bir sigara içme salonu sigara tiryakisine hizmet verecektir.sigara … · 2018. 3. 28. · •...

1‐ Bir sigara içme salonu 15 sigara tiryakisine hizmet verecektir.Sigara içme salonu için minimum taze hava ihtiyacı kişi başına 30 L/s olarak belirlenmiştir.Buna göre salona verilmesi gereken minimum taze hava debisini ve hava hızının 8 m/s ‘yi geçmesi istenmediğine göre ,kanalın çapını belirleyiniz.

15 SİGARA TİRYAKİSİ

30 L/s (kişi başına )

2‐Bir pompanın giriş ve çıkışına yerleştirilen bir diferansiyel termo‐eleman ,pompa içerisinden 42.48 L/s debi ile geçerken suyun sıcaklığının 0.048 °C arttığı gösterilmektedir.Pompaya verilen mil gücü 23 BG olduğuna göre pompanın mekanik verimini belirleyiniz.

3‐Bir yağ pompası 860 kg/ m3 yoğun luktaki yağı 0.1 m3 /s ‘lik debi ile basarken 25 kW elektrik gücü çekmektedir.Pompanın giriş ve çıkış borularının çapları sırasıyla 8 cm ve 12 cm ‘dir.Pompada meydana gelen basınç 250 kPa ve motor verimi % 90 olduğuna göre ,pompanın mekanik verimini belirleyiniz.Kinetik enerji faktörünü 1.05 alınız.

4‐ 2m*3m*3m ebatlarında bir banyoyu havalandırmak üzere bir fan seçilecektir.Titreşim ve gürültüyü en aza indirmek için hava hızı 8 m/s ‘yi geçmemesi istenmektedir.Kullanılacak fan‐motor grubunun toplam verimi %50 alınabilir.Fanın odadaki tüm havayı 10 dakika ‘da değiştirilmesi istendiğine göre , a) Satın alınacak fanı‐motor grubunun watt cinsinden gücünü, b) Fanın dış çapını , c) Fanın girişi ve çıkışı arasındaki basınç farkını belirleyiniz .Hava yoğunluğunun 1.25 kg/ m3 alınız ve kinetik enerji düzeltme faktörlerinin etkisini göz ardı ediniz.

5‐Bir hidrolik türbine su 30 cm çapındaki borudan 0,6 m3 /s lik debi ile girmekte ve 25 cm çapındaki borudan çıkmaktadır .Türbinde meydana gelen basınç düşümü cıvalı manometre ile 1.2 m olarak ölçülmüştür.Bileşik türbin‐jeneratörün verimi % 83 olduğuna göre ,net elektrik gücü üretimini belirleyiniz.Kinetik enerji düzeltme faktörlerini göz ardı ediniz.

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Measurements and Units

Understanding Units for Dimensional Analysis

Dimensionsymbol

Name Unit

M mass Kilogram

L Length Meter

T time Second

Θ (theta) Absolute temperature C, K, F

I current Ampere

N Amount of substance Moles

J Intensity candela

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Unit name Symbol Definition Dimension analysis

Speed Ve Meter/second LT‐1

Acceleration A Meter/second2 LT‐2

Force Newton kilogram*meter/second2 LMT‐2

Energy Joule,Cal Newton*meter L2M2T‐2

Power Watt Joule/second L2M2T‐3

Pressure Pascal Newton/meter2 L‐1MT‐2

Frequency Hertz 1/second T‐1

Charge Coulomb Ampere*second IT

Potential Volt Joule/charge L2M2T‐2/IT=L2M2T‐3I‐1

Resistance Ohm Volt/ampere L2MT‐3I‐2

Measurements and Units

50

Measurements and Units‐Examples

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Measurements and Units‐Examples

52

Notations and prefixes

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Standard Scientific Engineering

123 GW 1.23x1011 W 123x109 W

0.0698 m 6.98x10‐2 m 69.8X10‐3 m

501,000,000,000 g 5.01x1011 g 501x109 g

Notations and prefixes

Scientific notation: One digit to the left of the decimalEngineering notation: Standard units of thousand

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In‐class exercise1. Determine the fundamental dimensions

2. Express the following values using scientific and engineering notation.

T=Tera=1012 f=femto=10-15

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The Buckingham ‐ Teorem

Step 1. List all parameters in the studied system (dimensional parameters, non‐

dimensional variables and dimensional constants). Let ‘n’ to be the total number of

parameters

Step 2. List the primary dimensions of each ‘n’ parameters.

Step 3. Guess the reduction ( ). K is equal to the expexted amount of ‘ ’s

Step 4. Choose ‘j’ repeating parameters that will be used to construct each

Step 5. Generate the s by grouping ‘j’ repeating parameters with the remaining

parameters.

Step 6. Check all s write a functional relation between all s such as:

, … .

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57


58


59


V

Lift coefficient !

