0808Query18

course Phy 202

.................................................??????????

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

assignment #018

018. `query 8

Physics II

08-08-2007

......!!!!!!!!...................................

15:38:37

prin phy and gen phy problem 15.19 What is the maximum efficiency of a heat engine operating between temperatures of 380 C and 580 C?

......!!!!!!!!...................................

RESPONSE -->

e = W/QH = 1- (QL/QH)

convert to K = 653K and 853K

e = 1 - (653K/853K) = .234

.................................................

......!!!!!!!!...................................

15:42:48

query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?

......!!!!!!!!...................................

RESPONSE -->

Response to 15.19: OK.

Answer to 15.26

.28 = 1- (TL/823K)

-.72 = -(TL/823K)

TL = 592.56 K

.35 = 1-(592.56K/TH)

-.65 = -(592.56L/TH)

TH = 911.6K

.................................................

......!!!!!!!!...................................

15:43:45

** Carnot efficiency is eff = (Th - Tc) / Th.

Solving this for Tc we multiply both sides by Th to get

eff * Th = Th - Tc so that

Tc = Th - eff * Th = Th ( 1 - eff).

We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature)

If Th = 550 C = 823 K and efficiency is 30% then we have

Tc =823 K * ( 1 - .28) = 592 K.

Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th:

Tc we multiply both sides by Th to get

eff * Th = Th - Tc so that

eff * Th - Th = -Tc and

Tc = Th - eff * Th or

Tc = Th ( 1 - eff) and

Th = Tc / (1 - eff).

If Tc = 576 K and eff = .35 we get

Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx.

This is (912 - 273) C = 639 C. **

......!!!!!!!!...................................

RESPONSE -->

I didn't convert back to C, but I arrived at the correct answer in the incorrect units.

.................................................

......!!!!!!!!...................................

15:43:54

univ phy problem 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C

At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water?

......!!!!!!!!...................................

RESPONSE -->

Not assigned

.................................................

......!!!!!!!!...................................

15:44:01

** work done / thermal energy required = .07 so thermal energy required = work done / .07.

Translating directly to power, thermal energy must be extracted at rate 210 kW / .07 = 30,000 kW. The cold water absorbs what's left after the 210 kW go into work, or 29,790 kW.

Each liter supplies 4186 J for every degree, or about 80 kJ for the 19 deg net temp change. Needing 30,000 kJ/sec this requires about 400 liters / sec, or well over a million liters / hour.

Comment from student: To be honest, I was suprised the efficiency was so low.

Efficiency is low but the energy is cheap and environmental impact in the deep ocean can be negligible so the process can be economical, if a bit ugly. **

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

"

&#

This looks good. Let me know if you have any questions. &#