• Chapter 2 – and the First Law of – Forms of energy – transfer –

Forms of energy

• Thermal, mechanical, kinetic, potential, electric, magnetic, chemical & nuclear • Sum is total energy E (kJ) of system or per unit e = E/m (kJ/kg) • Thermodynamics deals with changes in energy so reference state is arbitrary, e.g. Eref =0 • Useful to consider macroscopic “forms you can see” i.e. kinetic and potential, and microscopic “forms that are hidden” i.e. related to molecular structure and activity • Sum of microscopic forms is called U (kJ)

Macroscropic +Microscopic = Total Energy forms, KE+PE forms, U

Macroscopic forms of energy

- due to system w/rt reference frame – KE = mV 2/2 (kJ) – ke=V 2/2 (kJ/kg) – V is relative to fixed reference frame • - due to system elevation in gravitational – PE=mgz (kJ) – pe=gz (kJ/kg) – g is gravitational ; z elevation of cg relative to reference

1 Microscopic forms of energy

• KE of molecules called • Molecular . . . sensible energy (~ T) • Intermolecular bind molecules to each other • Strongest in solids, weakest in • Adding energy to overcome forces breaks bonds causing phase change e.g. latent energy • binds atoms into molecules; nuclear energy binds nucleus

Total energy and units

• Total energy and specific total energy

E=++=+ U KE PE U mV2 / 2 + mgz e=++ u ke pe =+ u V2 / 2 + gz

• Work has units of • KE and PE have same units as work • In SI, energy unit is -meter or N-m called J or kiloJoule kJ • In English, units of work, KE, or PE are foot- pound force, ft-lbf and (Btu)

Watch Energy Units

• Changes in KE and PE have SI units of kg-m2/s2 • 1 N = 1 kg-m/s2 • 1 N-m = 1 kg-m2/s2 • 1 J = 1 N-m • 1 kJ = 1000 J = 1000 N-m • Therefore, to convert changes in KE and PE to kJ you need to divide by 1000 • In English units, you need to (a) convert from lbm to lbf via 32.2 lbm-ft/s2/lbf and then convert from ft-lbf to Btu by dividing by 778 ft-lbf/Btu


• Recall mass cannot cross the boundaries of a , but energy can • Energy can cross the boundaries of a closed system by two modes: – Heat, Q – Work, W

Heat transfer (cont.)

• Heat is a form of energy that is transferred between a system and its surroundings by virtue of a temperature difference in a direction of higher T to lower T

TB T T Tb air A

Heat transfer (cont.)

• Heat is energy in transition and is only recognized as it crosses the boundary • Heat is form of energy so units are kJ or Btu (1 kJ=0.94782 Btu)

3 Heat transfer (cont.)

• Adiabatic process – no heat transfer • There are two situations when this can occur – Insulated system – No temperature difference (no driving force)

Quantifying heat transfer

• Amount of heat transferred during a process between two states 1 and 2 is (notation, tbd): 2 ==δ → == Q12 Q∫ Q q 12 qQ 12 / mkJkg ( / ) 1 • Heat transfer δQ Qɺ= →= Q QdtQt ɺ =∆ ɺ dt ∫ • Heat transfer occurs in a particular direction so we have sign convention – Heat transfer to a system is positive – Heat transfer from a system is negative • Q is usually given or is zero or is only unknown left in equations

Heat transfer sign convention

 Q > 0: heat transfer to the system  Q < 0: heat transfer from the system  Q = 0: adiabatic

4 Heat transfer modes • Heat transfer by 3 modes – conduction (solid/fluid; molecular collision) ~ derivative of T – convection (fluid; motion) ~ T difference – (no medium needed/wave vs. ) ~ T^4

Work • If energy crosses boundary is not heat (due to temperature difference) then it is work [kJ or Btu] • More specifically, work involves a force acting through a distance • Units are – N-m or J; usually kJ in SI and – lbf-ft or Btu with 1Btu=778.169 lbf-ft – 1Btu=1.055056kJ • Examples include a rising piston, rotating shaft, current carrying wire, etc. • Criteria: If the sole effect could be the raising of a

Work notation

2 ==δ → == W12 W∫ W w 12 wWmkJkg 12 / ( / ) 1 • Note special derivative symbol indicating path function (tbd) • Sign convection – Work done by a system is positive – Work done on a system is negative • Rate of doing is work is called with units of kJ/s or kW δW Wɺ= →= W WdtWt ɺ =∆ ɺ dt ∫

