Base Calculations

This chapter concentrates on the calculations involved when testing engines. A selection of base calculations is described, with appropriate usage and suitability of the calcula­tions. The chapter starts with some brief descriptions of the most common ones.

Torque Backup

Torque backup is best described as

(Maximum torque — Torque at maximum speed)

Torque at maximum speed

Development of advanced common rail systems and associated equipment has intro­duced additional freedom to vary fuel delivery over the speed range of an engine. For example, the turbocharger matching process must be linked closely with the fuel system matching, even after optimum injection rates, pressures, nozzle sizes, and swirl have been achieved.

Motoring Mean Effective Pressure

The motoring mean effective pressure (MMEP) is determined from the torque required to motor an engine at a pre-determined condition and can be calculated as

. „ . 1000 x Motoring torque (Nm) x Number of strokes

MMEP (bar) — —————————— —— ———— ———————————

Capacity (cc)

Volumetric Efficiency

Volumetric efficiency fj is a measure of the effectiveness of the induction and exhaust processes. It is convenient to define volumetric efficiency as

, _ Volume of ambient density air inhaled per cylinder per cycle v Swept volume per cylinder at ambient pressure and temperature

Assuming that air obeys the gas laws, this can be written as

Volume of ambient density air inhaled per cylinder per cycle

V

подпись: vT1

Cylinder volume (cc)

, va

Where

Va = volumetric flow rate of air with ambient density Vs = engine swept volume

N = revolutions per second for a two-stroke unit, or revolutions per second divided by 2 for a four-stroke engine

Specific Fuel Consumption

The specific fuel consumption (SFC) can be defined as

Mass flow rate of fuel

SFC ————————————-

Power output

Correction Factors

Pressure:

R-L

A

For example,

25 psig = 25 + 14.504 = 39.504 psia 25 psig = 25 x 6894.76 Pa

= 172,368 Pa (gauge)

= 172,368 Pa + 101,325 Pa (1 atmosphere)

= 273,693 Pa (absolute)

Phase

Phase is the nature of a substance. Matter can exist in three phases: solid, liquid, or gas.

Cycle

If a substance undergoes a series of processes and returns to its original state, then it is said to have been taken through a cycle.

Process

A substance can be said to have undergone a process if the state is changed by operation of that process having been carried out on it.

• Constant temperature process

• Isothermal process

• Constant pressure process

• Isobaric process

• Constant volume process

• Isometric process or isochoric process

Heat

Temperature t (Celsius) = T — 273.15 (Kelvin)

Q — heat energy in joules/kg Specific heat capacity:

Heat transfer per unit temperature

Dt

The unit is joules/kg K (joules per kilogram per Kelvin).

Calorific value is the heat liberated by burning the unit mass or volume of a fuel (e. g., gasoline: 43 MJ/kg).

Work done in a polytropic process:

V2 v2

Work done = | PdV = C J V~ndV

Vi V,

= _c_(vr“+1-vrn+1l = PlVl~P^2 —n +1 ‘ ‘ n — 1

Enthalpy

A themodynamic quantity equal to the internal energy of a system plus the product of its volume and pressure.

H = U + PV

Where

U = internal energy P = pressure

V = volume

Specific Enthalpy

The specific enthalpy of a working mass is the property of that mass used in terms of dynamics. The S. I. unit for specific enthalpy is joules per kilogram.

H = U/m = u + Pv

Principle of the Thermodynamic Engine

„ Work done

Power =———————

Time taken

Where the unit is J/s = watt.

подпись: where the unit is j/s = watt. подпись: „ work done
power = 
time taken

The engine converts chemical energy into usable power and torque.

Base Calculations

That which is transmitted through a mechanical system and used to perform work.

Electrical Power

The product of voltage and current.

W = I V

Where the unit is J/s = watt.

Laws of Thermodynamics

The Zeroeth Law: If bodies A and B are in thermal equilibrium, and bodies A and C are in thermal equilibrium, then bodies B and C must be in thermal equilibrium.

The First Law:

W = Q

This means that if some work W is converted to heat Q, or some heat Q is converted to work W, then W = Q. It does not mean that all work can convert to heat in a particular process. .

