Baixe Resolução Livro SMITH VAN NESS e outras Exercícios em PDF para Termodinâmica, somente na Docsity!
P 3000atm D 0.17in A S 4
D^2 A 0.023 in^2
F P A g 32.174 ft sec^2
mass F g
mass 1000.7 lbm Ans.
1.7 (^) Pabs =U g hPatm
U 13.535 gm cm^3
g 9.832 m s^2
h 56.38cm
Patm 101.78kPa Pabs U g hPatm Pabs 176.808 kPa Ans.
1.8 (^) U 13.535 gm cm^3
g 32.243 ft s^2
h 25.62in
Patm 29.86in_Hg Pabs U g hPatm Pabs 27.22 psia Ans.
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).
Guess solution: (^) t 0 Given t = 1.8t 32 Find t() 40 Ans.
1.5 By definition: (^) P F A
= F = mass g Note: Pressures are in gauge pressure.
P 3000bar D 4mm A S 4
D^2 A 12.566 mm^2
F P A g 9.807 m s^2
mass F g
mass 384.4 kg Ans.
1.6 By definition: (^) P F A
= F =mass g
FMars K x FMars 4 u 10 ^3 mK
gMars
FMars mass
gMars 0.01 mK kg
Ans.
1.12 Given: z
d P d
= Ug and:^ U M P R T
= Substituting: z
d P d
M P
R T
= g
Separating variables and integrating: Psea
PDenver (^1) P P
´μ μ¶
d 0
zDenver M g z R T
´μ μ¶
= d
After integrating: (^) ln PDenver Psea
M g R T
= zDenver
Taking the exponential of both sides and rearranging: (^) PDenver Psea e
Mg R T zDenver
Psea 1atm M 29 gm mol
g 9.8 m s^2
1.10 Assume the following: (^) U 13.5 gm cm^3
g 9.8 m s^2
P 400bar h P U g
(^) h 302.3 m Ans.
1.11 The force on a spring is described by: F = Ks x where Ks is the spring constant. First calculate K based on the earth measurement then gMars based on spring measurement on Mars. On Earth:
F = mass g = K x mass 0.40kg g 9.81 m s^2
x 1.08cm
F mass g F 3.924 N Ks^ F x
Ks 363.333 N m On Mars: x 0.40cm
Ans.
(b) (^) Pabs^ F A
Pabs 110.054 kPa Ans.
(c) (^) 'l 0.83m Work F 'l Work 15.848 kJ Ans.
'EP mass g' l 'EP 1.222 kJ Ans.
1.18 (^) mass 1250kg u 40 m s
EK 1
mass u ^2 EK 1000 kJ Ans.
Work EK Work 1000 kJ Ans.
1.19 (^) Wdot mass g'^ h time
Wdot 200W g 9.8 m s^2
'h 50m
Patm 30.12in_Hg A S 4
D^2 A 1.227 ft^2
(a) (^) F Patm Amass g F 2.8642 u 103 lbf Ans.
(b) (^) Pabs F A
Pabs 16.208 psia Ans.
(c) (^) 'l 1.7ft Work F 'l Work 4.8691 u 103 ft lb (^) f Ans.
'PE mass g' l 'PE 424.9 ft lb (^) f Ans.
1.16 (^) D 0.47m mass 150kg g 9.813 m s^2
Patm 101.57kPa A S 4
D^2 A 0.173 m^2
(a) (^) F Patm Amass g F 1.909 u 104 N
mdot Wdot g 'h 0.910.
(^) mdot 0.488 kg s
Ans.
a) (^) cost_coal
ton 29 MJ kg
(^) cost_coal 0.95 GJ 1
cost_gasoline
gal 37 GJ m^3
(^) cost_gasoline 14.28 GJ ^1
cost_electricity 0. kW hr
(^) cost_electricity 27.778 GJ ^1
b) The electrical energy can directly be converted to other forms of energy whereas the coal and gasoline would typically need to be converted to heat and then into some other form of energy before being useful.
