METBD 330: Thermodynamics

Chapter 5: SECOND LAW OF THERMODYNAMICS

Determine if a process is possible
Defines the "direction" a process must follow

EXAMPLE: Hot coffee at 50^oC in air at 20^oC. Heats up to 70^oC ??

First Law Analysis:

Q - W = DU (for a closed, stationary system)
W = 0, so Q = DU
if U_coffee@50C = 500 kJ and U_coffee@70C = 600 kJ
Q = U₂ - U₁ = 600 - 500 kJ = +100 kJ
and the First Law of Thermodynamics is satisfied !

BUT, we know heat transfer (Q) occurs FROM a warmer body TO the cooler surroundings !!!!

We need a way to test the FEASIBILITY of a process which will come from the SECOND Law of Thermodynamics.

First, consider where Q comes from or goes to for a process.
Thermal Energy Reservoirs: can absorb Q without changing temp. and can supply Q without changing temp.
Remember the coffee cup. It actually cools off. The surrounding air in the room absorbs the heat but remains at 20^oC
Thermal Energy Reservoirs need to be "large" with respect to the system being studied.
SOURCE: a thermal energy reservoir which SUPPLIES heat
SINK: a thermal energy reservoir which ABSORBS heat

HEAT ENGINES: Convert heat (Q) into work (W)

Reference: Problem 3-46C: A fan in a closed room running all day raises the air temperature (Work is converted to Heat).
If you unplugged the fan and heated the room, could you add heat and get the fan to turn ????

Steam Power Plant

Fig5_10.gif (3784 bytes)

W_IN < W_OUT

Q_NET = Q_IN - Q_OUT

W_NET,OUT = W_OUT - W_IN

HEAT ENGINES:

Receive Q from a high temperature SOURCE
Convert some Q to W
Reject some Q to a low temperature SINK
Operate in a CYCLE (DU = 0)

The Working Fluid is the substance used in the cycle (steam, R-134a, air, etc.)

NOTE: Sign convention (all values are positive, subscript shows direction)

Q_IN = heat from the SOURCE
Q_OUT = heat to the SINK
W_OUT = work delivered BY the turbine
W_IN = work delivered TO the pump.

For the working fluid, Q - W = DU (for a cycle, DU = 0)

Q - W = 0

Q_IN - Q_OUT - (W_OUT - W_IN) = 0

W_OUT - W_IN = Q_IN - Q_OUT

W_NET,OUT = Q_IN - Q_{OUT

[Eqn 5-2]}

THERMAL EFFICIENCY, h_{th
[Eqn. 5-3, 5-4 and 5-5]}

IF the SOURCE is at temperature T_H, and the SINK at temperature T_L,

we define:

Q_H = the magnitude of heat from the SOURCE at T_H
Q_L = the magnitude of heat from the SINK at T_L

(magnitude = always positive numbers)

SO, [Eqn. 5-6, 5-7, 5-8]

SECOND LAW OF THERMODYNAMICS:

A device which operates in a cycle cannot receive heat from a single source (Q_H) and convert it completely to work (W_NET,OUT).

If it could, W_NET,OUT = Q_H and h_th = W_NET,OUT/Q_H = 1.0
also, with W_NET,OUT = Q_H and W_IN = 0 (NO Pump)
NO CYCLE can occur !!!

Section 5-4: REFRIGERATION

Characteristics:

Moves Q from a low temperature reservoir to a high temp. reservoir
Operates on a cycle

Refrigeration Cycle: Pg. 193

Fig5_20a.gif (5134 bytes)

NOTICE:

Q_L is heat from the T_L reservoir

Q_H is heat to the T_H reservoir

(opposite of heat engines)

Figure 5-20

EFFICIENCY OF REFRIGERATOR: COP_R

(a different measure than for heat engines)

[Eqn. 5-9]

From a First Law Analysis, W_IN = Q_H - Q_L [Eqn. 5-10]

SO,

[Eqn. 5-11]

HEAT PUMPS

Run on the same cycle as a refrigerator
Have a different objective
Objective of Refrigeration: Maintain a cold temperature in a cold space by removing Q_L from it.
Objective of Heat Pumps: Maintain a high temperature in a warm space by supplying Q_H to it.

EFFICIENCY OF REFRIGERATOR: COP_HP

(a different measure than for heat engines and refrigeration)

[Eqn. 5-12]

From a First Law Analysis, W_IN = Q_H - Q_L

SO,

[Eqn. 5-13]

NOTICE:

for heat engines: h_th < 1
for refrigeration: COP_R may be > 1
For heat pumps: COP_HP > 1
and COP_HP = COP_R + 1

EER = Energy Efficiency Rating (a consumer measure)

EER = 3.412 COP_R
Typical air conditioner: 8 < EER < 12
(which is 2.3 < COP_R < 3.5)

SECOND Law of Thermodynamics (from a refrigeration perspective) (Clausius Statement)

It is impossible for a cyclic device to simply move heat from Q_L to Q_H without work, W_IN, occurring.

