PH08 Notes

version 0.0.1

8.1.1 Impedance

Definition: Impedance is the resultant of the reactance and resistance of the circuit.

Z = V0 / I

Reactance of a capacitor, XC = VC / I

Reactance of an inductor, XL = VL / I

8.1.2 Phase Relationships

• Inductor: VL leads I by 90^O

• Capacitor: VC lags behind I by 90^O (i.e. I leads VC)

• Resistor: No phase difference between VR and I.

The maximum voltage, V0 across an a.c. circuit with inductive, capacitative and resistive loads is the resultant of the voltages across each of them.

VL and VC cancel out because they work against each other (at 180^O on a phasor diagram). The resultant is found by Pythagoras from the phasor diagram.

V0 = †( ( VL - VC )² + VR² )

For an LC circuit, V0 = | VL - VC |

As X is in phase with V and I for any particular component,

Z = †( ( XL - XC )² + R² )

Resonance

When XC = XL, so Z = R. The resonant frequency f = 1 / ( 2p †( LC ) )

8.1.3 Power Factor

Power loss in a.c. circuits is due to the resistance. Ploss = VRI

Ideally, a circuit will have no resistance: R = 0. Therefore the phasor diagram will have no horizontal component. However this is never possible, and VR > 0.

f is the phase angle, the angle between V0 and VR. Therefore, VR = cos f and:

Ploss = V0 cos f I

cos f is called the power factor. An ideal circuit has cos f = 0, as this means no power loss. This is called unity power factor and occurs when the phase angle f = 90^O.

Power definitions

Indicated Power Pid = area of pV loop x cycles/second x no of cylinders

Input Power Pip = calorific value of fuel x fuel flow rate

Output Power Po = (total mechanical power output) = Tw

Friction Power Pf = (power lost due to friction) = Pid - Po

Efficiency

Three types: overall, thermal and mechanical.

Overall Eff. = Po / Pip

Thermal Eff. = Pid / Pip

Mechanical Eff. = Po / Pid

Thermodynamics

Law Zero: If A is in thermal equilibrium with C, and B is also in thermal equilibrium with C, then A and B must be in thermal equilibrium with each other.

Law One: DQ = DU + DW

DQ = the (change in the) amount of heat put INTO the gas.

DU = change in the gas' internal energy. Increasing U raises the gas temperature.

DW = change in external work done by the gas. Positive DW means gas expands.

DU = C DT (C = heat capacity of this sample of gas)

DW = p DV for system at constant pressure.

Changes in Q, U, W

Isothermal process (gas T does not change): U constant

Isovolumetric process (gas V does not change): W constant

Adiabatic (no heat put in or taken out): Q constant

Isobaric process (pressure constant).. a bit more complex.

DW = p DV. p = constant.

So for an isobaric process: DQ = DU + p DV

Law Two: pV loop and the impossibility of 100% efficiency

All the heat applied to an engine can never be transferred into mechanical energy during a complete cycle. => an engine can never be 100% efficient.

Petrol engine (V+ = volume increases, V- = volume decreases)

V+ Induction (air/fuel in)

V- Compression (mixture compressed)

Spark- ignition

V+ Power

V- Exhaust (valve opens, gas pushed out)

The idealised cycle does not show this. It shows the adiabatic Compression phase, followed by the isovolumetric Ignition phase, followed by the adiabatic Power phase, and the exhaust phase is represented as a single isovolumetric change.

Diesel engine

Typically, hot air is drawn in, and fuel is injected later. Ignition is caused by the increase in internal energy of the gases due to the compression phase. The gases are compressed more strongly. The ideal cycle is different because the ignition phase is shown as an isobaric change - horizontal on a pV graph (the air expands at constant pressure).

Second law diagrams

Energy comes from the source, Qin and goes to the sink Qout. In doing so, some of the energy can be used by an engine to do work.

The total power in from the source, Qin, is the total area under and within the pV curve for a heat engine.

The total power going out to the sink, Qout, is the area under the curve only.

The work done by the engine is the area within the loop, ie.

W = Qin - Qout

(thermal) efficiency = Useful work / Energy in = W / Qin

But Q is thermal energy: Q = CT (C = heat capacity).

So efficiency = CTin - CTout / CTin = TH - TC / TH = 1 - ( TC / TH )

TC = temperature of sink (Tout)

TH = temperature of source (Tin)

For efficiency = 100%, TC / TH must be zero. But this is not possible. However efficiency tends to 100% for high source temperatures and low sink temperatures. But the greater the difference is, the faster heat is lost, so efficiency is still reduced.

TC < TH because if it were not, there would be no temperature gradient, or energy would move from the wrong place.

Heat pumps

Same heat engine cycle, direction reversed.

Fridge cooling effect caused by constriction in pipe. When the coolant is forced through the constriction the pressure suddenly decreases on the other side. The coolant does work in a near-adiabatic expansion: the energy to do so coming from it's internal energy. Thus the coolant's temperature drops.

Heat pumps

• work better in summer

• air has little mass and is therefore not a very good source of heat

• cost of pump is high

• pump is noisy

Fluids

A fluid is anything that can flow: a liquid or a gas.

An ideal fluid is:-

• incompressible

• non-viscous (no friction within liquid)

• streamlined

  (streamlines parallel (particles in the fluid travel along streamlines))

• non-turbulent

  (all particles in the fluid travel in the same direction with constant velocity)

Gauge pressure

The difference between the pressure in a system and atmospheric pressure. Needs to be corrected to absolute pressure by adding or subtracting atmospheric pressure. Manometers show gauge pressure.

hrg = static pressure ½rv² = dynamic pressure

Equation of Continuity

The rate of flow through a pipe (m³/s) is constant throughout that pipe.

Cross-sectional area x velocity = Av = constant

Change in height

If there is no change in A and v, then:

the work done in raising the fluid = gain in the p.e. of the fluid.

=> the pressure of the liquid increases or drops by Dhrg.

Change in cross sectional area

work done in pushing fluid into the constriction = gain in k.e. of the fluid.

=> pressure of the liquid increases or drops by ½rv²

Bernoulli's equation

hrg + ½rv² + p = constant

Aerofoil

Above the aerofoil the air travels faster than below it. As ½rv² + p = constant, this means that the pressure is lower above the aerofoil than below it. So a force acts upward.

Carburetor

Fuel is forced through a narrow tube, which causes an increase in v (continuity eqn) and a resulting decrease in p (Bernoulli's eqn). This tube expands suddenly, but v remains high, as air is drawn in by the low pressure. The result is an air/fuel mixture that is piped to the cylinder.

Spinning ball

The spin causes friction with the air, resulting in an air velocity differential across the ball, which is pushed in one direction by the pressure difference that results. This can create uplift, and can make a ball curve through the air.

Pitot-static tube and Venturi tube

change in height is due only to the dynamic pressure of the liquid.