Dimensional Analysis Std

7/28/2019 Dimensional Analysis Std

1/47

Reminders / Notes:

Class web site: Lecture notes and assignments updated

each Monday

http://www.if.uidaho.edu/~gunner/

Students in Moscow can also submit and collect their homework

assignments with Becky Schoenberg, UI ME Dept, Engr. Physics Bldg.

324K, PO Box 440902, Moscow, ID 83844-0902, 208.885.5024.

HW #1 IS DUE NEXT MONDAY FOR MOSCOW STUDENTS

(1-page, neat)
http://www.if.uidaho.edu/~gunner/http://www.if.uidaho.edu/~gunner/


2/47

DIMENSIONAL

ANALYSIS

TurbomachineryDr. Fred Gunnerson

2012


3/47

DIMENSIONAL ANALYSIS

Scale model to prototype design and analysis

Used to define, compare and predict

performance parameters

Used to select proper turbomachine

(axial, radial or mixed flow,)


4/47

UNITS

SI Units: F = ma

Newton (N) = Mass (kg) x Acceleration (m/s2)

one kilogram mass weighs 9.8 Newton force under standard gravity

USCS Units: F = ma / gc

Pound force (lbf) = Mass (slug) x Acceleration (ft/s2) / gc

1 slug = 32.2 lbm

gc = 32.2 (lbm/lbf) (ft/s2)

one pound mass weighs one pound force under standard gravity

MLT Fundamental Units

mass (M) length (L) time (T)


5/47

Work is one form of Energy

Work= Force x Distance W = F d

F = ma = m(dV/dt) Force (N, dyne, lbf, )

W = Fdx Work (J, btu, ft-lbf, )

Mass (kg, gm, lbm, mol, amu, )

1 amu = 1.66 (10)-27 kg

~ mass of hydrogen atom


6/47

Potential Energy

Energy possessed by matter within a gravity field by

virtue of its height

PE = mgh


7/47

Kinetic Energy

Energy possessed by matter due to

its speed or spin

KE = mV

KER = I


8/47

TYPES OF ENERGY (cont)

Thermal Energy or Heat

Energy possessed by matter

due to its temperaturesensible heat

Q = mCpTThermal energy transfer from hot to cold via:

ConductionConvectionRadiation

Q = thermal energy (J)

Cp = heat capacity (J/kgK)

Gases

Atomic E avg tot = kT

Molecular E avg tot = (3/2)kT

k= Boltzmanns constant = 1.38(10)-23 J/K

Porthole view into

coal fired power plant boiler

~2500F ~1400C


9/47

LATENT HEAT vs. SENSIBLE HEAT for water

One gram of ice at 0oC

1.09 cc +334 J of heat (latent heat of fusion)

One gram of water at 0oC

1 cc+

418 J of heat (sensible heat)(heat capacity = 4.18 J/C)

(Q = mCpT)

One gram of water at 100oC

1.04 cc


10/47

One gram of water at 100oC1.04 cc +

2258 J of heat(latent heat of vaporization)

One gram of steam at 100oC

1,673 cc


11/47

SENSIBLE HEAT Q=mCpTTemp increases with addition of heat

LATENT HEAT Q=mLf or Q=mhfgTemp constant with addition of heat

T

STEAM100

WATER

0ICE

Q


12/47

Nuclear Energy

Energy possessed by matter to hold atomic nuclei together

(binding energy)

Fission (and fusion)

E = mc2

m = mass converted to energyc = speed of light 3(10)8 m/s

One gram of mass = 9(10)13 J


13/47

Other types of energy.

Solar

Electrical

Chemical / combustionGeothermal

Photovoltaic

Electromagnetic

Human, ..etc.

Power (watt) = Energy (Joule) / time (second)


14/47


15/47

FUNDAMENTAL DIMENSIONS

Length: L Mass: M Time: T


16/47

Dimensional analysis is an important S&E tool

For example.assuming that the KE of an object is somehowrelated to its mass (m) and velocity (V)

Its easy to show, (just by playing with the units) that:

KE (N-m)

= dimensionless-term ( = from observation )

M (kg) [ V (m/s) ]**2

Physical Quantities and Dimensions:

m(kg) V(m/s) KE( kgm2/s2 = N-m = J)

Butif you have many physical quantities, its very difficult to determine the pi-terms.

Heres how you do it..


17/47

SUGGESTED PROCEDURE FOR DIMENSIONAL ANALYSIS

1. List the n physical quantities (Qn) with dimensions and the kfundamental dimensions. There will be (n-k) -terms.

2. Select k of these quantities, none dimensionless and no two having the

same dimensions. All fundamental dimensions must be included

collectively in the quantities selected.

3. The first -term can be expressed as the product of the selected

quantities each to an unknown exponent and one other quantity to a

known power (usually taken as unity).

4. Retain the selected quantities chosen as repeating variables and

choose one of the remaining variables to establish the next -term.Repeat the procedure for each successive -terms.

5. For each -term, solve for the unknown exponents by dimensional

analysis.


