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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

4

D

2

A

12.566 mm

2F

PA

g

9.807

m s
2

mass

F g

mass

384.4 kg

Ans.

1.6 By definition:
P 3000atm

P=
D

F A
0.17in

F = mass g
A

4

D

2

A

0.023 in

2

F

PA

g

32.174

ft sec
2

mass

F g

mass

1000.7 lbm

Ans.

1.7 Pabs =

gh

Patm

13.535

gm cm
3

g

9.832

m s
2

h

56.38cm

Patm

101.78kPa

Pabs

ghPatm

Pabs

176.808 kPa Ans.

1.8

13.535

gm cm
3

g

32.243

ft s
2

h

25.62in

Patm

29.86in_Hg

Pabs

gh

Patm

Pabs

27.22 psia

Ans.

1

1.10

Assume the following:

13.5

gm cm
3

g

9.8

m s
2

P

400bar

h

P g

h

302.3 m

Ans.

1.11

The force on a spring is described by: F = Ks x where Ks is the springconstant. 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
4 10

Ks
3

363.333

N m

On Mars:

x

0.40cm
FMars mass

FMars

Kx

FMars

mK

gMars

gMars

0.01

mK kg

Ans.

1.12 Given:

d P=dz

g

and:

=

MP RT

Substituting:

d P= dz

MP g RT

PDenver

Separating variables and integrating:
Psea

1 dP = P

zDenver

Mg dz RT

0

After integrating:

ln

PDenver Psea

=

Mg RT

zDenver

Taking the exponential of both sides and rearranging: Psea 1atm

PDenver = Psea e gm mol

Mg zDenver RT

M

29

g

9.8

m s
2

2

R

82.06cm atm mol K

3

T

( 10

273.15)K

zDenver

1 mi

Mg zDenver RT
Mg

0.194

PDenver

Psea e

RT

zDenver

PDenver

0.823 atm

Ans.

PDenver 1.13 The same proportionality applies as in Pb. 1.11.

0.834 bar

Ans.

gearth

32.186

ft s
2

gmoon

5.32

ft s
2

lmoon

18.76

learth

lmoon

gearth gmoon

learth

113.498

M
wmoonlearth lbm
M gmoon

M

113.498 lbm
18.767 lbf

Ans.

wmoon

Ans.

1.14

costbulb

hr 5.00dollars 10 day 1000hr

costelec

hr 0.1dollars 70W 10 day kW hr

costbulb

18.262

dollars yr

costelec

25.567

dollars yr

costtotal

costbulb

costelec

costtotal

43.829

dollars yr Ans.

1.15

D

1.25ft

mass

250lbm

g

32.169

ft s
2

3 Patm

30.12in_Hg

A

4

D

2

A
3

1.227 ft

2

(a) F

Patm A

mass g

F

2.8642 10 lbf

Ans.

(b) Pabs

F A

Pabs

16.208 psia

Ans.
3 4.8691 10 ft lbf Ans.

(c)

l

1.7ft

Work

F l

Work

PE

mass g l

PE

424.9 ft lbf

Ans.

1.16

D

0.47m

mass

150kg

g

9.813

m s
2

Patm

101.57kPa

A

4

D

2

A4

0.173 m

2

(a) F
(b) Pabs

Patm A
F A

mass g

F
Pabs

1.909 10 N
110.054 kPa

Ans.

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 2

mass u

2

EK

1000 kJ

Ans.

Work

EK

Work

1000 kJ

Ans.

1.19

Wdot =

mass g h 0.91 0.92time

Wdot

200W

g

9.8

m s
2

h

50m

4

mdot

Wdot g h 0.91 0.92
25.00 ton

mdot

0.488

kg s

Ans.

1.22 a) cost_coal

29

MJ kg
2.00 gal

cost_coal

0.95 GJ

1

cost_gasoline

37

GJ m
3

cost_gasoline

14.28 GJ

1

cost_electricity

0.1000 kW hr

cost_electricity

27.778 GJ

1

b) The electrical energy can directly be...
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