Phys211 Week 07: 3/04, 3/06

Front 1

Entropy

Back 1

physical quantity providing a measure of the degree of molecular disorder in a system

Front 2

S

Back 2

entropy

Front 3

S formula

Back 3

S = k_B ln(P)

S: entropy
k_B: Boltmann's constant
P: thermodynamic probability

Front 4

Clausius equation

Back 4

to compute ∆S

Front 5

∆S for heat reservoir

Back 5

∆S = Q / T

S: entropy
Q: heat
T: in Kelvin

Front 6

∆S for well-defined c

Back 6

∆S = cm ln(T_f / T_i)

S: entropy
c: specific heat
m: mass
T_f & T_i: final & initial temperatures, Kelvin

Front 7

Principle of Entropy Increase

Back 7

total entropy of a closed system can't decrease

S_f ≥ S_i

Front 8

What does the principle of entropy increase imply about ∆S?

Back 8

∆S > 0

because S_f ≥ S_i

Front 9

Given a house at 20°C with a window open to the outdoors at 0°C, can 100,000 J of Q be transferred to the house (i.e. can the house get warmer)?

Back 9

high temp heat reservoir: the house
cold temp heat reservoir: the outdoors
T_h = 293.15, T_c = 273.15 (remember to convert to K)

∆S_hot = Q / T or 100,000 J / 293.15 = 341.12 J/K

∆S_cold = Q / T or -100,000 J / 273.15 K = -366.10 J/K

∆S = ∆S_hot + ∆S_cold or 341.12 - 366.10 = -21.98

∆S = -21.98, < 0 ∴ violates principle of entropy increase

Front 10

heat engine

Back 10

device that converts E_th to useful energy

e.g. combustion engine, turbo generator

Front 11

heat engine & heat reservoirs

Back 11

a heat engine is designed to work between high & low temp heat reservoirs

Front 12

heat engine cycle

Back 12

the engine runs a process, returns to the same state of thermal equilibrium, repeat

S_f_engine = S_i_engine ∴ ∆S_engine = 0

Front 13

Heat engine & waste

Back 13

the conversion of high temperature thermal energy to work necessarily loses additional thermal energy to a low temperature reservoir

Q_h ≠ W

Front 14

What restricts Q_h ≠ W?

Back 14

Principle of entropy increase

Front 15

Conservation of total E applied to W & Q

Back 15

W_out = Q_h - Q_c

a waste of E_th is required when converting Q_h into W

Front 16

permutations of W_out = Q_h - Q_c

Back 16

W_out = Q_h - Q_c

Q_h = W_out + Q_c

Q_c = Q_h - W_out

Front 17

Q, absorbed or liberated?

Back 17

liberated < 0 < absorbed

Front 18

efficiency of heat engine

Back 18

e

(class using an uppercase e, book uses lower case)

Front 19

e formula

Back 19

e = W_out / Q_h

given W_out = Q_h - Q_c

e = (Q_h - Q_c) / Q_h

Front 20

e formula, convert to %

Back 20

* 100

Front 21

range of e of typical real-life processes

Back 21

40-45%

Front 22

Carnot's equation

Back 22

describes maximum efficiency

Front 23

Carnot's formula

Back 23

e_max = 1 - (T_c / T_h)

Front 24

analyze Carnot's

Back 24

Given that e_max ranges 0-1

T_c / T_h must be resolve to 0-1

∴ T_h must be in the denominator

Front 25

Heat Pump

Back 25

transfers heat from cold reservoir to hot reservoir

requires taking work in to do so

Front 26

COP

Back 26

Coefficient of Performance

applies to heat pumps

Front 27

heat pump uses

Back 27

heat pump: used to warm

air conditioner, refrigerator: used to cool

Front 28

AC or refrigerator

Back 28

pull Q_c from house or refrigerator, dump Q_h to reservoir (outdoors for AC, house for refrigerator)

Front 29

COP formula

Back 29

COP = what you get / what you paid

what you get
Q_h, if being used to warm
Q_c, if being used to cool (AC, fridge)

what you paid
W_in

Front 30

COP_max

Back 30

maximum Coefficient of Performance

applies to heat pumps

Front 31

COP_max formula

Back 31

COP_max = temp you want / (T_h - T_c)

temp you want
T_h, if being used to warm
T_c if being used to cool (AC, fridge)

Front 32

BTUH

Back 32

BTUs / hour

1 w = 1 J / s = 3.41 BTUH

Front 33

W_in

Back 33

measured in J / s = w or kW

Front 34

EER

Back 34

energy efficiency rating

EER = 3.41 COP

an appliance is generally deemed efficient if EER > 10

Front 35

convert between EER & COP

Back 35

EER = 3.41 COP

COP = EER / 3.41

Front 36

Between a house maintained at 70° F and a solar collection system with a rock reservoir maintained at 60°F lies a heat pump with a COP of 4 that vents 5000 BTUH into the house.

Find EER, # watts to run heat pump, COP_max

Back 36

EER = 4 COP * 3.41 EER / COP = 13.64

# watts
used to heat, so Q_h / W_in
Q_h in BTUH, convert to watts → 5000 BTUH * 1 w / 3.41 BTUH = 1466.28 watts
W_in = Q_h / COP → 1466.28 w / 4 = 366.57 w

COP_max
T_h / (T_h - T_c) = 294.3 / (294.3 - 288.7) = 52.55

Front 37

Find power to run an AC of EER =12 that extracts 18000 BTUH to the outdoors at 95°F to maintain a house at T_c 70°F.

Back 37

this is a heat pump used for cooling
Q_c = 18000 BTUH * 1 w / 3.41 BTUH = 5278.59 watts
COP = EER / 3.41 → 12 / 3.41 = 3.52

COP = Q_c / W_in → W_in = Q_c / COP
W_in = 5278.59 watts / 3.52 = 1499.6 watts

alternatively, the conversion from BTUH to watts could have been done after calculating W_in in BTUH

Front 38

Fluid

Back 38

system in which molecules move relatively freely

for this class, a liquid or gas

Front 39

hydrostatics

Back 39

aka fluid statics

study of fluids at rest

Front 40

hydrodynamics

Back 40

aka fluid dynamics

study of fluids in motion

Front 41

Density

Back 41

measures amount of fluid per unit volume

Front 42

ρ (rho)

Back 42

density as mass / volume

ρ = m / V in kg / m³

Front 43

D

Back 43

density as weight / volume

D = w / V in N / m³

Front 44

convert between D & ρ

Back 44

given w = mg and g=9.8

D = ρ * g
ρ = D / g

Front 45

P

Back 45

pressure

force exerted per unit area