must be provided.Step 3: Thermally Balanced Design A thermally balanced design means that the thermal resistances on both sides (air and gas) should be approximately equal. Here, we see that the air side has a much larger thermal resistance (89.8%), suggesting the design might not be well-balanced. This imbalance means the heat transfer on the air side is more limited than on the gas side. To balance the design, you might need to increase the heat transfer area for the gas side (e.0., by adjusting fin geometry). Step 4: Relative Pressure Losses for Each Stream The pressure loss due to friction in a heat exchanger is given by: oe AP= - (A. + KK.) 2p Where: * G =mass flow rate (in Ibm/hr), * = density (in lb/ft), J, and A, = pressure loss coefficients for the air and gas sides, respectively. To calculate the pressure loss, we will need: 1. The flow rates and density for both the air and gas streams (calculated next using the ideal gas law). 2. The A, and A, values from the charts in the reference book. Once we extract the Reynolds numbers (Re) for both the air and gas flows, we can find the A, and A, values from the provided charts. Step 5: Ke and Ke Values The values of AY, and A, depend on the flow regime (laminar, transitional, or turbulent) and the geometry of the heat exchanger. Using the Reynolds number and the flow canditions from the problem setup, we would extract A, and A', values from the provided charts. These values are necessary to calculate the pressure losses in the heat exchanger. Step 6: Inlet and Exit Densities Calculation Using the Ideal Gas Law The ideal gas law is: PRT Where: * p= density, P =pressure (in lbm/fi*-s*), = ideal gas constant for air (1717 it js? Ri, J = temperature in Rankine. Given the inlet pressure and temperatures, we can calculate the densities at the inlet and exit for both the air and gas streams. Step 7: Pressure Drops Balance The pressure drop is the difference between the inlet and exit pressure, calculated using the pressure drop equation above. We will check if the pressure drops on both the air and gas sides are balanced (i_., similar in magnitude). Step 8: Fraction of Total Pressure Drop Due to Entrance and Exit Losses The entrance and exit losses are typically calculated as: Vw A Pontrance = Loss coefficient: Where WV is the velocity at the entrance and exit. Once the entrance and exit losses are calculated, we will determine the fraction of the total pressure drop caused by these losses. Step 9: Heat Exchanger Suitability for the Application To determine if the heat exchanger is suitable: 1. Thermal Effectiveness: Should be close to 0.75 (or whatever is specified). 2. Pressure Drop: Should not be excessively high. 3. Surface Area: Should be sufficient to meet the heat transfer requirements. lf the heat exchanger fails to meet these criteria, a redesign is required. Step 10: Design Features for a Better Design lf the current heat exchanger does not meet the required effectiveness and pressure drop values: |ncrease the surface area: More fins or larger fins. Optimize flow arrangement: Consider counterflow, crossflow, or parallel flow almangements. Step 11: Effect on Heat Exchanger Size and Cost lf the surface area is increased, the size of the heat exchanger will increase, which will increase manufacturing and operational costs. The increase in cost will depend on the materials and the change in surface area required. Step 12: Comparison of Available and Required UA Values We will compare the available overall heat transfer coefficient (UA) with the required UA value. If the available UA is insufficient, the heat exchanger size will need to increase to meet the requirements. Step 13: Fin Length for Air and Gas Sides The fin length can be calculated using the surface area and fin pitch: Surface Area Fin Length = Fin Pitch x Number of Fins Step 14: Fin Efficiency Calculations The fin efficiency is calculated using: tanh(ml) Of = mL Where: 2h . L = fin length, . t = fin thickness