18/47

NOTES:

Fundamental dimensions are typically MLT (mass, length, time)or FLT (force, length, time).

Any dimensionless quantity is already a -term.

If two quantities have the same dimensions, their ratio will be a -term.

Example: L/D.

Any -term may be replaced by any power of that term.

Example: 2, , 1/.

Any

-term may be replaced by multiplying it by a constant.

Any -term may be expressed as a function of the other-terms.

Example: 1 = 2 x 3 / 4.


19/47

GENERIC TURBOMACHINE (turbine, compressor, pump, .)

Rotor / ImpellerDiameter = D

Power shaft

In = compressor, pump

Out = turbine

fluid in

fluid out

What physical quantities do or may effect the efficiency (output / input)?

shaft power, volumetric flow rate, impeller diameter, RPM, fluid properties

(density, viscosity, compressibility, heat capacity, inlet & outlet head pressure)

others?


20/47


1. List the n physical quantities (Qn) with dimensions and the k

fundamental dimensions. There will be (n-k) -terms.2. Select k of these quantities, none dimensionless and no two having the

same dimensions. All fundamental dimensions must be included collectively

in the quantities selected.


quantities each to an unknown exponent and one other quantity to a knownpower (usually taken as unity).

4. Retain the selected quantities chosen as repeating variables and choose

one of the remaining variables to establish the next -term. Repeat the

procedure for each successive -terms.


analysis.


21/47

PHYSICAL QUANTITY (Qn) SYMBOL UNITS FUNDAL DIMENSIONS

1. Efficiency - -

2.ShaftPower Psft-lbf/hr ML2T-3

3.Volumetricflow Qft3/s L3T-1

4.Impellerdiameter D ft L

5.Specificenergy gHft-lbf/lbm L2T-2

6.Rotatingspeed N, rpm,rad/s T-1

7.Fluiddensity lbm

/ft3 ML-3

8.Abs.viscosity lbfs/ft2 ML-1T-2

9.Fluidelasticity(v/)e lbf/ft2 ML-1T-2

n = 9 (number of physical quantities) k = 3 (number of fundamental dimensions)n k = 9 3 = 6 (number of independent, dimensionless, pi-terms)

1. List the n physical quantities (Qn) with

dimensions and the k fundamental dimensions.

There will be (n-k) -terms.


22/47


1. List the n physical quantities (Qn) with dimensions and the k fundamental

dimensions. There will be (n-k)

-terms.

2. Select k of these quantities, none dimensionless and no two having

the same dimensions. All fundamental dimensions must be included


3. The first -term can be expressed as the product of the selectedquantities each to an unknown exponent and one other quantity to a known

power (usually taken as unity).





analysis.

Select: k = 3 quantitiesN (rotational speed ) T-1

D (impeller diameter) L

(fluid density) ML-3

Note: all quantities can be physically measured


23/47


1. List the n physical quantities (Qn) with dimensions and the k

fundamental dimensions. There will be (n-k) -terms.


same dimensions. All fundamental dimensions must be included


3. The first -term can be expressed as the product of the selectedquantities each to an unknown exponent and one other quantity to a

known power (usually taken as unity).





analysis.

1 = Na Dbc Q

M0L0T0 = (T-1)a (L)b (ML-3)c (L3T-1)

THUS..MASS: 0 = c TIME: 0 = -a 1 LENGTH: 0 = b 3c +3

a = -1 b = -31 = Q / ND3 = Flow Coefficient / Capacity Coefficient


24/47

1 = Q / ND3 = Flow Coefficient or Capacity Coefficient ()

= Q / ND3 thedimensionlessswallowingcapacityofthemachineVolumetric flow rate (Q) can be related to the fluid velocity (V) via Q = V A

A particular value of impliesa specific relationship between

fluid velocity and blade/impeller velocity.

A particular value of yieldsa specific velocity triangle.

At a different set of operating conditions,with the same value of, the velocity triangles will be similar. = Q/ND3 ~ VA/ND3 ~ VD2/ND2D ~ V/ND ~ Vx/U

Vout Vin

U


25/47

EFFICIENCY () VS FLOW COEFFICIENT ()

max

max

For any turbomachine with fixed

geometry, negligible viscous and

cavitation effects, there is a

unique relationship between

efficiency and flow coefficient


26/47

Flow Coefficient


27/47


28/47


1. List the n physical quantities (Qn) with dimensions and the k fundamentaldimensions. There will be (n-k) -terms.


same dimensions. All fundamental dimensions must be included collectively

in the quantities selected.


quantities each to an unknown exponent and one other quantity to a known

power (usually taken as unity).

4. Retain the selected quantities chosen as repeating variables and

choose one of the remaining variables to establish the next -term.Repeat the procedure for each successive -terms.5. For each -term, solve for the unknown exponents by dimensional

analysis.