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Reynolds #

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Another simple exercise on Buckingham – Teorem can be found in the following link

http://www‐mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/dimension/node9.html

62

Dimensional Analysis of Turbomachines

Treating a turbomachine as a pump

N: Rotational speed (can be adjusted by the current)Q: Volume flow rate (can be adjusted by external vanes)

For fixed values of N and Q

Torque ( ), head (H) are dependent on above parameters (Control variables)Fluid density ( ) and dynamic viscosity are ( ) specific to the utilized fluid (Fluid properties)Impeller diameter (D) and length ratios (l1/D1 and l2/D2 are geometric variables for the pump (geometric variables)

Dimensional analysis can now bemade by considering above terms

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1. Incompressible Fluid Analysis

The net energy transfer (gH), pump efficiency ( ), and pump power (P) requirement are functions of aforementioned variables and fluid properties:

Using three primary dimensions (mass, length, time) or three independent variables we can form 5 dimensional groups by selecting ( , N, D) as repeating parameters. Using these groups it is possible to avoid appearance of fluid terms such as and Q

The work coefficient (energy transfer coefficient ‐ ) :

(i)

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Efficiency (already non‐dimensional): (ii)

Power coefficienct ( ) : (iii)

Here, (Q/(ND3)) is also regarded as a volumetric flow coefficient and ( ) is referred to as Reynolds

number. The volumetric flow coefficient is also called velocity or flow coefficient ( ) and can also be defined in terms of velocities:

Since the independent variables are complex, some assumptions are made for simplification. Effect ofgeometric variables by assuming similar values of these ratios are constant. In addition another

assumption is made by assuming the effect of Reynolds number ( ) is neglected for the flow. Now

the functional relationships are simpler:

(iv)

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• Using these dimensionless groups, it is now possible to write a relationship between the power, flow, head coefficients and the efficiency.

• Since the new hydraulic power for a pump is , and efficiency is the ratio of net power to the actual power / . Than one can use (i), (ii), and (iii) to form a relation between these parameters:

• Which yields to :

• For an hydraulic turbine (since / ):

(i)

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2. Compressible Fluid Analysis

• For an ideal compressible fluid, mass flow rate is used instead of volumetric flow rate and twoadditional parameters are required specific to the incompressible fluid, namely the stagnation soundspeed ( ) and the specific heat ratio ( ). Total power produced, efficiency, and the isentropicstagnation enthalpy change is considered as functions for non‐dimensioning

• The subscript (1) represents the inlet conditions since these parameters vary through theturbomachine. This 8 dimensional groups may be reduced to 5 by considering the stagnation density,rotational speed, and the turbomachine diameter as repeating parameters:

• Taking ND/a01 as the Mach number, a rearrangement can be made

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• For an ideal compressible fluid the stagnation enthalpy can be written as and knowing that:

we can write

And knowing the new definition is

With the ideal gas law, the mass flow can bemore conveniently explained

And the power coefficient:



Collecting all definitions together, and assuming the specific heat ratio ( ) and the Re # effect are dropped for simplification

The first term in the function is referred to as flow capacity and is the most commonly used form of non‐dimensionalmass flow. For fixed sized machinery D, R and the specific heat ratio are dropped. Here, independent variables are no Longer dimensionless.

A compressor efficiency can be written in terms of common performance parameters as:

There is still more parameters that are needed to fix the problem in variation of density and flow Mach number which are variables. Therefore two new parameters, namely flow coefficient ( ) and stage loading (work coefficient ‐ ) are defined.


3. Specific Speed and specific diameter

There may be a direct relation between three dimensionless parameters in a hydraulic turbine when Re # effects and cavitation is absent.

For specific speed, “D” are cancelled and thus,

For power specific speed :

Ratio of above definitions provide:

Note: D is the only common parameter for all dimensionless parameters.

By eliminating the speed from flow and work coefficients one may obtain the specific diameter:

Dimensional Analysis of Turbomachines3. Specific Speed and specific diameter

Selection of pumps based on

dimensionless parameters

Selection of turbines based on

dimensionless parameters

Note: Here, N is replaced by and in rad/sec, instead of rev/min.


3. Specific Speed and specific diameter

For compressible flow:

Design of Axial Flow Turbines1. The Velocity diagram (Courtesy of Dr. Damian Vogt)

Axial turbine stage comprises a row of fixed guide vanes or nozzles (often called a stator row) and a row of moving blades orbuckets (a rotor row). Fluid enters the stator with absolute velocity c1 at angle α1 and accelerates to an absolute velocity c2at angle α2

From the velocity diagram, the rotor inlet relative velocity w2, at an angle β2, is found by subtracting, vectorially, the bladespeed U from the absolute velocity c2.

The relative flow within the rotor accelerates to velocity w3 at an angle β3 at rotor outlet; the corresponding absolute flow(c3, α3) is obtained by adding, vectorially, the blade speed U to the relative velocity w3.

