Mathcad Solutions to Fluid Mechanics Problems

P 3000atm D 0.17in A S

4˜D² A 0.023 in² F P A˜ g 32.174 ft

sec²

mass F

g mass 1000.7 lb_m Ans.

1.7 P_abs = U˜g˜hP_atm

U 13.535 gm cm³

˜ g 9.832 m

s²

˜ h 56.38cm

P_atm 101.78kPa P_abs U˜g˜hP_atm P_abs 176.808 kPa Ans.

1.8 U 13.535 gm cm³

˜ g 32.243 ft

s²

˜ h 25.62in

P_atm 29.86in_Hg P_abs U˜g˜hP_atm P_abs 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= 1.8t32 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˜D² A 12.566 mm² F P A˜ g 9.807m

s²

mass F

g mass 384.4 kg Ans.

1.6 By definition: P F

= A F = mass g˜

1

F_Mars K x˜ F_Mars 4u10³m K

g_Mars F_Mars

mass g_Mars 0.01m K

kg Ans.

1.12 Given:

z d P d

U ˜g

= and: U M P˜ R T˜

= Substituting:

z d P d

M P˜ R T˜ ˜g

=

Separating variables and integrating:

P_sea P_Denver

1 P P

´µ µ¶

d

0 z_Denver

M g˜ z R T˜

§¨ ©

·

¹

´µ µ¶

= d

After integrating: ln P_Denver

P_sea

§ ¨

©

·

¹

M˜g

R T˜ ˜z_Denver

=

Taking the exponential of both sides

and rearranging: P_Denver P_sea e

M˜g

R T˜ ˜z_Denver

©§¨ ·

˜ ¹

=

P_sea 1atm M 29 gm

mol g 9.8m

s² 1.10 Assume the following: U 13.5 gm

cm³

g 9.8m s²

P 400bar h P

U˜g h 302.3 m Ans.

1.11 The force on a spring is described by: F = K_s x where K_s is the spring constant. First calculate K based on the earth measurement then g_Mars based on spring measurement on Mars.

On Earth:

F = mass g˜ = K x˜ mass 0.40kg g 9.81m s²

x 1.08cm

F mass g˜ F 3.924 N K_s F

x K_s 363.333N m On Mars:

x 0.40cm

2