HW3_Solution(2)

pdf

School

University of Massachusetts, Lowell *

*We aren’t endorsed by this school

Course

3820

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by PrivateWater4873

Homework 3: ANSYS Tube Banks In this assignment you will compare empirical relations for calculating heat transfer from tube banks against solutions using ANSYS Fluent. The tube bank has the following relevant dimensions and configuration: Inlet conditions are u ∞ = 3 m/s, and T i = 290 K with air as the fluid. The tube surface temperature is held constant at 350 K. 1 Pre-analysis Mathematical model The actual mathematical model used by ANSYS Fluent to solve the given problem is not something we’ve covered in detail because it is very difficult/not possible to solve by hand and thus we’ve focused on empirical correlations. It is still good practice to identify the mathematical model, including boundary conditions and assumptions, since that will inform your interpretation of results. The equations solved are called the Reynolds Averaged Navier-Stokes (RANS) equations, including the energy equation, with the k-omega SST turbulence model and the intermittency transition model. Assumptions include steady state, 2D planar geometry with an infinite (i.e., very large) number of tubes in each row (remembering rows are normal to the incoming fluid direction), and that air properties are constant with the following values: Density ( ρ ) 1.225 kg/m 3 Specific Heat ( c P ) 1006.43 J/kg-K Thermal Conductivity ( k ) 0.0242 W/m-K Dynamic Viscosity ( µ ) 1 . 7894 × 10 − 5 kg/m-s The boundary conditions are: uniform velocity inlet at 3 m/s, inlet temperature of 290 K, pressure outlet at 0 gauge, constant pipe surface temperatures at 350 K. Symmetric boundary conditions on the top and bottom surfaces allow us to approximate an infinite number of tubes in each row. Hand Calculations 1. What is the Reynolds Number for this tube bank? 2. Select the appropriate Nusselt Number correlation from Table 7-2 and 7-3 in your text. Write down which correlation you choose and how many rows the current tube bank has. 3. Calculate the Nusselt Number (please use the same (constant) air properties as defined in the Mathe- matical Model) and average convective heat transfer coefficient. 4. What is the outlet (a.k.a. exit) temperature of the fluid from the tube bank? 1

5. What is the total rate of heat transfer for the given tube bank? You should calculate your heat transfer rate for the same geometry as given in the ANSYS Fluent mesh, i.e. your surface area will be for a single tube per row and to get the true overall rate of heat transfer you would need to multiple your answer by however many tubes per row you actually had in the real system. Geometry, Mesh, Mathematical model setup, Numerical solution A mesh file is provided to you. You will need to import this mesh file by creating a new Fluent project in workbench, and then importing the mesh. After importing the mesh, double-click on Setup to load Fluent (accept the default launch parameters), and then follow the directions in the following YouTube video link to set up and solve the given problem in ANSYS Fluent: https://www.youtube.com/watch?v=ebiIkOdZtHU . The video was recorded using an older version of ANSYS Fluent and the intermittency model options have changed from a check box to a drop- down menu. Select the gamma-transport-eqn model from the drop down, and make sure that Production Kato-Launder is checked if it isn’t automatically selected. 2

2 Post-processing 1. Create contour plots of the temperature and velocity (2 different plots) in the fluid both using the cyan-yellow color map. 2. What is the mass-weighted average fluid temperature at the outlet in the simulation? 3. What is the total rate of heat transfer summed over all rows in the tube bundle. 4. Using the total rate of heat transfer and temperatures from the ANSYS simulation it is possible to reverse out an average convective heat transfer coefficient and Nusselt Number. What are the average convective heat transfer coefficient and Nusselt Number based on the ANSYS predicted temperatures and overall rate of heat transfer? 3 V&V 1. The original source for the correlations listed in the textbook gives their accuracy as ± 15%. Using the values derived from the correlation as the baseline values, are the Nusselt Numbers calculated with the correlation and the CFD results consistent to within the stated correlation accuracy (i.e. is their relative difference less than ± 15%)? 2. List 2 changes to the mathematical model you would make to improve the accuracy of the CFD simulations and explain why you think they would improve the simulation accuracy. 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Tables 4

Homework 3 Solution: ANSYS Tube Banks Pre-analysis Hand Calculations The first step is to find the fluid properties. The ANSYS calculations assume constant fluid properties with specific given property values, an assumption we will carry over into our hand calculations. The Prandtl Number is not explicitly provided, but can be calculated from the given values as Pr = c P µ k = 1006 . 43 J/kg-K · 1 . 7894 × 10 − 5 kg/m-s 0 . 0242 W/m-K = 0 . 744176 The next step is to find u max for the given inlet flow velocity and tube bundle geometry. Calculating the two possible u max values u 1 = u ∞ S T S T − D = 3 m/s 0 . 033 m 0 . 033 m − 0 . 015 m = 5 . 5 m/s u 2 = u ∞ S T / 2 q ( S T 2 ) 2 − S 2 L − D = 3 m/s 0 . 033 / 2 m q ( 0 . 033 m 2 ) 2 + (0 . 034 m) 2 − 0 . 015 m = 2 . 17180 m/s u 1 is obviously larger and will be used as u max . With u max and the fluid properties determined the Reynolds Number and Nusselt Number correlation may be determined. Reynolds Number is defined as Re D = ρu max D µ = 1 . 225 kg/m 3 · 5 . 5 m/s · 0 . 015 m 1 . 7894 × 10 − 5 kg/m-s = 5648 . 84 Based on this Reynolds Number, the third correlation listed in Table 7-2 for the staggered tube bank configuration should be used Nu D = 0 . 35 S T S L 0 . 2 Re 0 . 6 D Pr 0 . 36 Pr Pr s 0 . 25 Because the fluid properties are assume constant, the surface Prandtl Number is the same as the bulk Prandtl Number = 0 . 35 0 . 033 m 0 . 034 m 0 . 2 (5648 . 84) 0 . 6 (0 . 744176) 0 . 36 0 . 744176 0 . 744176 0 . 25 = 55 . 7709 5