Axial Flow Turbines

1. The Velocity diagram

Axial Flow Turbines2. First design Parameter: Stage Reaction

Axial Flow Turbines3. Second design Parameter: Stage Loading Coefficient (Work Coefficient)

Axial Flow Turbines3. Second design Parameter: Stage Loading Coefficient

Axial Flow TurbinesConsidering the sign convention:

Axial Flow Turbines4. Third design Parameter: Flow Coefficient

Axial Flow Turbines5. The Normalized Velocity Triangle

Axial Flow Turbines5. The Normalized Velocity Triangle

Another common way to represent a velocity triangle for axial turbines is:

ψ

φ

Axial Flow Turbines6. Special Cases

Axial Flow Turbines7. Thermodynamics of axial flow turbines

From the Euler’s work eq.

Since there is no work done through the nozzle row:

Writing above Eqs together :

From the velocity triangle , than we can re‐arrange as

In terms of relative stagnation enthalpy :

Axial Flow Turbines7. Thermodynamics of axial flow turbines

Mollier diagramFor a turbinestage

Axial Flow Turbines8. Repeating Stage turbines

Substituting main Reaction and rothalpy definitions:

…….(i)

……..(ii)

Substituting (ii) into (1): or

And the work coefficient – reaction relation yields to:

Axial Flow Turbines9. Stage loss coefficients

Losses can be defined in terms of exit kinetic energy from each blade row:

Adapting this into total‐to‐total and total‐to‐static efficiencies of the stage with velocitycomponents:

Axial Flow Turbines10. Preliminary axial turbine design

Number of stages:

With the continuity equation and the flow coeffMean radius can be defined:

Where, here, t and h stand for tip and hub.

The blade height requirement for a flow is related with flow coefficient and the mean radius as:

1. An isentropic turbine with 110 mm tip and 430 mm hub radius and with 4500 rpm produces 6 MW power while the axial velocity is constant through the turbine and rates as 150 m/s. Fluid is deflected by 58° at the stator outlet and while stator inlet angle at every stage is 18°. If the fluid density at the inlet is 0.85 kg/m3 calculate:

a) Stage reactionb) The work (stage loading) coefficientc) Minimum number of stages

SOLVED PROBLEMS

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Stator inlet angle (°)

Stag

e Re

actio

n (-

)

NOTE: Stage having reaction less than 0.5 suggest that pressure drop or enthalpy drop in the rotor is less than the pressure drop in the stator for the turbine. Therefore the stator has a larger contribution to the total work extracted or work done. Therefore, zero reaction turbines are called Impulse turbines that the power is completely produced via stators.

Effect of control variables on turbine characteristics

1000 2000 3000 4000 5000 6000 70000

1

2

3

4

5

6

Rotational Speed (rev/min)

Flow

and

wor

k co

effic

ienc

ts (-

)

SOLVED PROBLEMSDrawing the Normalized velocity diagram:

SOLVED PROBLEMS

SOLVED PROBLEMS

2. An axial turbomachine with 47 cm mean radius has 3000 rev/min rotational speed at its rotor. Fluid enters the stator at 120 m/s axial velocity and leaves the stator at 67° absolute flow angle. For 50% reaction, calculate:

a) Calculate the rotor inlet relative angle.b) Calculate the rotor inlet absolute and relative velocities.c) Calculate the rotor outlet absolute and relative velocities.d) Calculate the flow and work coefficients.e) Calculate the power produced from this stage.f) Draw the normalized velocity triangle. g) Calculate the specific speed and the specific diameter

SOLVED PROBLEMS

Bonus Homework (Get a 5% added to your midterm if you submit first)

Solve the previous problem when the reaction is assumed to be 60%.

SOLVED PROBLEMS

3. Below data is given for a three stage axial turbine. For 50% reaction:

a) Calculate the flow and work coefficients for 50% reaction.b) Calculate the absolute and relative flow angles, as well as fluid deflection angles.c) Calculate the exit Mach number of the turbine (Courtesy of R.I. Levis)

SOLVED PROBLEMS4.

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5. A four - stage axial turbine with 0.5 m mean radius has 2500 rev/min rotational speed at itsrotor. 15 kg/s fluid enters the stator and 120 m/s axial velocity and leaves the stator at 56°absolute flow angle. For 50% reaction:

a) Draw the normalized velocity diagram and show the stage loading and flow coefficients on the diagram

b) Calculate the power produced from this turbine.

c) Calculate the specific speed and the specific diameter

123

6. For an axial compressor stage, prove that higher reaction reduces the stage loading with a case study. (Make reasonable assumptions to specify the angles, frame velocity etc.). Show this claim on the velocity diagram

bir sigara içme salonu sigara tiryakisine hizmet verecektir.sigara … · 2018. 3. 28. · •...

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