Homework/Project
Homework #1 - due Thursday, Sept 15
Solutions
In Task 3, the value of H as defined by Eq. (1) can be calculated by
first defining F ≡ VnρCpT using the "Custom Field Function" tool, then
integrating F using the "Surface integral" function. The first two examples
of solutions posted below use this approach. The mass flow rate, M, can also
be obtained by the same method.
Solutions - Example 1 (thanks to Ram Mohan Telikicherla)
Solutions - Example 2 (thanks to Rylie Lodes)
The calculations for Task 3 can also be done using tools in CFD-post. The
next example provides the detail. (For brevity, only the solution for
Task 3 is shown.)
Solutions - Example 3 (thanks to William Burk)
Many have used the tools under "Flux Report" to calculate H and M. It
produces the same result for the mass flow rate as the approaches shown
in Examples 1-3. On the other hand, the report of "Total heat transfer
rate" gives a very different value which is not the "H" defined by Eq. (1).
Since instructor did not discuss this subtle point beforehand, those
who used this approach are still given full credit.
The following example of solution provides some insight into the possible
definition of "Total heat transfer rate" in "Flux Report". It is found
that, if Cp*T in the Custom Field Function is replaced by "enthalpy",
the approach used in Examples 1-2 would produce the same value as the
"Total heat transfer rate" from "Flux Report". This suggests that the
latter is defined by Eq. (1) but with VnρCpT in the integrand replaced
by VnρE, where E is enthalpy.
Solutions - Example 4 (thanks to Leighann Ngo)
The remaining question is how is the "enthalpy" (as a built-in variable)
defined in Fluent. Another student provided an explanation in the following.
Essentially, it is not defined as CpT but Cp(T−Tref), where Tref is a constant
reference temperature. This also clarifies the issue with the signs at
the inlet/outlet.
Further explanation (thanks to Vijay Jangid)
Project #1 + Challenges #1-2, due Thursday, October 13th
Release note: § 1 From the discussion on HW1, the Tout in Eq. (1) can be evaluated using the approaches in Examples 1-3 of the solutions of that homework. For Task 4, the heat flux coming from the bottom plate is not the same as the "H" defined in HW1 since no fluid flows through the bottom plate. Instead, it arises from heat conduction. More precisely, the thermodynamic energy equation reads ∂T/∂t = −V•∇T + κ∇2T + ... The 1st term in the r.h.s. represents heat transfer by "convection" (not necessarily buoyancy-driven but also the effect of horizontal fluid flow carrying heat around). The 2nd term represents heat conduction which can happen even when the fluid velocity is zero. Integrating the equation over the volume of the system and applying Gauss divergence theorem, the 1st term would lead to the boundary flux "H" defined in HW1. At the bottom plate, the heat flux into the water tank comes from the 2nd term. It is part of your task to find how to calculate it in Fluent. § 2 In case it might be useful, the following is a link to a paper that provides the empirically determined relation of ρ as a function of T for water. (See Eq. (1) or Eq. (3) in the paper.) NIST 1992 formula for ρ(T) By taking the derivative of ρ(T) one can readily obtain the thermal expansion coefficient. Note that the thermal expansion coefficient (as used in Fluent) is defined by β ≡ α−1(∂α/∂T) where α ≡ 1/ρ is specific volume.
Solutions Instructor's remarks Regular tasks: Solutions - Example 1 (thanks to Abir Kumar Deb) Solutions - Example 2 (thanks to Amogh M. Gadagi) Challenge #1: Solution - Example 1 (thanks to Eduardo Aguirre) Solution - Example 2 (thanks to Benjamin Williams) Challenge #2: Solution - Example 1 (thanks to Swapnil Nimse) Solution - Example 2 (thanks to Gargi Kailkhura)
Project #2 (revised) + Challenges #3-4, due Thursday, November 10th
Solutions Regular tasks: Remarks: (1) In Task 2, with a proper setup (even without further mesh refinement) the water jet should not "stick to the wall" even for the case with u = 0.3 m/s. Such a feature was produced due to three potential problems: (i) Water was put in the left pipe at t = 0. (This is incorrect, as clearly instructed in the handout.) This hinders the development of the jet because the incoming water flow has to overcome a pre-existing body of water with zero velocity. (ii) Coarse mesh resolution, when "coarse" or "medium" resolution was used instead of "fine" as required. The coarse resolution "chokes" the relatively "friction free" region in the left pipe and, again, hinders the development of the jet. (iii) Large "time step size". This increases the numerical error which could also damp the jet. (2) See reference solutions below for the correct answer to Task 4b. The major problems with this task are insufficient mesh resolution (which is critical at large time) and inadequate choice of domain/boundary condition. As already discussed, the use of high resolution mesh will actually not be costly because flow velocity keeps