The obvious advantage of coal is that it is cheap if it is used as a heat source. Otherwise it is messy to handle and bulky for tranport and storage.
Gasoline is an important transportation fuel. It is more convenient to transport and store than coal. It can be used to generate electricity by burning it but the efficiency is limited. However, fuel cells are currently being developed which will allow for the conversion of gasoline to electricity by chemical means, a more efficient process.
Electricity has the most uses though it is expensive. It is easy to transport but expensive to store. As a transportation fuel it is clean but batteries to store it on-board have limited capacity and are heavy.
T t 273.15 lnPsat ln Psat( )
Array of functions used by Mathcad. In this case, a 0 = A, a 1 = B and a 2 = C.
Guess values of parameters
F T a( )
exp a 0
a 1 T a 2
exp a 0
a 1 T a 2
T a 2
exp a 0
a 1 T a 2
a 1 T a 2 2
exp a 0
a 1 T a 2
guess
Apply the genfit function
A B C
genfit T Psat( guessF)
A
B
C
2.29 u 103 69.
Ans.
Compare fit with data.
(^0240 260 280 300 320 340 )
50
100
150
200
Psat f T( A BC)
T To find the normal boiling point, find the value of T for which Psat = 1 atm.
This is an open-ended problem. The strategy depends on age of the child, and on such unpredictable items as possible financial aid, monies earned by the child, and length of time spent in earning a degree.
c)
The salary of a Ph. D. engineer over this period increased at a rate of 5.5%, slightly higher than the rate of inflation.
i Find i( ) i 5.511 %
C 2
C 1
( 1 i) t 2 t 1 Given =
C 2 80000 dollars yr
C 1 16000 dollars yr
b)t 1 1970 t 2 2000
The increase in price of gasoline over this period kept pace with the rate of inflation.
C 2 1.513 dollars gal
C 2 C 1 ( 1 i) t 2 t 1
C 1 0.35 dollars i 5% gal
a)t 1 1970 t 2 2000
Tnb 273.15K 56.004 degC Ans.
Tnb B Tnb 329.154 K A ln Psat kPa
§ ¨ C
Psat 1atm K
Q 34 800 J W 34 300J
'Ut 34 Q 34 W 34 'Ut 34 500 J Ans.
Step 1 to 2 to 3 to 4 to 1: Since 'Ut^ is a state function, 'Ut^ for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the 'Ut^ values for all of the steps must sum to zero.
'Ut 41 4700J 'Ut 23 ''Ut 12 ' Ut 34 Ut 41
'Ut 23 4000 J Ans.
Step 2 to 3: (^) 'Ut 23 4 u 103 J Q 23 3800 J
W 23 'Ut 23 Q 23 W 23 200 J Ans.
For a series of steps, the total work done is the sum of the work done for each step. W 12341 1400 J
2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor.
i 9.7amp E 110V Wdotmech 1.25hp
Wdotelect iE Wdotelect 1.067 u 103 W
Qdot Wdotelect Wdotmech Qdot 134.875 W Ans.
2.5 Eq. (2.3): 'Ut^ =Q W
Step 1 to 2: (^) 'Ut 12 200 J W 12 6000 J
Q 12 'Ut 12 W 12 Q 12 5.8 u 103 J Ans.
Step 3 to 4:
'U 12 kJ Q 'U Q 12 kJ Ans.
2.13Subscripts: c, casting; w, water; t, tank. Then
mc 'Uc mw 'Uw mt 'Ut= 0 Let C represent specific heat, (^) C = CP=CV Then by Eq. (2.18) mc 'C (^) c tc mw 'C (^) w tw mt 'C (^) t tt= 0
mc 2 kg mw 40 kg mt 5 kg
Cc 0.50 kJ kg degC
Ct 0.5 kJ kg degC
Cw 4.18 kJ kg degC
tc 500 degC t 1 25 degC t 2 30 degC (guess)
Given m (^) c Cc t 2 tc = mw Cw mt Ct t 2 t 1
t 2 Find t 2 t 2 27.78 degC Ans.
