MEEG 4483 - Thermal Systems Analysis and Design
Homework 01 - Fall 2012
08/22/12. Due 08/31/12
Create Visual Basic function ffactor(Re, rr) that calculates the DarcyWeisbach friction factor in a pipe
02-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 02
Examples, Fluid Flow Resistance
Example
1
Water at 20 C flows through the piping
system shown. The ells are short radius, the
entrance at th
15-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 15
Friction Factors
Heat Exchanger Design Process
Equations for UA and P Recalled
R fc
1
1
R fh
1
=
+
+ R wall +
+
UA (ThA)h (T A)h
(T A)c (ThA
MEEG 4483 - Thermal Systems Analysis and Design
Homework 02 - Fall 2012
08/31/12. Due 4:00 PM, 09/07/12.
1.
The chilled water system for a 3 story building with an
nd
exhibit hall is shown. All pipes
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MEEG 4483 - Thermal Systems Analysis/Design
Lecture 14
Heat Exchanger Analysis Procedure
Heat Transfer Coefficients and Fin Efficiency
Equations for UA and P Recalled
1
1
R fh
R fc
1
R wall
UA
MEEG 4483 - Thermal Systems Analysis and Design
Homework 03 - Fall 2012
09/07/12. Due 09/14/12
A power plant condenser cooling water pump is designed to operate at 250 rpm with a 10 ft diameter impell
MEEG 4483 - Thermal Systems Analysis and Design
Homework 04 - Fall 2012
09/14/12, Due 09/21/12, 4:00 PM
Obtain the curves for a Goulds Model 3196 2x3 -10 pump at 1750 rpm from the Goulds site at
http:
MEEG 4483 - Thermal Systems Analysis and Design
Sample Exam 1, In-class Part - Open Book, Note
approx. 60 minutes
Use your own paper. Turn this sheet in with your work before starting lab part.
1.(20)
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MEEG 4483 - Thermal Systems Analysis/Design
Lecture 13
Heat Exchanger Analysis
Thermodynamic Requirement
If there is no loss from one of the fluids to the environment, the rate of energy removal
03-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 03
Branching Pipe Flows
Consider the three reservoirs shown connected by pipes as shown.
Writing the energy equation form node 1 to node J, ass
04-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 04
Piping Networks and Hardy - Cross Solution
Consider the piping network shown. This network consists of
2 loops, 5 lines, and 4 nodes. The fo
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MEEG 4483 - Thermal Systems Analysis/Design
Lecture 05
Generalized Hardy - Cross Method
1 cfs
A heat exchanger is added to line 4. Its head loss can be
written as hL = 50 Q2 (hL in ft, Q in cfs).
06-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 06
Pumps
System Curve
The system curve is the relationship between the pump head required and the flow rate through the
piping system. The curv
07-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 07
Pump Affinity Laws, Specific Speed
Predicting Pump Performance at Different Speeds and with Different Impeller Diameters
Often a pump compan
08-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 08
Pumps - Viscous Correction, Cavitation, Parallel/Series Pumps
Pump Reynolds No.
Most pump performance theory is based on the assumption that
09-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 09
Closed Loops, Power Savings with Variable Speed
HX
Closed Loop System Curve
Writing the energy equation from 1 to 2 around the
heating/cooli
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MEEG 4483 - Thermal Systems Analysis/Design
Lecture 10
Ductwork Systems
Energy Equation Applied to Fans
Consider the fan/duct system shown. A fan delivers ambient air at weight density through
du
11-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 11
Fan Performance
Fan Characteristics
P=
Q * FSP Q * FTP
=
.
S
T
FTP
S , T
FSP
FSP, FTP
A typical fan curve looks a lot like a typical pump
cu
01-1
MEEG 4483 - Thermal Systems Analysis/Design
Lecture 01
Fluid Mechanics Energy Equation and Head Loss
2
Energy Equation
Consider a typical pump/piping system
P1 V12
P
V2
+
+ z1 + H hL = 2 + 2 + z