fundamentals of electric circuits 3rd edition solutions manual chapter 10
Fundamentals of Corporate Finance 8th edition: Solutions Manual
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Solutions Manual
Fundamentals of Corporate Finance 8
th
edition
Ross, Westerfield, and Jordan
Updated 03-05-2007
B-28 SOLUTIONS
Market-to-book ratio ... liabilities and equity
EFN = $10, 890 – 12,925
EFN = –$2,035
3. An increase of sales to $5,967 is an increase of:
Sales increase = ($5,967 – 5 ,100 ) / $5 ,100
Sales increase = .17 or ... dark because of the costs of
compliance. The costs to comply with Sarbox can be several million dollars, which can be a large
percentage of a small firms profits. A major cost of going dark...
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fundamentals of electric circuits
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... Menu
Textbook Table of Contents
Problem Solving Workbook Contents
1
DC CIRCUITS
PART 1
Chapter
1
Basic Concepts
Chapter
2
Basic Laws
Chapter
3
Methods of Analysis
Chapter
4
Circuit Theorems
Chapter
5
Operational ... 191
Comprehensive Problems 200
Contents
xi
Chapter 2 Basic Laws 27
Chapter 3 Methods of Analysis 75
PART 1 DC CIRCUITS 1
Chapter 1 Basic Concepts 3
Chapter 4 Circuit Theorems 119
Chapter 5 Operational Amplifiers ... Synthesis
15 .10
Summary 694
xiii
CONTENTS
PART 3 ADVANCED CIRCUIT ANALYSIS 643
Chapter 15 The Laplace Transform 645
Chapter 12 Three-Phase Circuits 477
Chapter 13 Magnetically Coupled Circuits 527
Chapter...
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fundamentals of heat and mass transfer solutions manual phần 1 docx
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08/08/2014, 17:20
... s/h 24h/d 365 d/y 5.80 10 J
== × × × = ×
With a furnace energy consumption of
11
E E/ 6.45 10 J,
ff
==ì the annual cost of the loss
is
5
C C E 0.01 $/MJ 6.45 10 MJ $6450
gf
== ìì =
... P
T.
A
ss
SS
4
sss
SS
1/4
s
SS
s
s
α
αεσ
α
εσ
′′
−+
′′
−+
′′
+
=
In the shade
()
q0,
S
′′
=
100 0 W
T 364 K.
1 m 1 5.67 10 W/m K
1/4
s
2824
==
ìì ì
<
In the sun,
0.25 1 m 750 W/m 100 0 W
T 380 K.
1 m 1 5.67 10 W/m K
1/4
22
s
2824
ìì +
==
ìì ... advantage of the double pane construction resides with the low thermal conductivity
of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use
of the double...
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fundamentals of heat and mass transfer solutions manual phần 2 docx
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... K
q=
ln 1.04 1
2 2m 1.4 W/m K
1.04m 2m 10 W/m K
400 300 K
+
11
1 0.962 m
2 1.4 W/m K
1.04m 10 W/m K
100 K 100 K
q=
2.23 10 K/W + 15.30 10 K/W 4.32 10 K/W + 29.43 10
+
+
+
ìììì
q = 5705W + ... K 10. 2 12.7
10
12W/m K4 10. 2 10 m
=+
ì
ì
()
2
2-3
3
191.2 K/W+13.2 K/W+246.7 K/W=451.1 K/W
6 W/m K4 12.7 10 m
+=
ì
With lens:
t,w
3
3111
R 191.2 K/W+13.2 K/W+ m
4 0.80 W/m K 12.7 16.5
10
=
ì
()
2
2-3
3
191.2 ... and thermal conductivity of steel tubes. Temperature of steam
flowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath.
Temperature of ambient air and surroundings....
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fundamentals of heat and mass transfer solutions manual phần 3 pdf
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... numerical values, find
(
)
(
)
(
)
(
)
( ) ( ) ( )( )
q50 W/mK[1/2120.5 5100 120.6 4100 121.2 9100
123.8 9100 134.5 7100 1/2147.1 4100 ]
′
=⋅−+−+−
+−+−+−
q6711 W/m.