W 41 W 12341 W 12 W 23 W 34 W 41 4.5 u 103 J Ans.
Step 4 to 1: (^) 'Ut 41 4700J W 41 4.5 u 103 J
Q 41 'Ut 41 W 41 Q 41 200 J Ans. Note: (^) Q 12341 =W 12341
2.11 The enthalpy change of the water = work done.
M 20 kg CP 4.18 kJ kg degC
't 10 degC
Wdot 0.25 kW 'W
M C' (^) P t Wdot
'W 0.929 hr Ans.
2.12 (^) Q 7.5 kJ 'U 12 kJ W 'U Q
W 19.5 kJ Ans.
mdot C (^) p T 3 T 1 mdot 2 CP T 3 T 2 =Qdot T 3 C (^) P mdot 1 mdot 2 =Qdot mdot 1 CPT 1 mdot 2 CPT 2
mdot 1 1.0 kg s
T 1 25degC mdot 2 0.8 kg s
T 2 75degC
CP 4.18 kJ kg K Qdot 30 kJ s
T 3
Qdot mdot 1 CPT 1 mdot 2 CPT 2 mdot 1 mdot 2 CP
T 3 43.235 degC Ans.
2.25By Eq. (2.32a): (^) 'H 'u
2 2
= 0 'H =CP 'T
By continuity, incompressibility u^2 u^1
A 1
A 2
= CP 4.18 kJ kg degC
2.22 (^) D 1 2.5cm u 1 2 m s
D 2 5cm
(a) For an incompressible fluid, U=constant. By a mass balance, mdot = constant = u 1 A 1 U = u 2 A 2 U
u 2 u 1
D 1
D 2
2 u 2 0.5 m s
Ans.
(b) (^) 'EK^1 2
u 22 1 2
u 12 'EK 1.875 J kg
Ans.
2.23 Energy balance: (^) mdot 3 H 3 mdot 1 H 1 mdot 2 H 2 =Qdot
Mass balance: (^) mdot 3 mdot 1 mdot 2 = 0 Therefore: (^) mdot 1 H 3 H 1 mdot 2 H 3 H 2 =Qdot or
u 2 3.5 m s
molwt 29 kg kmol
Wsdot 98.8kW ndot 50 kmol hr
CP^7
R
'H CP T 2 T 1 'H 6.402 u 103 kJ kmol By Eq. (2.30):
Qdot 'H
u 22 2
u 12 2
molwt
ndotWsdot Qdot 9.904 kW Ans.
2.27By Eq. (2.32b): (^) 'H 'u
2 2 g c
= also^
V 2
V 1
T 2
T 1
P 1
P 2
By continunity, constant area u^2 u^1
V 2
V 1
= u 2 u 1
T 2
T 1
P 1
P 2
= 'u^2 =u 22 u 12
'u^2 u 12
A 1
A 2
2 1
= » ¼ 'u
(^2) u 1
2 D^1
D 2
4 1
SI units: (^) u 1 14 m s
D 1 2.5 cm D 2 3.8 cm
'T
u 12 2 C (^) P
D 1
D 2
4
» ¼ 'T 0.019 degC Ans.
D 2 7.5cm
'T
u 12 2 C (^) P
D 1
D 2
4
» ¼ 'T 0.023 degC Ans.
Maximum T change occurrs for infinite D2: D 2 f cm
'T
u 12 2 C (^) P
D 1
D 2
4
» ¼ 'T 0.023 degC Ans.
2.26 (^) T 1 300K T 2 520K u 1 10 m s
By Eq. (2.23): (^) Q n C (^) P t 2 t 1 Q 18.62 kJ Ans.
