Flash Steam and Steam Condensates in Return LinesVijay Sarathy
In power plants, boiler feed water is subjected to heat thereby producing steam which acts as a motive force for a steam turbine. The steam upon doing work loses energy to form condensate and is recycled/returned back to reduce the required make up boiler feed water (BFW).
Recycling steam condensate poses its own challenges. Flash Steam is defined as steam generated from steam condensate due to a drop in pressure. When high pressure and temperature condensate passes through process elements such as steam traps or pressure reducing valves to lose pressure, the condensate flashes to form steam. Greater the drop in pressure, greater is the flash steam generated. This results in a two phase flow in the condensate return lines.
ECONOMIC INSULATION FOR INDUSTRIAL PIPINGVijay Sarathy
Thermal Insulation for Industrial Piping is a common method to reduce energy costs in production facilities while meeting process requirements. Insulation represents a capital expenditure & follows the law of diminishing returns. Hence the thermal effectiveness of insulation needs to be justified by an economic limit, beyond which insulation ceases to effectuate energy recovery. To determine the effectiveness of an applied insulation, the insulation cost is compared with the associated energy losses & by choosing the thickness that gives the lowest total cost, termed as ‘Economic Thickness’.
The following tutorial provides guidance to estimate the economic thickness for natural gas piping in winter conditions as an example case study.
Flash Steam and Steam Condensates in Return LinesVijay Sarathy
In power plants, boiler feed water is subjected to heat thereby producing steam which acts as a motive force for a steam turbine. The steam upon doing work loses energy to form condensate and is recycled/returned back to reduce the required make up boiler feed water (BFW).
Recycling steam condensate poses its own challenges. Flash Steam is defined as steam generated from steam condensate due to a drop in pressure. When high pressure and temperature condensate passes through process elements such as steam traps or pressure reducing valves to lose pressure, the condensate flashes to form steam. Greater the drop in pressure, greater is the flash steam generated. This results in a two phase flow in the condensate return lines.
ECONOMIC INSULATION FOR INDUSTRIAL PIPINGVijay Sarathy
Thermal Insulation for Industrial Piping is a common method to reduce energy costs in production facilities while meeting process requirements. Insulation represents a capital expenditure & follows the law of diminishing returns. Hence the thermal effectiveness of insulation needs to be justified by an economic limit, beyond which insulation ceases to effectuate energy recovery. To determine the effectiveness of an applied insulation, the insulation cost is compared with the associated energy losses & by choosing the thickness that gives the lowest total cost, termed as ‘Economic Thickness’.
The following tutorial provides guidance to estimate the economic thickness for natural gas piping in winter conditions as an example case study.
OPERATING ENVELOPES FOR CENTRIFUGAL PUMPSVijay Sarathy
The following tutorial provides a step by step procedure to predict the allowable operating range or “Operating Envelope” for a centrifugal pump’s range of operation.
A QUICK ESTIMATION METHOD TO DETERMINE HOT RECYCLE REQUIREMENTS FOR CENTRIFUG...Vijay Sarathy
Turbomachinery Engineers often conduct studies to determine if a hot gas bypass is required for a given centrifugal compressor system. This would mean building a process model and simulating it for Emergency Shutdown conditions (ESD) & Normal Shutdown conditions (NSD) to check if the compressor operating point crosses the surge limit line (SLL). A quick estimation method that uses dimensionless number called the inertia number can be used to check prior to the study, if a Hot gas bypass (a.k.a. Hot Recycle) is required in addition to an Anti-surge line (ASV or a.k.a Cold Recycle).
OPERATING ENVELOPES FOR CENTRIFUGAL PUMPSVijay Sarathy
The following tutorial provides a step by step procedure to predict the allowable operating range or “Operating Envelope” for a centrifugal pump’s range of operation.
A QUICK ESTIMATION METHOD TO DETERMINE HOT RECYCLE REQUIREMENTS FOR CENTRIFUG...Vijay Sarathy
Turbomachinery Engineers often conduct studies to determine if a hot gas bypass is required for a given centrifugal compressor system. This would mean building a process model and simulating it for Emergency Shutdown conditions (ESD) & Normal Shutdown conditions (NSD) to check if the compressor operating point crosses the surge limit line (SLL). A quick estimation method that uses dimensionless number called the inertia number can be used to check prior to the study, if a Hot gas bypass (a.k.a. Hot Recycle) is required in addition to an Anti-surge line (ASV or a.k.a Cold Recycle).