2 = Na Dbc (gH)

M0L0T0 = (T-1)a (L)b (ML-3)c (L2T-2)

THUS..MASS: 0 = c TIME: 0 = -a 2 LENGTH: 0 = b 3c + 2

a = -2 b = -2

2 = (gH) / N2D2 = Head Coefficient ()Energy Transfer Coefficient

Work Coefficient

Loading Coefficient

(gH) corresponds to the specific energy of the fluid

N2D2 relates to the rotor or impeller velocity, rotor K.E.


29/47

: Loading coefficientHead coefficient

Energy transfer coefficient

= gH / N2D2Dimensionless

Some texts divide the coefficient by two (watch out!)Positive for turbines (hydraulic too)

Negative for pumps, compressors, fans, The numeratorcorresponds to specific energy of the fluid

The denominatoris related to the rotor / impeller velocity (U):ND U rotor K.E.

Thus, relates a ratio of

FLUID HEAD / K.E. OF ROTOR

In terms of enthalpy, is defined:

= - gc12ho / U2= - 12(UVu) / U2= - 12 Vu / U constant U from inlet to outlet


30/47

3 = NaDbc P (P = power)

M0L0T0 = (T-1)a(L)b(ML-3)c(ML2T-3)

a = -3 b = -5 c = -1

3 =P / N3 D5 = Power Coefficient (, Cp)

4 = NaDbc ( = absolute viscosity)

M0L0T0 = (T-1)a(L)b(ML-3)c(ML-1T-1)

a = -1 b = -2 c = -1

4 = / ND2 = 1/Re = Inverse Reynolds Number (ND2/)

5 = NaDbc e (e = fluid elasticity)

M0L0T0 = (T-1)a(L)b(ML-3)c(ML-1T-2) a = -2 b = -2 c = -1

5 = e / N2D2 = Compressibility Factor(related to Mach Number)

6 = (sometimes called 0) = efficiency alreadydimensionless

Continuing / repeating the

dimensional analysis

method for the other

physical quantities..


31/47

Pi-Terms can be combined to form a new pi-term:

One of particular interest (since it eliminates diameter dependence)

7 = (1) / (2)

= NQ / (gH) = SPECIFIC SPEED (Ns)

(shape number)

1 = flow coefficient 2 = head coefficient

There are several variations of Specific Speed (see: Peng, p 25) for pumps, fans,

compressors and turbines (hydraulic) in USCS units.


32/47

EFFICIENCY () VS SPECIFIC SPEED (Ns)

NsNs,max

max

For any turbomachine with fixed

geometry, negligible viscous and

cavitation effects, there is a

unique relationship between

efficiency and specific speed


33/47

NQ / (gH)


34/47

Specific Speed (shape number)

Ns = N (Q)1/2

/ (gH)3/4

Where: Ns = specific speed (dimensionless)

N = rotational speed, radians/s

Q = flow, in m3/sH = head, in m

Turbine Type Ns Range

Impulse (Pelton) 0.03

0.3Francis 0.32

Kaplan (propeller) 2 - 5

Used in turbomachinery selection and performance evaluation.

Combines the flow coefficient, with the loading coefficient, in such

a way as to eliminate geometry dependence.

Several different forms, some with dimensions, others dimensionless.

One common form used in the U.S. / Europe


35/47

Specific speed and performance

of compressors


36/47

Specific Speed and Cavitation


37/47

Specific Diameter (s)

Another non-dimensional number of interest:

8 =s =

/ = D(gH) / Q ThereareseveralvariationsoftheSpecificDiameter

forcompressors,pumpsandfansutilizingUSCSunits.


38/47


39/47

SIMILITUDE

Small model data can be used to predict

the performance of a full-scale prototype

Predict machine performance at other operating conditions

GEOMETRIC (L/D)m = (L/D)pLinear dimension ratios are the same everywhere.

Photographic enlargement

KINEMATIC (m = p)Same flow coefficients

Same fluid velocity ratios (triangles) are the same

DYNAMIC (m = p)Same loading coefficient

Same force ratios (and force triangles)

ModelFull-Scale

Prototype


40/47

Similarity Laws

COEFFICIENTS

Flow

Head

Power


41/47

Head Pressure (H)

Q

CV

PUMP (rpm)

EXAMPLE: Similitude and Pump Performance

The performance characteristics of a turbomachine may be predicted over aspectrum of operating conditions from a simple set of tests at a single speed


42/47

Q1

N1 = constant rpm

H

Q

H1

N2 < N1N3 < N2

Q3 Q2

H2

H3

Q

H

rpmExample (cont): Hydraulic Head vs Volumetric Flow Rate


43/47

N1 Q1 / N1D3= Q1 / N1 = Q2 / N2 = Qn / NnN2 Q2 / N2D3

Ratio of Flow Coefficients

Ratio of Head Coefficients

N1 H1 / N12 D2

= H1

/ N1

2 = H2

/ N2

2

N2 H2 / N22 D2


44/47


45/47


46/47

Flow coefficient

Head coefficient

Power coefficient


47/47

Assign Homework #2 End of Lecture

Documents

Dimensional Analysis Std