2.31 (a) (^) t 1 70 degF t 2 350 degF n 3 mol
CV 5 BTU
mol degF
By Eq. (2.19):
Q n C (^) V t 2 t 1 Q 4200 BTU Ans. Take account of the heat capacity of the vessel:
mv 200 lb m cv 0.12 BTU lbm degF
Q mv cv n C V t 2 t 1 Q 10920 BTU Ans.
(b) (^) t 1 400 degF t 2 150 degF n 4 mol
Continuity: (^) D 2 D 1
u 1 V 2 u 2 V 1 D 2 1.493 cm Ans.
2.30 (a) (^) t 1 30 degC t 2 250 degC n 3 mol
CV 20.
J
mol degC
By Eq. (2.19): (^) Q n C (^) V t 2 t 1 Q 13.728 kJ Ans.
Take into account the heat capacity of the vessel; then
mv 100 kg cv 0. kJ kg degC
Q mv c (^) v n C V t 2 t 1 Q 11014 kJ Ans.
(b) (^) t 1 200 degC t 2 40 degC n 4 mol
CP 29.
joule mol degC
Wdot W (^) smdot Wdot 39.52 hp Ans.
2.34 H 1 307 BTU
lbm
H 2 330 BTU
lbm
u 1 20 ft s
molwt 44 gm mol
V 1 9.25 ft
3 lbm
V 2 0.28 ft
3 lbm
D 1 4 in D 2 1 in
mdot
S
D 12 u 1
V 1
mdot 679.263 lb hr
u 2 mdot
V 2
S
D 22
u 2 9.686 ft sec
Ws 5360 BTU lbmol
Eq. (2.32a): (^) Q H 2 H 1 u 22 u 12 2
Ws molwt
Q 98.82 BTU
lbm
CP 7
BTU
mol degF
By Eq. (2.23):
Q n C (^) P t 2 t 1 Q^ ^7000 BTU^ Ans.
2.33 H 1 1322.6 BTU lbm
H 2 1148.6 BTU
lbm
u 1 10 ft s
V 1 3.058 ft
3 lbm
V 2 78.14 ft
3 lbm
D 1 3 in D 2 10 in
mdot 3.463 u 104 lb mdot sec
S
D 12 u 1
V 1
u 2 mdot
V 2
S
D 22
u 2 22.997 ft sec
Eq. (2.32a): (^) Ws H 2 H 1 u 22 u 12 2
Ws 173.99 BTU lb
Re
Re D^ ^ Uu P
o
u
m s
D
cm
Note: HD = H/D in this solution
P 9.0 10 ^4 kg HD 0. m s
U 996 kg m^3
'H 1.164 kJ Ans. mol
'H 'Ha 'Hb
'U 0.831 kJ^ Ans. mol
'U 'Ua 'Ub
'Ub 6.315 u 103
J
mol
'Ub 'Hb P 2 V 2 V 1
'Ha 7.677 u 103
J
mol 'Ha 'Ua V 1 P 2 P 1
V 2 0.
m^3 mol
V 2
R T 2
P 2
V 1 2.437 u 10 ^3 m
3 mol
V 1
R T 1
P 1
'Ua 5.484 u 103 J mol
'Ua CV 'Ta
'Hb 8.841 u 103 J mol
'Hb CP 'Tb
'Tb T 2 Ta2 'Tb 303.835 K 'Ta Ta2 T 1 'Ta 263.835 K
Cost (^15200) Cost 799924 dollarsAns. Wdot kW
Wdot mdot H 2 H 1 Wdot 1.009 u 103 kW
Assume that the compressor is adiabatic (Qdot = 0). Neglect changes in KE and PE.
H 2 536.9 kJ kg
H 1 761.1 kJ kg
mdot 4.5 kg s
'P'L Ans.
kPa m
'P'L ^2
D
§ ¨ U fF u^2
o
mdot^ Ans.
kg s
mdot U uS 4
§ ¨ D 2
o
fF
fF 0.3305 ln 0.27 H D^7 Re
2
o
