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Thermodynamics

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- Introduction
- First Law of Thermodynamics
- Calculation of Work
- PVT diagrams
- Thermodynamic processes
- Examples

- Governs theoretical energy & statistical limits, not engineering details
- (we’re not “Click and Clack” www.cartalk.com)

- If fuel releases 100 kJ energy
- cannot get more than 100 kJ mechanical work
- (Actually less since need to exhaust to cold)

- If heat pump does 100 kW mechanical work
- Can obtain at least 100 KW heat
- (Actually more since heat added from cold)

- 2 Fundamental Rules
- Energy must be conserved - First Law (ΔU = Q – W)
- Must be statistically probable - Second Law (disordered->ordered)

- First Law Thermodynamics
ΔU = Q – W

- U internal energy, ΔU change in internal energy.
- Q is heat, positive in.
- W is mechanical work, positive out.

- Don’t let W sign convention spook you!
- Negative work in, double negative U increase
- Positive work out, negative U decrease
- Energy diagram
- Convention used so expansion does positive work
- Example 15-1

- Work done on piston
- W = F * d (Physics 103)
- W = P * Area *d
- W = P * ΔV
- (in general integration)

- Volume on x, Pressure on y
- Work is area under curve
- Temperature are lines of constant PV (isotherms)
- Higher temperature, higher isotherm
- Same temperature, constant isotherm
- Return to same temperature, return to same isotherm
- Since U = 3/2 nRT, lines of constant U

- Q is the fudge factor that makes up difference

- Isovolumetric (temp increase)
- Pressure increase, constant volume
- Work zero
- Temperature increase
- Heat in Q = ΔU

- Isobaric (expansion)
- Pressure constant
- Work = PΔV
- Temperature increase
- Heat doubly in ΔU = Q -W

- Isothermal (expansion)
- Constant temperature, internal energy ΔU =0
- Work out W = ∫ P dV
- Heat in equals work out Q = W

- Adiabatic (expansion)
- Q = 0
- ΔU = - W
- As expands, falls to lower isotherm

- Example 15-5
- Calculate the works
- Calculate ΔU
- Know BA
- Know BD and DA if more information U = 3/2 nRT

- Calculate Q
- Calculate isothermal work (Net work out)

- Example 15-6

- One mole of an ideal gas undergoes the series of processes shown in the figure. (a) Calculate the temperature at points A, B, C and D. (b) For each process A B, B C, C D, and D A, calculate the work done by the process. (c) Calculate the heat exchanged with the gas during each of the four processes. (d) What is the net work done during the entire process? (e) What is the net heat added during the entire process? (f) What is the thermodynamic efficiency?

- For paths ab and cd pressure is constant
- For paths bcand da volume is constant

- Net work during entire cycle 2000J
- Net heat absorbed/expelled during cycle 2000 J
- Change in internal energy during cycle 0 J
- Efficiency (common sense)

- Problem 11
- Know P, V, and n, can find T
- Note “ideal gas”, can use U = 3/2 nrT
- Calculate ΔU directly, or shortcut

- Problem 12
- If you get Uab by any path, know for all paths
- Uab = - Uba
- U’s along path must add