Download Exam 1 Solutions - Thermodynamics - Fall 2008 | ME 201 and more Exams Thermodynamics in PDF only on Docsity! ME 201 Exam No. 1 February 11, 2008
Name: Se | uk Loy \Student Number:
Instructions
There are three problems. Allocate your time and effort to maximize your points.
Layout your work clearly and show your reasoning.
This is an open-book (Text Book) examination. One 84x11 paper with your notes allowed.
Problem 1 (20 points)
(a) What is the difference between intensive and extensive properties? (5 points) fe
Intensive properties a¥e tu dz pende nt of the mass of asys
Extewsive pvoperties are Pro povoral to the mass of asystes
(b) How many independent intensive properties are required to completely specify the state of a
simple compressible system? (5 points)
2
(©) A rock is dropped from an elevation of 10 m (10 m above the ground). What is the velocity of
the rock when it is 5 m from the ground? (5 points)
Ahectmy” 4Ke = 4Pe
Ape = mah phe SVv=gh , Van2gh ajaxages
= 7.9 (m/e)
(d) Can quality be expressed as the ratio of the volume occupied by the vapor phase to the total
volume? Explain. (5 points)
No,
The gucdity ts defined. as the. metro of tho mess of -
He. Vapor phase +0 the totall mass ,
Problem 2 (40 points)
The volume of 1 kg of air in a piston-cylinder device is initially 1.722 m?. Now the air is
compressed to 0.861 m? while its pressure is maintained constant at 200 kPa. Determine
(a) the initial and final temperatures of the air, (10 points)
(b) the work required to compress the air, (10 points)
(c) the net heat transfer during the process, is the system losing heat or gaining heat? (15 points)
(a) Draw the process ona P—v diagram. (5 points)
S. Z
Sibel _ S fochel2)
O mat*4. pag 257K ‘
4
=200|K
P =200kfPa. A Pew
Aty » Vz = 0.86 | m3
Vj =h7 220? Aw ane,
& PVYsmRT Ideal gas €fnatun of Hote Bo
722
-- PLY 200 KAT
{= 27> mf 200\!K
i WR iX 0.287
- Poy _ 200 %aFel L
Th = ee = “Trooe7 7 600K
) Wwe fr V =P CU-V,) = 200 (0.86 1—[.-722)= - 172.2 (KT)
The werk repudied -+o Compress the ody iS (72.27. f
©). The first low for a closed systenr 4U =Q,,-Qig TW, ih,
Qin Bonk = Quiet, attioy = M(ita-U,) ~ Wy
Ue= 434 79 Keg U, = 733.33 Teg. WIT 2K
Oh wets tu = (434-47 33.33) — 172.2 = - 670,75 as)
The systenn os losing heck
Pra @
wv) <4 %
