Solutions Manual for Munson, Young and Okiishi-s Fundamentals of Fluid Mechanics, 9th Edition.pdf

Solutions Manual for Munson, Young and
Okiishi's Fundamentals of Fluid Mechanics,
9th Edition
By Andrew Gerhart, John Hochstein, Philip
Gerhart


All Chapters 100% Original Verified A+ Grade


Chapter 1: Introduction


Problem 1.1
Given: A fluid has a viscosity of 0.005 Pa·s
and a specific gravity of 0.85.
Find: Its absolute viscosity in lb·s/ft² and its
kinematic viscosity in m²/s and ft²/s.
Solution:
Absolute viscosity in lb·s/ft²:


1

,μ = 0.005 Pa·s × 0.020885 = 1.04 × 10⁻⁴
lb·s/ft²
Kinematic viscosity:
ρ = 0.85 × 1000 = 850 kg/m³
ν = μ/ρ = 0.005/850 = 5.88 × 10⁻⁶ m²/s
ν = 5.88 × 10⁻⁶ m²/s × 10.764 ft²/m² = 6.33 ×
10⁻⁵ ft²/s


Problem 1.2
Given: A fluid has a viscosity of 0.0030 Pa·s
and a specific gravity of 1.20.
Find: Its absolute viscosity in lb·s/ft² and its
kinematic viscosity in m²/s.
Solution:
μ = 0.0030 Pa·s × 0.020885 = 6.266 × 10⁻⁵
lb·s/ft²
ρ = 1.20 × 1000 = 1200 kg/m³


2

,ν = 0.0030/1200 = 2.50 × 10⁻⁶ m²/s


Problem 1.3
Given: A fluid has a kinematic viscosity of 1.2
× 10⁻⁵ m²/s and an absolute viscosity of 0.010
Pa·s.
Find: Its density and specific gravity.
Solution:
ρ = μ/ν = 0.010/(1.2 × 10⁻⁵) = 833.33 kg/m³
SG = ρ/1000 = 0.833


Problem 1.4
Given: A fluid has a specific gravity of 0.95
and a viscosity of 0.008 Pa·s.
Find: Its kinematic viscosity in ft²/s.
Solution:


3

, ρ = 0.95 × 1000 = 950 kg/m³
ν = 0.008/950 = 8.421 × 10⁻⁶ m²/s
Convert to ft²/s: ν = 8.421 × 10⁻⁶ × 10.764 =
9.06 × 10⁻⁵ ft²/s


Problem 1.5
Given: A fluid has an absolute viscosity of
0.0015 lb·s/ft² and a kinematic viscosity of
0.0008 ft²/s.
Find: Its density and specific gravity.
Solution:
ρ = μ/ν = 0.0015/0.0008 = 1.875 slugs/ft³
ρ_water = 1.94 slugs/ft³
SG = 1.875/1.94 = 0.966


Problem 1.6


4