Chapter 2 - Section A - Mathcad Solutions
2.1 (a) Mwt := 35 kg Work := Mwt g z (b) Utotal := Work g := 9.8 m s
2
z := 5 m Work = 1.715 kJ Ans. Utotal = 1.715 kJ dU + d ( PV) = CP dT Ans.
(c) By Eqs. (2.14) and (2.21): Since P is constant, this can be writ
Chapter 9 - Section B - Non-Numerical Solutions
9.1 Since the object of doing work |W | on a heat pump is to transfer heat | Q H | to a heat sink, then: What you get = | Q H | What you pay for = |W | Whence For a Carnot heat pump, = |Q H | |W |
|Q H | TH
Chapter 8 - Section B - Non-Numerical Solutions
8.12 (a) Because Eq. (8.7) for the efciency Diesel includes the expansion ratio, re VB / V A , we relate this quantity to the compression ratio, r VC / VD , and the Diesel cutoff ratio, rc V A / VD . Since V
Chapter 8 - Section A - Mathcad Solutions
8.1 With reference to Fig. 8.1, SI units, At point 2: Table F.2, At point 4: Table F.1, At point 1: At point 3: Table F.1, x3 := 0.96 H2 := 3531.5 H4 := 209.3 H1 := H 4 Hliq := H 4 H3 := H liq + x3 H lv Sliq := 0.
Chapter 7 - Section B - Non-Numerical Solutions
7.2 (a) Apply the general equation given in the footnote on page 260 to the particular derivative of interest here: T S T = P S S P P T The two partial derivatives on the right are found from Eqs. (6.17) and
Chapter 7 - Section A - Mathcad Solutions
7.1 u2 := 325 m sec R := 8.314 J mol K molwt := 28.9 gm 7R CP := mol 2 molwt
With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to H
Chapter 6 - Section B - Non-Numerical Solutions
H S =T
P
6.1 By Eq. (6.8),
and isobars have positive slope T S
Differentiate the preceding equation: 2 H S2
2 H S2 =
P
=
P
P
Combine with Eq. (6.17):
T CP
and isobars have positive curvature.
6.2 (a) Applica
Chapter 6 - Section A - Mathcad Solutions
6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS = V dP and dH = ( 1 T) V dP
For an estimate, assume properties independent of pressure. T := 270 K V := 1.551 10 P1 := 381 kPa
3 3 m
P2 := 1200
Chapter 5 - Section B - Non-Numerical Solutions
5.1 Shown to the right is a P V diagram with two adiabatic lines 1 2 and 2 3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce
Chapter 10 - Section B - Non-Numerical Solutions
10.5 For a binary system, the next equation following Eq. (10.2) shows that P is linear in x1 . Thus no maximum or minimum can exist in this relation. Since such an extremum is required for the existence of
Chapter 11 - Section A - Mathcad Solutions
11.1 For an ideal gas mole fraction = volume fraction CO2 (1): N2 (2): S := R 11.2 x1 := 0.7 x2 := 0.3 i := 1 . 2 J mol K
i
xi ln ( xi)
S = 5.079
Ans.
For a closed, adiabatic, fixed-volume system, U =0. Also, for
Chapter 11 - Section B - Non-Numerical Solutions
11.6 Apply Eq. (11.7): (nT ) Ti ni =T n ni =T (n P ) Pi ni =P n ni =P
P ,T ,n j
T , P ,n j
P ,T ,n j
T , P ,n j
11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: mi Let Mk be t
Chapter 1 - Section B - Non-Numerical Solutions
1.1 This system of units is the English-system equivalent of SI. Thus, gc = 1(lbm )(ft)(poundal)1 (s)2 1.2 (a) Power is power, electrical included. Thus, Nm kgm2 energy [=] [=] time s s3 (b) Electric current
Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this
equation by setting t(F) = t(C).
t := 0
Guess solution:
Given
t = 1.8t + 32
1.5 By definition:
P=
Find ( t ) = 40
F
A
F = mass g
P
Chapter 16 - Section B - Non-Numerical Solutions
16.1 The potential is displayed as follows. Note that K is used in place of k as a parameter to avoid confusion with Boltzmanns constant.
Combination of the potential with Eq. (16.10) yields on piecewise in
Chapter 14 - Section B - Non-Numerical Solutions
14.2 Start with the equation immediately following Eq. (14.49), which can be modied slightly to read: ln i = (nG R/ RT ) (n Z ) ln Z +n +1 ni ni ni
where the partial derivatives written here and in the foll
Chapter 13 - Section A - Mathcad Solutions
Note: For the following problems the variable kelvin is used for the SI unit of absolute temperature so as not to conflict with the variable K used for the equilibrium constant 13.4 H2(g) + CO2(g) = H2O(g) + CO(g
Chapter 12 - Section B - Non-Numerical Solutions
12.2 Equation (12.1) may be written: yi P = xi i Pi sat . Summing for i = 1, 2 gives: P = x1 1 P1sat + x2 2 P2sat . Differentiate at constant T : d 1 d 2 dP = P1sat x1 + 1 + P2sat x2 2 d x1 d x1 d x1 d 2 =0
Chapter 5 - Section A - Mathcad Solutions
5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) = TC Work = 1 QH TH TH := 798.15 K TC QH := 250 kJ s kJ s
TC := 323.15 K Work := Q H 1
TH
or
Work = 148.78
Work = 148.78 kW which is t
Chapter 4 - Section B - Non-Numerical Solutions
4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as: CP = A + B C T1 ( + 1) + T12 ( 2 + + 1) 2 3
where T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperatur
Solutions, Chapter 11
_
11.1
f
pipe Rpipe = 16
2
See the discussion below Eq. 11.10, which shows that f pipe = 3 f pm and that Rpipe = Rpm
3
.
Substituting these, we find
3 R pm 3 fpm
f pm Rpm = 16
= 72
2 Rpipe f pipe
_
11.2 From Eq. 11.14 we see that i
Solutions, Chapter 10
_
AL
1
in 3
gal
60 s
m
= 10 in 2 5 in = 50
= 13.0 gpm = 8.2 10 -4
3
t
s
s 231 in min
s
_
10.1* Q =
gal lbf 231 in 3 ft
hp s
min
Po = Q P = 500
25 2
min in
gal 12 in 550 ft lbf 60 s
= 7.29 hp = 5.44 kW
This answer is independent of
Solutions Chapter 9
_
9.1 (a) Inertia and surface forces, hence the Weber number
(b) Inertia and gravity forces, hence the Froude number. See Bird et. al., p 108.
(c) Here we add viscous forces, so in addition to the Froude number, the Reynolds
number is
Solutions Chapter 8
_
P 4 f V 2
=
8.1*
x
D2
ft 2
1000
2
P 4(0.005)
lbm
ft 2
psi
MPa
s lbf s
(a )
=
62.3 3
= 18.2
2 = 807
2
x
ft
2
32.2 lbm ft 144 in
ft
m
ft
12
P
0.075
psi
kPa
(b )
= above answer
= 0.97
= 22.0
x
62.3
ft
m
The point of this pr