5 Work sign convention

• Note: It is opposite of heat transfer sign convenction  W < 0 is work done on the system  W > 0 is work done by the system • Both heat and work sign conventions are related to whether or not the interaction increases or decreases the energy content of the system

More on heat and work

• Direction of work and heat depicted via arrows or subscripts in and out • For given system, both Q and W depend – Process path – Initial and final states • Point functions only • These are called path depend on initial/final functions state • They have inexact • They have exact differentials differentials δQ, δ W⇔ dQ , dW dE, dV

Illustration of point vs. path functions ∆=3 = ∆=3 = VAA3 mW , 8 kJ V BB 3 mW , 12 kJ 2 2 =−=∆δ =≠∆ ∫dVVVV2 1 (for A or B) ∫ WWW12 1 1

6 Mechanical forms of work

• Recall from physics that mechanical work involves a force acting through a distance

2 W= Fs if force is constant, else δ W =Fi d s , W = ∫ F(s) ds 1

Moving boundary work

• Form of mechanical work • Interaction between system and surroundings . . . – Must have a force acting on boundary – Boundary must move e.g. force must act through a distance – If either is absent, then no boundary work

Moving boundary work

• Expansion or compression of a gas in a piston-cylinder device • We can only evaluate work analytically if we have a QE process (slow) • This is because work is a path function and we need to know path • Formula at right δW= Fds = PAds = Pd V consistent with sign b 2 convection ∴ = Wb ∫ P( V ) dV 1

7 Moving boundary work and its relation to process path • under process path on P-v diagram represents QE boundary work • Gas can follow several different paths as it expands from state 1 to 2 • Different so different work • Work is a path function 2 2 == = = Area A∫ dA ∫ PdV W b 1 1

Work is a path function

Work for cyclic devices

• If work was not a path function, no work producing cyclic devices could exist i.e. zero net work output • Net work for a cycle is difference between work done by system and work done on system • If cycle proceeds clockwise, net work is positive (output); if counterclockwise the net work is negative (input)

W= W + W = W cycle12B 21 A net > 0 CW  Wnet  < 0 CCW 

8 Moving boundary work – special cases and derivation of formulas 2 = = Wb PdV 0 ∫ V= const 1 1

2 = = −=∆ →=∆ υ Wb PdV PVV(2 1 ) PVkJ () wb P (/) kJkg ∫ P= const ÷m 1

2 22 2 = =C === dV V2 V 2 = V 2 Wb PdV dVC Cln PV11 ln PV 22 ln ∫PV= const ∫ ∫ 1 1V 1 VV1 V 1 V 1

2 2 −+n1 −+ n 1 − V− V  PV − PV W== PdV CVn dV = C 2 1 = 2211 b ∫n ∫   PV= const −n +1n≠1 1 − n 1polytropic 1  

Moving boundary work – special cases on a P-v diagram P



V1 V2 V


• In what forms can energy cross the boundaries of a closed system? • When is the energy crossing the boundaries of a closed system heat and when is it work? • What is an adiabatic process? • A gas in a piston-cylinder device is compressed, and as a result its temperature rises. Is this a heat or work interaction? • What are point and path functions? Give some examples.

9 Summary

• On a P-v diagram, what does the area under the process curve represent? • Show that 1 kPa-m3 = 1 kJ


• Other forms of mechanical work • Examples

Other forms of mechanical work

• Recall mechanical work involves a force acting through a distance • We can determine whether or not a boundary interaction involves mechanical work by the following test:

Could the sole effect of the interaction be the raising or lowering of a weight?

10 Examples of other forms of mechanical work: Shaft work • If a rotating shaft crosses the boundary of a system, then this is a mechanical work interaction • Why? Because you could tie a string around the shaft and attach a weight to end of the string e.g. as shaft turns, weight is raised and lowered TFr= →= FTr/ →= s( 2 π rn) ∴ = = ()π Wsh Fs( Tr / )2 rn =π →ɺ = π 2nTkJ () Wsh 2 nTɺ ( kW ) • T is and n is revolutions

Examples of other forms of mechanical work: Spring work • Compression or expansion of a spring involves a force acting through a distance • We consider only linear elastic springs e.g. Hooke’s law δ = → =→ Wspring Fdx F kx linearspring x= 0,0 F = 1 W= kx(2 − x 2 ) ( kJ ) spring 2 2 1 • How did I get the final work expression?

Rotating shaft and spring work

11 Other mechanical forms of work

• Work done on elastic bars • Work associated with stretching a film • Work done to raise or to accelerate a body • Non-mechanical forms of work – electrical, magnetic, etc. (see text) • Read about above briefly in textbook


To be done in class