decreasing with time. At large time, a large time step size can be used without violating the numerical stability condition. Solutions - Example 1 (thanks to William Klemaszewski) Solutions - Example 2 (thanks to Sharan Kishore) Solutions - Example 3 (thanks to Thomas Chengattu) Challenge #3: Remarks: (1) The UDF should be applied to the volume fraction of either methane or air. It is incorrect to apply the UDF to the inlet velocity (i.e., to turn the inlet velocity on and off). Note that as the volume fraction of methane drops to zero, the VF of clean air rises to one. The inlet should keep "puffing" clean air as clearly instructed in the handout. (2) In reference solution #1, the UDF was defined only over the interval up to t = 20s, which nevertheless is sufficient for this task. The UDF in reference solution #2 works for all time (i.e., the alternating pattern of the VF of methane will go on indefinitely.) Solution - Example 1 (thanks to Tobie Miller) Solution - Example 2 (thanks to Yousif Alansari) Challenge #4: Remarks: It is easier to summarize what types of boundary conditions do not work: Essentially, a b.c. that does not allow two-way traffic cannot be used for the simulation of this problem. "Wall" is not allowed because it prohibits any traffic in and out of the container. "Velocity inlet" is not allowed because it prohibits air to go out. "Outflow" is not allowed because it prohibits air to come in. Most of the combinations of pressure inlet, pressure outlet, and "vent" type boundary conditions will work with a proper setup. Solution - Example 1 (thanks to Shane Dombrowski) Solution - Example 2 (thanks to Steven Gomez)
Project #3 + Challenges #5-6, due Wednesday, November 23rd, at 5 PM
Hard copy reports will be collected in class on November 22nd, and in office
hours: 3-5 pm on Nov 22nd, 10:45-12:00 and 4-5 pm on Nov 23rd.
Profile data for the "half fish"
For reference only: matlab code for generating the "half fish" profile
Solutions
Regular tasks:
Remarks:
(1) The solution for Task 1(a) is slightly sensitive to the detail of
initialization. A solution with a less dramatic "meandering tail" (compared
to the reference solutions), as produced by some students, is also acceptable.
A possible source of the difference is in whether the initialization used
default ("do nothing", thereby leaving truly zero velocity in the domain) or
"compute from inlet". On the other hand, the solution for Task 1(b) is
not sensitive to such detail. The solution should be one without a
meandering tail.
(2) For task 2, the typical range of lift is from -7 to -9 N while that of
drag is from 2 to 3 N. For task 3, the typical range of drag is from 0.1 to 0.3 N.
The drag for task 4 is several times its counterpart for task 3. The greater
drag force is expected since the impending velocity is higher for task 4.
If drag coefficient is considered instead, the values for the two cases
will be much closer.
(3) For the 3-D simulations (task 3 and 4), the flow pattern is somewhat
sensitive to the setup of mesh and solution procedure (e.g., number of
iteration). The solutions shown in the following two examples are typical.
Solutions - Example 1 (thanks to Alex Kozlowski)
Solutions - Example 2 (thanks to Andrew Perez)
Challenge #5:
Remarks:
The case with elongation of the cylinder in y-direction produces a higher
amplitude and longer period of oscillation compared to the circular-cylinder
case. The case with elongation in the x-direction is the opposite of the
case with elongation in y-direction. This is consistent with the
experiments in a video discussed in class.
Solution - Example 1 (thanks to Adalberto Campos)
Challenge #6:
Remarks:
Although the absolute values of lift and drag depend on the detailed setup,
some specific trends are robust: (i) Both lift and drag increase with the
tilt angle (theta) and both are positive (except for lift at theta = 0).
(ii) The lift at theta = 0 is zero. This is expected because of the axial
symmetry of the system. (This is the reason that we did not require the
computation of lift for task 3 and 4.) The typical ranges of lift and
drag are shown in the two reference solutions.
Solution - Example 1 (thanks to Chris Wellons)
Solution - Example 2 (thanks to Sankaranand Ramasamy)
Project #4, due Wednesday, December 7th, at 5 PM
Hard copy reports will be collected at instructor's office from 3:00-5:00 pm on Wednesday, December 7th.
Slides (typed)
Slides from the first lecture (8/18)
More slides from the first lecture (8/18)
From the mini symposium (12/1):
Part I. Survey and comparison of CFD software Autodesk CFD (presented by Joe Tichacek) OpenFOAM (presented by Girish Nigamanth Raghunathan) COMSOL Multiphysics (presented by Vaasavi Sundar) Solidworks Flow Simulation (presented by Alexander Wilson) Part II. Examples of simulations using alternative software Contributed by Aditya Vipradas, Girish Nigamanth Raghunathan, Mohammed Mehdi, Sujal Tipnis, and Vijay Jangid OpenFOAM, Solidworks Flow Simulation, and COMSOL
Useful links
Basic Matlab programming