EXAMPLE: (similar to problem 5-44) (Energy in terms of rate)

Ice machine (i.e., refrigeration system) has a COP_R of 2.2
Water enters at 15^oC, ice at -5^oC is produced at a rate of 8 kg/hr
(Energy to form ice at -5^oC from H₂O at 15^oC, 384 kJ/kg)

Example 5-46: Refrigeration A/C

COP_R = 2.5, 800 kg of air in a house at 32^oC
for the air: C_V = 0.72 kJ/kg ^oC and C_P = 1.0 kJ/kg ^oC
after 15 min. (900 sec.), air temp. is 20^oC
Determine the power used by the A/C unit.

First Law Analysis on Air: (Closed, Stationary System with No Work)

Q = DU = m C_V (T₂ - T₁)

= (800 kg) (0.72 kJ/kg ^oC) (20 -32) ^oC

Q = -6912 kJ or 6912 kJ

from the system over 15 minutes

REVERSIBLE and IRREVERSIBLE Processes Section 5-6

Reversible Processes:

can be reversed
leaving NO trace on the surroundings
are idealized (not real)

Irreversible Processes

cannot be reversed without leaving the surroundings changed
are real processes

Why do we care about REVERSIBLE Processes (that are not real) ?

Recall, for heat engines, h_th < 1 (typically, 20% < h_th < 40%)
A heat engine that runs with REVERSIBLE processes, produces the MAXIMUM amount of Work
Similarly, Refrigerators and Heat Pumps which operate with REVERSIBLE processes require the MINIMUM Work input.
We evaluate these systems with respect to the ideal, reversible processes.

WHAT MAKES A PROCESS IRREVERSIBLE ?

Friction
Fast expansion or compression
Heat transfer

- ALL REAL PROCESSES are IRREVERSIBLE !

Internally reversible processes = no irreversibilities in the system

Externally reversible process = no irreversibilities outside the system

CARNOT Cycle Section 5-7

a hypothetical, reversible cycle
a cycle of 4 processes (heat engines)

Carnot Cycle Processes:

Expansion (Dv) at constant T (Q_H is supplied to maintain T)
Expansion (Dv), adiabatic (T drops)
Compression (Dv) at constant T (Q_L is released to maintain T)
Compression (Dv), adiabatic (T increases)

On a Pv diagram, it looks like this:

Fig5_39.gif (3841 bytes)

Notice:

processes 1-2 and 2-3 are expansion (v increases)
processes 3-4 and 4-1 are compression (v decreases)

Running the same cycle IN REVERSE gives the Carnot Refrigeration cycle:

Expansion (Dv), adiabatic (T drops)
Expansion (Dv) at constant T (Q_L is supplied to maintain T)
Compression (Dv), adiabatic (T increases)
Compression (Dv) at constant T (Q_H is released to maintain T)

On a Pv diagram, it looks like this:

Fig5_40.gif (3366 bytes)

Notice:

processes 1-2 and 2-3 are expansion (v increases)
processes 3-4 and 4-1 are compression (v decreases)

Differences between the CARNOT heat engine and refrigeration cycles:

Direction of Q_L and Q_H
CW vs. CCW (based on direction of first process, i.e., expansion at constant T or adiabatic)

CARNOT PRINCIPLES Section 5-8

Between two given thermal energy reservoirs, an IRREVERSIBLE heat engine will have a LOWER thermal efficiency (h_th) than that of a reversible heat engine.
Between two given thermal energy reservoirs, all REVERSIBLE heat engines must have the SAME thermal efficiency (h_th).

ABSOLUTE TEMPERATURE SCALES Section 5-9

FOR REVERSIBLE PROCESSES … ONLY !

(Use ABSOLUTE T) [Eqn. 5-18]

THE CARNOT HEAT ENGINE Section 5-10

Since the Carnot cycle is reversible,

Remember … for ANY heat engine

So … for a reversible heat engine [Eqn 5-20]

The Carnot efficiency, h_th, is the highest possible efficiency for a heat engine operating between thermal energy reservoirs at T_H and T_L.

CARNOT Refrigeration and Heat Pumps Section 5-11

[Eqns. 5-22, 23]

These COP’s are the highest possible values for refrigeration cycles operating between thermal energy reservoirs at T_H and T_L.

EXAMPLE 5-72: Carnot Heat Engine (Carnot = REVERSIBLE) !

Given: Q_H = 500 kJ, Q_L = 200 kJ, T_L = 17 ^oC = 290 ^oK

Find: T_H and h_th Fig5_72.gif (3965 bytes)

Since it is reversible:

T_H = T_L(Q_H / Q_L)_REV

T_H = 290 ^oK (500 kJ / 200 kJ)

T_H = 725 ^oK = 452 ^oC

and

using both formulae:

h_th = 1 - (200 kJ / 500 kJ) and h_th,REV=1 - (290 ^oK / 725 ^oK)

h_th = 0.6 = 60 % and h_th,REV= 0.6 = 60 %