The turbo machine is an energy conversion device which converts mechanical energy to kinetic/pressure energy or vice versa. The conversion is done through the dynamic interaction between a continuously flowing fluid and rotating machine component. Turbo machines comprise various types of fans, blowers, compressors, pumps, turbines etc. More and more experimental research work is available in the field of turbo machine design and its evaluation. Literature review has revealed that a few literatures are available on three dimensional numerical analysis of a centrifugal fan/blower. Literature review in present work is highly focused on centrifugal blower and use of CFD techniques in turbo machines. In this course of work, input parameters and design parameters of centrifugal blower is obtained as per church and Osborne design methodology developed by Kinnari Shah, PROF. NitinVibhakar. Fluid model is made as per this design data in PRO-E SOFTWARE. And this fluid model is simulated using computational fluid dynamics (CFD) approach in ANSYS (CFX). Numerical analysis carried out in this work is to understand the flow characteristics at design and off-design conditions under varying mass flow rates, varying rotational speeds and number of blades in both design methodology. This numerical analysis is under consideration of steady flow and for rotational domain (frozen rotor interference) is used. Performance curves are obtained under different variable inlet parameters like volume flow rate, rotational speed and number of impeller blades. Here mass flow rate as a inlet boundary condition and static pressure as a outlet boundary condition. Volume flow rate is changed by changing the mass flow rate at inlet. Overall work carried out on flow behaviour and performance graphs for different cases are discussed in length in results and discussions chapter. Comparative evaluation of two design method indicates that error in static pressure gradient is higher in Osborne design rather than church design, and performance parameters are better for church design than the Osborne design.
Hardy cross method of pipe network analysissidrarashiddar
Hardy Cross Method of pipe network analysis has revolutionized the municipal water supply design. i.e., EPANET, a public domain software of water supply, uses the Hardy cross method for pipe network analysis. It is an iterative approach to estimate the flows within the pipe network where inflows (supply) and outflows (demand) with pipe characteristics are known.
1. Fan Speed Design Project
Design Project for ME 449
By Dillon O’Connor
1
2. Duct System
Technical Details
The given dimensions are shown on in
the Table 1.
Other relevant information:
1) Duct recirculates air within the
house
2) The ducts are made of galvanized
iron sheeting
3) Outlet velocities may not be the
same.
4) Outlet velocities should be close to
25 ft/s without exceeding that
amount.
Description Cross Section Length
Duct 1 24 in. X 18 in. 25 ft
90o Turns (1 to 2; and 1 to 3) - -
Duct 2 18 in. X 12 in. 12 ft
Duct 3 12 in. X 12 in. 7 ft
90o Turns (2 to 4; and 2 to 5) - -
Duct 4 w/ one 90o Turn 8 in. diameter (circular) 6 ft
Duct 5 w/ one 90o Turn 8 in. diameter (circular) 6 ft
90o Turns (3 to 6; and 3 to 7) - -
Duct 6 w/ one 90o Turn 8 in. diameter (circular) 6 ft
Duct 7 w/ one 90o Turn 8 in. diameter (circular) 6 ft
Return Duct w/ one 90o Turn 24 in. X 18 in. 22 ft
Table 1: Given duct system dimensions
2
3. Duct Schematic
• Figure 1 shows an isometric view of
the system.
• Figure 2 shows a dimensioned top
down view (left) and a dimensioned
side view (right) of the system.
Figure 1. Isometric view of the duct system. Figure 2. Dimensioned Schematic of the duct system.
3
4. Fan Curve
• The fan has already been selected.
• However the required operating
speed needs to be determined.
• The fan curve for the given fan has
been provided in Figure 3.
• The curve fit equation to this fan
curve will be used in later
calculations.
Figure 3. Fan curve for the given system’s fan.
4
5. Circuit Diagram
• Figure 4 shows the equivalent
circuit diagram for the system.
• Major loss sections are shown as
long rectangular boxes.
• Minor loss sections are shown as
short rectangular boxes.
• Each will have a hf that
contributes to the resulting
pressure loss.
Figure 4. Equivalent circuit diagram of the duct system.
5
6. Minor Loss Coefficients
The minor loss coefficients used are
shown on in Table 2.
Tee and Elbow K values are taken from
Table 6.5 of the Fluid Mechanics (5th
edition) text by F. M. White. Flanged
connections were assumed.
K values for sudden expansion and
sudden contraction come from
equations 6.80 and 6.81 respectively
from the same textbook.
𝐾 = 1 −
𝑑2
𝐷2
2
𝐾 = 0.42 1 −
𝑑2
𝐷2
Duct Minor
Loss Coefficient
Value Description
K1 0.41 Tee (d = 20 in.)