The Conservation of Energy

For a system,

Initial energy + Energy entering = Final energy + Energy leaving

Potential energy = gZ

• 1 /~i2 Kinetic energy = — mC

Joule’s Law

Internal energy of gas is the function of temperature only and independent of changes in volume and pressure,

PlVf = P2V2n

Pl_

1

N

’V

N

And

Vi _

>1*

P2

L vi.

V2

1

NT

The specific heat capacity at constant volume is Cv.

U2 — Uj = mcv (T2 — Tj) (change in internal energy) The specific heat capacity at constant pressure is Cp.

U2-U1+P(V2-V1) = mcp(T2-T1)

From the characteristic equation

H

T2

подпись: h
t2
PlVl _ P2V2

PlVl _

N

Vi _

P2V2

IvJ

V2

IVJ

N-1

And

-1

N-l

/ ^

Z’

——

I 11

N

I 11

N

Lp2J

L?2J

Ti P. V. Pi

T2 P2V2 P2

подпись: ti p.v. pi
t2 p2v2 p2

That is,

II

FV

T2

N-l

N

II

Й-^l

Fpll

T2

Ip2;

Base Calculations

Entropy

A measure of the unavailability of a system’s energy to do work.

Base Calculations

DQrev =Tds

подпись: dqrev =tds

Revisable heat transfer is Qri

Mi­ll -1 T2

Cr,

S2 _ S1 — c

= — cp‘n|-+(cp-c»)^:

N-lnlL

1 t,

— n

In—= Cp — Cv" In —

N — 1 Tt n — 1

= c.

Qrev = / Tds

 

An isolated system can change to states only of equal or greater entropy. Entropy for gas:

Polytropic process = PVn = C

 

Heat transferred = dQ = —- Pdv c = —

C — 1 c„

 

, dQ c — nP, c-nR,

Ds = — =————— dv =————- dv

T c — IT c — lv

 

(combined characteristic, e. g., where s minus the specific entropy v is a unit of V)

S2 _ v2 , _

F A c “ nD f dV c “ n Tl 1 V2

Ds = 3——— R — s2 — Sj = ———————— R In —

J C-1JV C — 1 Vi

Sl V! 1

 

Base Calculations
Base Calculations

~Cp(n~1) + Cpn~CvnlnT1

 

N — 1

 

To

 

Base Calculations

N-1

 

T,

подпись: t,

(cp — cv = R)

подпись: (cp - cv = r)= cnln — — Rln p T,

That is,

T2 di„P2

S7 — Si = cnln—— — Rln p T.

For example, the air at T = 288 K and standard atmosphere is compressed to P = 5 times of standard atmospheric pressure with a temperature of 456.1 K. Calculate the entropy values.

For air, Cp = 1005 J/kgK and R = 287 J/kgK.

Before compression,

S2 = 1005 In 288 -2871n^^- = 53

273.15 TOC o "1-5" h z 0.101

After compression,

I™* . 456 1 „о-,, 5×0.101 „

S2 = 1005 In ———— 287 In—:—:—— =53

0.101

подпись: 0.101273.15

There is no change in entropy.

The pressure, humidity, and temperature of the ambient air inducted into an engine, at a given speed, affect the air mass flow rate and hence the potential power output. Correc­tion factors are used to adjust measured wide open throttle power to standard atmospheric conditions to provide a more accurate basis for comparison among engines.

The basis for these correction factors is the equation for one-dimensional steady com­pressible flow through an orifice.

1.2

0.6

99

T + 273

298 1.01325 bar 20°C (293°K) ps= 1.205 kg/m3

Correction Formulae

• 88/195/EEC

• EEC80/1269

• ISO 1585

• JISD 1001

• SAEJ1349

• DIN 70 020

Correction formulae: 88/195/EEC =

Pressure: Standard atmosphere Temperature:

Corresponding density: where

Base Calculations

P = barometric pressure (atmospheric) — partial vapor pressure (kPa) T = ambient intake air temperature in degrees Celsius (°C) ps = density of air

The standard conditions are considered to be a barometric pressure of 99 kPa and a temperature of 298°K (25°C).

The partial vapor pressure required to calculate the dry barometric pressure term P in the equation 88/195/EEC can be obtained in two ways.

