Fundamentals of Chemical Engineering Thermodynamics Solutions Manual | Dahm & Visco | Step-by-Step Answers

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, Chapter 1: Introduction

1-16) The value g = 9.81 m/s2 is specific to the force of gravity on the surface of the

 
earth. The universal formula for the force of gravitational attraction is:


= G

Where m1 and m2 are the masses of the two objects, r is the distance between the
centers of the two objects, and G is the universal gravitation constant,
G = 6.674 × 10-11 N(m/kg)2.
A) Research the diameters and masses of the Earth and Jupiter.
B) Demonstrate that F = m(9.81 m/s2) is a valid relationship on the surface of the
earth.
C) Determine the force of gravity acting on a 1000 kg satellite that is 2000 miles
above the surface of the Earth.
D) One of the authors of this book has a mass of 200 lbm. If he was on the surface
of Jupiter, what gravitational force in lbf would be acting on him?

Solution:
A) Measurements obtained from different sources will vary slightly.
DEarth~ 12,742 km DJupiter~ 142,000 km
Massearth= 5.97 × 1024 kg Massjupiter= 1.90 × 1027 kg


B) Massearth= 5.97 × 1024 kg RadiusEarth= 6.371 × 106 meters
/0 1
 !

= G

= m 6.674 × 10
  ! (*.+%×&,
) . (
) 5


"#.$%×& '  ( 234





6 = 7(8. 9: ;<=> )
7




C) 2000 miles = 3218.68 km = 3218680 m


= 1000kg 6.674 × 10
 ! (*+%&&& = EFE9 G


(#.$%×& ' AB)
C+D*D& )




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© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

,Chapter 1: Introduction


D) RadiusJupiter= 66854000 m MassJupiter= 1.898 × 1027 kg 200lbm = 90.7 kg


kg m
Nm (1.90 × 10 kg) K O
sec 
 %

= 90.7 kg .6.674 × 10

5 J P = >>9: G
kg  (7.10 × 10% m) (1 N)



1-17) A gas at T=300 K and P=1 bar is contained in a rigid, rectangular vessel that is 2
meters long, 1 meter wide and 1 meter deep. How much force does the gas exert on
the walls of the container?

Solution:
1 Bar = 100,000 Pa

AreaTUVWX = (2 × W × H) + (2 × W × L) + (2 × H × L)

Force = Pressure × Area
N
 !
Force = (100000Pa)(2 × 1m × 1m + 2 × 1m × 2m + 2 × 1m × 2m) b m c
Pa
Force = : × :de G


1-18) A car weighs 3000 lbm, and is travelling 60 mph when it has to make an emergency
stop. The car comes to a stop 5 seconds after the brakes are applied.
A) Assuming the rate of deceleration is constant, what force is required?
B) Assuming the rate of deceleration is constant, how much distance is covered
before the car comes to a stop?

Solution:

A) Force = mass × acceleration 60mph +*&&m!  
no ! = 88 m
lV #D&W W


velocityWnsXo − velocitynsnnXo
Acceleration =
time
ft
88 sec − 0
a=
5 sec
ft
a = 17.6
sec 



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© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

, Chapter 1: Introduction


ft
Force = 17.6 × 3000lb

sec 
z{7 |}
Force = y>ydd
;<= >


B) PositionWnsXo = + velocitynsnXo × time + positionnsnXo
XoVXns∗n


€

%.* !(#m)
PositionWnsXo = + 88 m ! (5sec) + 0
234 W


PositionWnsXo = >>d |}


1-19) Solar panels are installed on a rectangular flat roof. The roof is 15 feet by 30 feet,
and the mass of the panels and framing is 900 lbm.
A) Assuming the weight of the panels is evenly distributed over the roof, how much
pressure does the solar panel array place on the roof?
B) The density of fallen snow varies; here assume its ~30% of the density of liquid
water. Estimate the total pressure on the roof if 4 inches of snow fall on top of the
solar panels.

Solution:

A) Pressure = Area = Length × Width Force = Mass × Acceleration
V
‚VX

ft
(900lb
) 32.2 !
Pressure = sec  b 1lbW c
(15ft)(30ft) lb ft
32.2

sec
z{|
Pressure = >
|} >


B) Densityms† = (30%)63.3 = 19.0
o1 o1
Wˆ Wˆ

Forcems† = Volumems† × Densityms† × gravity
Forcems† + ForceŠXsom
Pressure =
Area
4inches = 0.333ft




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© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

, Chapter 1: Introduction


Pressure
ft lb ft
.(900lb
) 32.2 !5 + "(15ft)(30ft)(0.333ft)( 19.0
! 32.2 !
sec  ft + sec  1lbW
= b c
(15ft)(30ft) lb
ft
32.2
sec 
z{|
Pressure = 9. FF
|} >

1-20) A box has a mass of 20 kg, and a building has a height of 15 meters.
A) Find the force of gravity acting on the box.
B) Find the work required to lift the box from the ground to the roof of the
building.
C) Find the potential energy of the box when it is on the roof of the building.
D) If the box is dropped off the roof of the building, find the kinetic energy and
velocity of the box when it hits the ground.


A) Force = (20kg) 9.81 m ! = :8e. > G


Solution:



B) Work = Force × Distance
Work = (196.2N)(15m) = >8EFG7 = >8EF ‹


C) Potential Energy = Mass × Height × Gravity
m
Potential Energy = (20kg)(15m) 9.81 ! = >8EFG7 = >8EF ‹
sec 

D) We know that Energy is conserved, so if the box is dropped from a height of 15
meters, its Kinetic Energy at Height = 0 meters will be the same as its potential energy at
Height = 15 meters.

Kinetic Energy = >8EFG7

To find the object’s velocity as it hits the ground:

Kinetic Energy = ! Mass × Velocity 




2943Nm = ! (20kg) × V 


kg m
Nm sec  m
K294.3 Ob c = 294.3 .  5 = V 
kg 1N sec


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© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.