′
=
<
COMMENTS: For nodes a through ... finite-difference solution. Using these values
as a guide, try sketching a few isotherms.
- 100 100 100 100 100 -
50 86.0 105 .6 119 131.7 151.6 200
50 88.2 117.4 138.7 156.1 174.6 200
50 99.6 137.1 162.5 ... ( )
2
10
fc 0 10 0
TT
kD
qkA hPx2TT TT 2DhxTT
x4x
π
π
∞∞
−
=− − ∆ − =− − − ∆ −
∆∆
Substituting numerical values, with
∆
x = 0. 010 m, find
()
()
()
2
f
2
15W m K 0.012m
q 93.4 100 C
4 0.010m
0.012m...
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fundamentals of heat and mass transfer solutions manual phần 4 docx
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... independent of time.
Hence,
()
(
)
p1 p p
o
11 2
T T 1 2Fo Bi Fo Fo T T Bi FoT
+
∞
=−−⋅+ ++⋅
.(5)
End Nodal Point 10:
p1 p
10 10
abc c p
TT
x
qqq A c
2t
ρ
+
−
∆
++=⋅ ⋅
∆
(
)
(
)
pp p1p
pp
910 10 10
cc ... outline of the slab, the boundary conditions,
and the triangular mesh before using the Reduce-mesh option.
PROBLEM 5.96 (Cont.)
() ()
73
x
q L,t 10W m K 149.80 100 C 0.010m 2 10 W m 0.010m 2
= ... or Eq. 5.77,
(
)
( )
p+1pp
010
T2 FoTBiT12 Fo2 BiFoT.
∞
=+⋅+−−⋅ (1)
The numerical values of Fo and Bi are
( )
( )
62
22
2-3
t5.610m/s18s
Fo0.448
x
0.015m
100 W/mK1510m
hx
Bi0.075.
k20 W/mK
ìì
===
ìì
===
Recognizing...
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fundamentals of heat and mass transfer solutions manual phần 5 ppsx
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... (
°
C)
56.87 57.00
b (
C/s)
-0.1472 -0.2641
c (
C/s
2
)3
ì
10
-4
9
ì
10
-4
d (
C/s
3
)-4
ì
10
-7
-2
ì
10
-6
e (
C/s
4
)2
ì
10
-10
1
ì
10
-9
: (1) Airstream over the
test plate
approximates ... emissivity of Nichrome wire. Electrical
current. Temperature of air flow and surroundings. Velocity of air flow.
(a) Surface and centerline temperatures of the wire, (b) Effect of flow velocity and electric
current ... 10 as the local coefficient at x = 95 mm,
10x0.095m
hh.
=
=
1/21/3
10x
k
h0.453RePr
x
=
5
x
-62
ux40 m/s0.095 m
Re2.06 510
18. 410 m/s
ì
===ì
ì
(
)
( )
1/2
1/3
52
10
0.0282 W/mK
h0.4532.06 5100 .70354.3W/mK
0.095
=ì=
(...
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Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
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...
5.2
A
650 W
(e)
The efficiency of this
motor
at
full load is
(
)(
)
⎜
OUT
P
100 %
=
⋅
2100
=
0 hp
746 W/hp
100 %
⋅
97.6%
=
IN
P
16.05 MW
... and a synchronous reactance of 0.50 per unit.
Ignore all
losses.
(a)
What
is the speed of
rotation of this motor?
(b)
What
is the output
torque of this motor at the rated ...
are
168
reactance
of 0.90 and
a
per-unit resistance of 0.02.
(a)
What
is the rated input power of this motor?
(b)
What
is the magnitude of
E
A
at
rated conditions?
...
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Fundamentals of Ecological Modelling, Third Edition pot
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... model of the
Lagoon of Venice by application of the steady-state software ECOPATH. Response
models are also presented. The Vollenweider model for temperate lakes is used as
an illustration of ... in Chapter 5. Parameter estimation of eco-
toxicological parameters is particularly demanding and a number of methods are
available which are briefly discussed in this chapter. Early in the chapter, ... determine the optimum number of
subsystems to be included in the model for an acceptable level of accuracy defined by
the scope of the model. Due to lack of data, it will often become necessary at...
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