This is for a tube bank with over 16 rows, but the current configuration only has 4, so a correction factor from Table 7-3 must be applied Nu D = F · Nu D,N L > 16 = 0 . 89 · 55 . 7709 = 49 . 6361 The convective heat transfer coefficient can be determined from the definition of the Nusselt Number h = Nu k D = 49 . 6361 · 0 . 0242 W/m-K 0 . 015 m = 80 . 0796 W/m 2 -K To calculate the outlet temperature the surface area and mass flow rate must be calculated. Because we are calculating for just a single column across 4 rows of the tube bank, the surface area is just four times the surface area of a single tube. A s = N L · πD = 4 · π · 0 . 015 m = 0 . 188496 m 2 / m The mass flow rate can be found from the inlet conditions ˙ m = ρu ∞ S T = 1 . 225 kg/m 3 · 3 m/s · 0 . 033 m = 0 . 121275 kg/s-m The extra meter in the denominator of the units of both A s and ˙ m are because we haven’t specified a tube length, but they will cancel each other out in the calculations of the outlet temperature. T e = T s − ( T s − T i ) e h − Ash ˙ mc P i = 350 K − (350 K − 290 K) exp − 0 . 188496 m 2 /m · 80 . 0796 W/m 2 -K 0 . 121275 kg/s-m · 1006 . 43 J/kg-K = 296 . 980 K Typically you would use this new exit temperature to re-calculate the bulk temperature and re-evaluate the fluid properties, but in this case to remain consistent with the ANSYS simulations we are leaving the properties and solution as-is. With the exit temperature now known, either the log mean temperature difference or a simple energy balance can be used to calculate the total rate of heat transfer. For the LMTD approach ∆ T LM = ( T s − T e ) − ( T s − T i ) ln T s − T e T s − T i = (350 − 296 . 980) K − (350 − 290) K ln 350 − 296 . 980 K 350 − 290 K = 56 . 4382 K ˙ Q = hA s ∆ T LM = 80 . 0796 W/m 2 -K · 0 . 188496 m 2 · 56 . 4382 K = 851 . 915 W/m 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Where the per meter units is because we haven’t specified a tube length. Using the energy balance approach ˙ Q = ˙ mc P ( T e − T i ) = 0 . 121275 kg/s-m · 1006 . 43 J/kg-K(296 . 980 − 290) K = 851 . 915 W/m Post-processing ANSYS results After following the directions of the ANSYS-provided video with the minor tweaks specified in the assignment directions you should have temperature and velocity contours like the following Of the requested quantities, 2 are directly available from Fluent itself. The mass-averaged fluid temperature at the outlet is 296.03 K, and the total heat transfer from all 4 rows is 735.745 W/m (ANSYS reports 7

it as Watts, but this is misleading since it is actually per unit length of tube). To calculate the average convective heat transfer coefficient and Nusselt number we need to inverse the calculation we performed for the hand-calculations in the pre-analysis. The first step is to calculate the Log Mean Temperature Difference ∆ T LM = ( T s − T e ) − ( T s − T i ) ln T s − T e T s − T i = (350 − 296 . 03) K − (350 − 290) K ln 350 − 296 . 03 K 350 − 290 K = 56 . 9318 K Then we can rearrange Newton’s Law of Cooling to solve for the average convective heat transfer coefficient, h ˙ Q = hA s ∆ T LM ⇒ h = ˙ Q A s ∆ T LM = 735 . 745 W/m 0 . 188496 m 2 /m · 56 . 9318 K = 68 . 5601 W/m 2 -K The Nusselt Number can then be calculated from its definition Nu = hD k = 68 . 5601 W/m 2 -K · 0 . 015 m 0 . 0242 W/m-K = 42 . 4959 V&V The values calculated with through hand-calculations and the ANSYS-derived values are not particularly close. This, however, may not mean there is a mistake in either calculation. The Nusselt Number correlation used has an uncertainty of ± 15%, and the actual percent difference (using the correlation as the baseline value) is Nu Fluent − Nu hand calc Nu hand calc = 42 . 4959 − 49 . 6361 49 . 6361 = − 0 . 14385 or about 14.4% smaller. This is within the ± 15% accuracy range for the given correlation and so it is not possible to say that either calculation is wrong (though ideally we’d still like them to be closer to one another). That being said, there are several ways in which the simulations, or the hand-calculations, could be im- proved. First and foremost is to use variable fluid properties in the ANSYS Fluent simulation. For the hand-calculations, using interpolated fluid properties and iterating on the bulk temperature would also likely improve the calculation accuracy. Other possible areas to investigate are the residual tolerances, the mesh cell sizing and mesh quality. The current mathematical model also uses models to approximate the fundamental flow physics and using a more fundamental mathematical model could also improve prediction accuracy (at the expense of much higher computational costs). 8