K2 0.7035
Sudden Contraction (d1 to d2)
+ Tee (d = 14.4 in.)
K3 0.8004
Sudden Contraction (d1 to d3)
+ Tee (d = 12 in.)
K4 1.5504
Sudden Contraction (d2 to d4)
+ Elbow (90 degree, regular, d = 8 in.)
+ Sudden Expansion (d4 to large d)
K5 1.5504
Sudden Contraction (d2 to d5)
+ Elbow (90 degree, regular, d = 8 in.)
+ Sudden Expansion (d5 to large d)
K6 1.4933
Sudden Contraction (d3 to d6)
+ Elbow (90 degree, regular, d = 8 in.)
+ Sudden Expansion (d6 to large d)
K7 1.4933
Sudden Contraction (d3 to d7)
+ Elbow (90 degree, regular, d = 8 in.)
+ Sudden Expansion (d7 to large d)
K8 0.63
Sudden Contraction (large d to d8)
+ Elbow (d = 20 in.)
Table 2: Minor loss coefficients
(6.80)
(6.81)
6
7. Constants
• The other constants used in this
design are shown in Table 3.
• Air properties, including density,
were assumed constant throughout
the duct system.
• Air properties were evaluated at
standard room temperature (70 ºF)
and pressure (1 bar).
• Acceleration due to gravity was
taken at sea level.
symbol Description Value
g Acceleration due to gravity 32.2 ft/s²
ε Duct roughness 0.0005 ft
ν Kinematic viscosity of air 1.64e-4 ft²/s
γ Specific weight of air 7.492e-2 lb/ft³
Table 3: Essential constants.
7
9. Method • Guess a Q1 value.
• Calculate total pressure lost through the top line for Q2 = 0 to 100% of Q1.
• Calculate total pressure lost through the bottom line for Q3 = 0 to 100% of
Q1.
• Plot ΔPtop leg vs. increasing% of Q1 and ΔPbottom leg vs. decreasing % of Q1.
• Independent axis value at intercept = % of Q1 to Q2.
• Q2 = %*Q1 and Q3 = (1oo-%)*Q1.
• Q4 = Q5 = 0.5*Q2 and Q6 = Q7 = 0.5*Q3
• Intercept can be better tuned by comparing resulting
ΔPtop leg and ΔPbottom leg values until they are equivalent.
• Find exit velocities using v = Q/A
• Change Q1 until max v (of v4-v7) is close to 25 ft/s
9
10. Q1 And Where It Goes
• A Q1 guess of 34 ft³/s yielded the
plot shown in Figure 5.
• The intercept in this case can be
seen at about 51% of Q1 to Q2 and
the remaining 49% of Q1 to Q3.
• Fine tuning the % by ΔP values
yields a 51.26% of Q1 to Q2. The
remaining 48.74% of Q1 goes to
Q3.
Figure 5. Plot of losses in each line as a function of decimal
fraction of Q1 that travels into Q2 for a guessed Q1 of 34
ft³/s.
10
11. Verifying Accuracy Of Guess
• A guess of Q1 = 34 ft³/s yields a maximum exit velocity of 24.96 ft/s in
ducts 4 and 5.
• The lesser exit velocity is 23.74 ft/s in ducts 6 and 7.
• This satisfies the given criteria of an exit velocity close to 25 ft/s without
exceeding it.
• The resulting TP = ΔP = 0.2761 in.H2O
11
12. Equations (2nd Part)
𝑄1 =
𝛼
𝑇𝑃2
1
𝑄2
2
2
+
𝛽
𝑇𝑃2
𝑁2 =
𝑄2
𝑄1
𝑁1
For these equations:
• Q2 is our Q1 from the previous slide
converted to ft³/min.
• TP2 is our TP from the previous slide.
• N2 is the operating speed to achieve it.
• Q1 is the Q at N=2000 RPM as shown
on the Fan Curve.
• N1 = 2000 RPM
• α = 5.95 in.H2O
• β = 8.5e-7 (in.H2O/CFM²)
12
13. Final Results
• The required Q1 = 2040 cfm
• The required N = 1607 RPM
• The resulting TP = ΔP = 0.2761 in.H2O
• The maximum exit velocity is 24.96 ft/s in ducts 4 and 5.
• The lesser exit velocity is 23.74 ft/s in ducts 6 and 7.
13
14. REFERENCES
White, F.M., Fluid Mechanics, 5th Edition, McGraw-Hill, New York, 2003
Department of Mechanical, Aerospace, and Biomedical Engineering, ME
449 A Preliminary Design Problem, University of Tennessee, Knoxville,
2015
14
