HW 3 SOLUTION
1.
Qreforming
H1
Reformer
CH4+2H2O(L)
300K
H2
CO,CO2,H2,H2O,CH4
T=1100K
Using Equil one can calculate the mole fractions of the final stage and enthalpies of
the initial and final stages. In the "Gas Chemistry input" (chem.inp) file in Equil
3.15 Electrical, Optical, and Magnetic Materials and Devices
Caroline A. Ross
Fall Term, 2005
Final Exam (6 pages)
Closed book exam. Formulae and data are on the last 4 pages of the exam.
This takes 180 min and there are 180 points total. Be brief in your
MASSACHESUTS ISNTITUTE OF TECHNOLOGY
FUNDAMNETALS OF ADVANCED ENEFRGY CONVERSION
SPRING 04
HOMEWORK III
DUE DATE, March 1, 2004
Most fuel cells run on pure hydrogen, or a mixture of inert gases and hydrogen, as a
fuel; and pure oxygen, or air as an oxidiz
MASSACHESUTS ISNTITUTE OF TECHNOLOGY
FUNDAMNETALS OF ADVANCED ENEFRGY CONVERSION
SPRING 04
HOMEWORK II
DUE DATE, FEBRARY 23, 2004
The rising atmospheric concentration of carbon dioxide, a product of hydrocarbon
combustion and a known green house gas, is a
1. (20 ) For the telescope conguration shown below, where lenses L1 and L2 have
focal lengths f1 , f2 , respectively, the object plane and two stops S1 and S2 of
halfsizes a1 , a2 , respectively, are at the locations shown,
object
plane
L1
L2
S1
S2
a1
f1
Sample Final - Solutions
Problem 1
a.
Material is multidomain.
Domains in Co line up with c-axis. Hoop shape will depend on angle between
co-axis and eld.
b.
Add a linear term mu0 H.
c.
Assume thermal instability when kT energy needed to reverse particle.
2.710 Optics
Final Exam Solutions
Spring 09
1. Consider the following system.
(a) If we position an on-axis point source at the center of the object plane (front
focal plane of L1), a collimated ray bundle will emerge to the right of L1 and
its diameter i
3.032ProblemSet3Solutions
Fall2007
Due: Startof Lecture, 10.01.07
1.
Here, we will analyze the stress states of a solid block of aluminum under various
loading conditions using the Mohrs Circle construction. We are going to use the
following convention to
Homework II Solution
(1) The efficiency in the ideal case (Carnot cycle) is:
ideal =1-(300K)/(273K+1300K)=0.81.
Then, the second law efficiency becomes
0.5
II =
= 0.62
0.81
LHV of C6H6=40141kJ/kg=3131MJ/kmol (liquid) (Boiling point of C6H6 is 80 C)
Fuel