chapter1 a

P 3000atm:= D 0.17in:= A 4D2:= A 0.023 in2=

F P A:= g 32.174 ftsec2

= mass Fg

:= mass 1000.7 lbm= Ans.

1.7 Pabs g h Patm+=

13.535 gmcm3:= g 9.832 m

s2:= h 56.38cm:=

Patm 101.78kPa:= Pabs g h Patm+:= Pabs 176.808kPa= Ans.

1.8 13.535 gmcm3:= g 32.243 ft

s2:= h 25.62in:=

Patm 29.86in_Hg:= Pabs g h Patm+:= Pabs 27.22psia= Ans.

Chapter 1 - Section A - Mathcad Solutions1.4 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: PFA

= F mass g= Note: Pressures are ingauge pressure.

P 3000bar:= D 4mm:= A 4D2:= A 12.566mm2=

F P A:= g 9.807 ms2

= mass Fg

:= mass 384.4kg= Ans.

1.6 By definition: PFA

= F mass g=

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. 2005 The McGraw-Hill Companies, Inc. All rights reserved. Limited distribution permitted

FMars K x:= FMars 4 10 3 mK=gMars

FMarsmass

:= gMars 0.01 mKkg= Ans.

1.12 Given:zPd

d g= and: M P

R T= Substituting: zPdd

M PR T g=

Separating variables and integrating:

Psea

PDenver

P1P

d

0

zDenver

zM gR T

d=

After integrating: lnPDenverPsea

M gR T zDenver=

Taking the exponential of both sidesand rearranging: PDenver Psea e

M gR T zDenver

=

Psea 1atm:= M 29 gmmol:= g 9.8m

s2:=

1.10 Assume the following: 13.5 gmcm3

:= g 9.8 ms2

:=

P 400bar:= h P g:= h 302.3m= 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 gMarsbased on spring measurement on Mars.On Earth:

F mass g= K x= mass 0.40kg:= g 9.81 ms2

:= x 1.08cm:=

F mass g:= F 3.924N= Ks Fx:= Ks 363.333Nm

=On Mars:

x 0.40cm:=

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Ans.

wmoon M gmoon:= wmoon 18.767 lbf= Ans.

1.14 costbulb5.00dollars1000hr

10 hrday

:= costelec 0.1dollarskW hr 10hrday

70 W:=

costbulb 18.262dollarsyr

= costelec 25.567 dollarsyr=

costtotal costbulb costelec+:= costtotal 43.829 dollarsyr= Ans.

1.15 D 1.25ft:= mass 250lbm:= g 32.169 fts2

:=

R 82.06cm3 atmmol K:= T 10 273.15+( )K:= zDenver 1 mi:=

M gR T zDenver 0.194=

PDenver Psea e

M gR T zDenver

:= PDenver 0.823atm= Ans.

PDenver 0.834bar= Ans.

1.13 The same proportionality applies as in Pb. 1.11.

gearth 32.186ft

s2:= gmoon 5.32 ft

s2:= lmoon 18.76:=

learth lmoongearthgmoon:= learth 113.498=

M learth lbm:= M 113.498 lbm=

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Ans.

(b) PabsFA

:= Pabs 110.054kPa= Ans.

(c) l 0.83m:= Work F l:= Work 15.848kJ= Ans.

EP mass g l:= EP 1.222kJ= Ans.

1.18 mass 1250kg:= u 40 ms

:=

EK12mass u2:= EK 1000kJ= Ans.

Work EK:= Work 1000kJ= Ans.

1.19 Wdotmass g h

time0.91 0.92=

Wdot 200W:= g 9.8 ms2

:= h 50m:=

Patm 30.12in_Hg:= A 4 D2:= A 1.227 ft2=

(a) F Patm A mass g+:= F 2.8642 103 lbf= Ans.

(b) PabsFA

:= Pabs 16.208psia= Ans.

(c) l 1.7ft:= Work F l:= Work 4.8691 103 ft lbf= Ans.

PE mass g l:= PE 424.9 ft lbf= Ans.

1.16 D 0.47m:= mass 150kg:= g 9.813 ms2

:=

Patm 101.57kPa:= A 4 D2:= A 0.173m2=

(a) F Patm A mass g+:= F 1.909 104 N=

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mdotWdot

g h 0.91 0.92:= mdot 0.488kgs

= Ans.

1.22

a) cost_coal

25.00ton

29MJkg

:= cost_coal 0.95GJ 1=

cost_gasoline

2.00gal

37GJ

m3

:= cost_gasoline 14.28GJ 1=

cost_electricity0.1000kW hr:= cost_electricity 27.778GJ 1=

b)The electrical energy can directly be converted to other forms of energywhereas the coal and gasoline would typically need to be converted to heatand 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 heatsource. Otherwise it is messy to handle and bulky for tranport andstorage.

Gasoline is an important transportation fuel. It is more convenient totransport and store than coal. It can be used to generate electricity byburning it but the efficiency is limited. However, fuel cells are currentlybeing developed which will allow for the conversion of gasoline to electricityby chemical means, a more efficient process.

Electricity has the most uses though it is expensive. It is easy to transportbut expensive to store. As a transportation fuel it is clean but batteries tostore it on-board have limited capacity and are heavy.

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1.24 Use the Matcad genfit function to fit the data to Antoine's equation.The genfit function requires the first derivatives of the function withrespect to the parameters being fitted.

Function being fit: f T A, B, C,( ) eA

BT C+

:=

First derivative of the function with respect to parameter A

Af T A, B, C,( )d

dexp A

BT C+

First derivative of the function with respect to parameter B

Bf T A, B, C,( )d

d1

T C+ exp AB

T C+

First derivative of the function with respect to parameter C

Cf T A, B, C,( )d

dB

T C+( )2exp A

BT C+

t

18.59.50.2

11.8

23.1

32.7

44.4

52.1

63.3

75.5

:= Psat

3.18

5.48

9.45

16.9

28.2

41.9

66.6

89.5

129

187

:=

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T t 273.15+:= lnPsat ln Psat( ):=Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C.

Guess values of parameters

F T a,( )

exp a0a1

T a2+

exp a0a1

T a2+

1T a2+ exp a0

a1T a2+

a1

T a2+( )2 exp a0a1

T a2+

:= guess15

3000

50

:=

Apply the genfit function

A

B

C

genfit T Psat, guess, F,( ):=A

B

C

13.421

2.29 10369.053

= Ans.

Compare fit with data.

240 260 280 300 320 340 3600

50

100

150

200

Psat

f T A, B, C,( )

T

To find the normal boiling point, find the value of T for which Psat = 1 atm.

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This is an open-ended problem. The strategy depends on age of the child,and on such unpredictable items as possible financial aid, monies earnedby 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 5.511%=i Find i( ):=C2C1

1 i+( )t2 t1=Given

C2 80000dollarsyr

:=C1 16000 dollarsyr:=t2 2000:=t1 1970:=b)

The increase in price of gasoline over this period kept pace with the rate ofinflation.

C2 1.513dollarsgal

=C2 C1 1 i+( )t2 t1:=

i 5%:=C1 0.35 dollarsgal:=t2 2000:=t1 1970:=a)1.25

Tnb 273.15K 56.004degC= Ans.

Tnb 329.154K=Tnb BA ln

PsatkPa

C

K:=Psat 1atm:=

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