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 or duct, where Re is the Reynolds
No. and rr is the rel
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 are commercial steel. The 2
rd
and 3 floors have one ga
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)c
P =
G2
2 i
i
i
i
i A w
2
1(1 + ) + K c + K e + K +
13-1
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 from the hot
fluid is the rate of energy addition to th
14-1
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 (ThA)h (T A)h
(T A)c (ThA)c
P
G2
2 i
i
i A w
i
i
2
1
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 impeller. An
identically shaped 12 in. diameter impeller is t
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:/capsicum.me.utexas.edu/ChE354/files/Goulds_3196/3196_c
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) Data for a 2 ft diameter impeller operating at 440 rpm
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
curve. If only one of the two sets of curves is given, it
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 the tank is square edged, and the
valve is a fully open g
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, assuming no velocity
head at J yields
1
P
P1 V12
+
+ z 1 h
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 following two requirements
allow us to solve these kinds
05-1
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). The head loss in any line
with a device is written as
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 curve is generated by solving the energy equation for pump
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 company must design a very large impeller to deliver a very h
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 flow within the pump is fully
Most
turbulent and, as a
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/cooling loop shown yields
H=
cool
Cooling Loop
P2 -P1 Q 2 1
10-1
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
ductwork. The equation written in terms of pressures inst
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 2
2g
2g
1
or
P1
P
Q2
Q2
+
+ z1 + H hL = 2 +
+ z2
2
2
2