1. Using wet and dry air bulb temperature measurements and psychometric tables

2. Using relative humidity and ambient air temperature measurements and the follow­ing equations:

Dry barometric pressure = Atmospheric pressure — Partial vapor pressure

P = Pa — Ppv

When

Relative humidity (%) x Saturated vapor pressure (kPa)

Ppv =

100

The saturated vapor pressure Psv (kPa) is given by

Log10Psv =

30.59051 -8.2 xlog10(T) +(2.480E — 3) xT — (3.142 x 3l)

T

Where

T(k) = Ta(°C) + 273.15 Ta = ambient air temperature Correction formulae (DIN):

Correction factor = 760(273 + T/P)293

Where

760

подпись: 760273 + T P + 293

P = barometric pressure (mm Hg)

T = ambient intake air temperature (°C)

Examples of Calculations Required Within the Test Cell Environment Test Bed Fuel Flow Measurement

Many test beds are equipped with gravimetric fuel measuring devices that provide an output in the form of mass of fuel used per selected time interval (e. g., grams of fuel consumed in 30 seconds).

Test Bed Airflow Measurement

To calculate the air/fuel ratio, we need to know the fuel flow and the air used by an engine. This can be done chemically using gaseous emission measuring equipment, but a cross-reference is a prerequisite of intelligent testing.

The actual measurement of air mass flow rate is not carried out on a regular basis because it can be calculated from SPINDT AFR (exhaust gas) and from brake-specific fuel consumption. However, various airflow meters are available for measuring engine air consumption, namely, the following:

• Lucas Dawe hot wire corona discharge

• Hot wire wheat stone bridge balance system

• Alcock viscous airflow meter

• Base orifice plate

All types of measuring devices should be used upstream of any intake pressure pulsa­tions to allow for stable readings to be taken and logged. (This usually means prior to an air cleaner/resonator volume.)

Brake Specific Fuel Consumption

Having measured the fuel mass flow rate by one of the methods highlighted in this book, a more useful parameter—the brake specific fuel consumption (BSFC)—can be calculated. It is defined as

„„„„ Fuel mass flow rate mf

BSFC =—————————— =——

Measured brake power P P

With units of grams of fuel consumed to deliver 1 kW over 1 hour (g/kWh).

The BSFC provides a measure of how efficiently an engine is using fuel supplied to produce work. For spark ignition engines, typical values of BSFC at wide open throttle are between 250 and 260 g/kWh.

Figure 13.1 gives an example of a specific fuel consumption contour map for the complete operating range of an engine. The contour lines represent lines of constant specific fuel consumption; the lower the figure, the more efficient the running condition.

Brake specific fuel consumption also can be applied to other items whose output is compared with the brake power delivered by the engine. For example, in emissions, one will note brake specific hydrocarbons (BSHC).

/

/ /

^ і r

УПГгч

Ц—1—1—1

4-

ZL J’

‘ t

Arv 1 ^

-f — A i /

A (

^N4

ІГх]—1

VI K

Л-Vvjjj b1

V 245

1J J 1 [ і

F

Щ

Z-Щ

Wfff

/ Г7 1 /

Zip

AJ-K

L-L-l A

І Jo L7 !

Щ

—і ———- r^T^-ii—.

Ш±

Ш

ВМЕР

(bar)

Rev/Mis

Figure 13.1 Oyster curve example.

подпись: figure 13.1 oyster curve example.Brake Specific Air Consumption

Similarly, brake specific air consumption (BSAC) is defined as the air mass flow rate per unit power output

BSAC = Airflow/power (units kg/kWh)

Where power is the brake power output. This provides a measure of how efficiently an engine is using the air supplied to produce work.

Factors that affect the BSFC and BSAC are as follows:

• Compression ratio

• Air/fuel ratio and ignition settings

• Friction—Rubbing in the engine and accessories

• Pumping losses—Intake system restrictiveness/cylinder head design and exhaust system design

• Calorific value of the fuel

• Barrel swirl ratio

• Heat pickup through the induction system

• Heat transfer from the combustion chamber

• Mixture preparation

• Fuel/air mixture distribution

Добавить комментарий

Ваш e-mail не будет опубликован. Обязательные поля помечены *