231 Physic Lecture 31
Main points of last lecture: Heat engines and efficiency: Weng Qh Qc Qc e= = = 1 Qh Qh Qh Reversible and irreversible processes. Carnot cycle and Carnot engine. Main points of todays lecture: Engines and refrigerators. Entropy:
S =
Qreversible T
e=
Weng Qh
Th Tc Tc = = 1 Th Th
Entropy is the log of the probability of a state. The total entropy of all interacting systems can only increase or stay the same during a process. It can never decrease.
T is in Kelvin.
Hookes law:
Fs = kx PEs =
12 kx 2
Example
A power plant has been proposed that would make use of the temperature gradient in the ocean. The system is to operate between 20.0C (surface-water temperature) and 5.00C (water temperature at a depth of about 1 km). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant is 75.0 MW, how much energy is absorbed per hour? (c) In view of your answer to (a), do you think such a system is worthwhile (considering that there is no charge for fuel)?
a) emax = ecarnot
W W (75MW )(3600s ) b) e = Qh = = = 5.3x1012 J Qh e 0.051 c ) What is the energy required to pump the water?
Tc 278 = 1 = 1 = 0.051 Th 293
Story of Hawaiian deep water project
Keahole sits at a point where underwater land slopes sharply down into the sea, it was a place where warm water can be piped from the surface of the sea and cold water can be piped from depths of about a half-mile. A process called ocean thermal energy conversion, or OTEC, used the temperature difference between hot and cold sea water to produce 50 KW of electricity at Keahole in 1993. The process worked but it was uneconomical.
KAILUA, HAWAI'I Koyo USA Corp., a company selling deep-sea water from Keahole Hawai'i, is expanding its plant and has applied to sell the water in the United States The company is producing more than 200,000 bottles a day and says it can't keep up with demand in Japan, where it sells 1.5 liter bottles of its MaHaLo brand for $4 to $6 each.
Example
An engine does 20900 J of work and rejects 7330 J of heat into a cold reservoir at 298K. What is the smallest possible temperature of the hot reservoir?
Qh = W + Qc = 20900 J + 7330 J = 28230 J e= T W 20900 J = = 0.74 ecarnot = 1 c Qh 28230 J Th Tc T c 1 .74 = .26 Th Th
0.74 1
Tc T 298K .26 c Th Th Th .26 .26
1146 K Th
Quiz
A heat engine operating between a hot reservoir at 500 K and a cold reservoir at 200 K an has efficiency that is 70% of its maximum possible value. If it receives lx106 J heat energy from the hot reservoir in 25 minutes, it can do a quantity of work equal to a)6.3x105J. b)4.2x105J. 200 T c)3.lxl05J. a) emax = ecarnot = 1 c = 1 = 0.6 500 Th d)2.5x105J. e = 0.7emax = 0.7(0.6) = .42 e)1.7x105J.
e=
W W = e Qh = (.42 )(1x106 J ) = 4.2 x105 J Qh
Heat pumps and refrigerators
Heat engines can run in reverse Send in energy Energy is extracted from the cold reservoir Energy is transferred to the hot reservoir This process means the heat engine is running as a heat pump A refrigerator is a common type of heat pump An air conditioner is another example of a heat pump In the south, people often use heat pumps to heat homes
Example
A Carnot refrigerator maintains the food inside it at 276 K while the temperature of the kitchen is 298 K. The refrigerator removes 3.00x 104 J of heat from the food. How much heat is delivered to the kitchen?
As a Carnot heat engine, we know that Qc Tc Q c Tc = 1 = ecarnot = 1 Qh Th Q h Th The Carnot engine run in reverse takes mechanical energy W to move Q c from the inside of the refrigerator and deposit Q h in the kitchen. Q c Tc T 298 = Q h = Q c h = 3x10 4 J = 3.23x104 J Q h Th Tc 276
Qreversible S = T
Entropy can only be calculated from a reversible path, and must be done that that way even if the system actually follows an irreversible path To calculate the entropy for an irreversible process...