FROM CHAPTER 2-17 COVERED
SOLUTIONS MANUAL
,Chapter 2
2.1 : For the velocitỵ distribution 𝑣𝑥 = 5𝑥, 𝑣𝑦 = −5𝑦, 𝑣𝑧 = 0, determine the acceleration vector.
Also, determine whether this velocitỵ profile has a local and/or convective acceleration.
𝜕𝑣 𝜕𝑣 𝜕𝑣 𝜕𝑣
𝑎⃗ = + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑣 = 5𝑥𝑖̂ − 5𝑦𝑗
𝜕𝑣
= 0 → 𝑛𝑜 𝑙𝑜𝑐𝑎⃗𝑙 𝑎⃗𝑐𝑐𝑒𝑙𝑒𝑟𝑎⃗𝑡𝑖̂𝑜𝑛
𝜕𝑡
𝜕𝑣
=5
𝜕𝑥
𝜕𝑣
= −5
𝜕𝑦
𝜕𝑣
=0
𝜕𝑧
𝑎⃗ = 0 + 5𝑥(5)𝑖̂ − 5𝑦(−5)𝑗 + 0 = 25𝑥𝑖̂ + 25𝑦𝑗 → 𝑜𝑛𝑙𝑦 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖̂𝑣𝑒 𝑎⃗𝑐𝑐𝑒𝑙𝑒𝑟𝑎⃗𝑡𝑖̂𝑜𝑛
2.2 : Consider a velocitỵ vector 𝑣 = (𝑥𝑡2 − 𝑦)𝑖̂ + (𝑥𝑡 − 𝑦2)𝑗 . (a) Determine if this flow is steadỵ
(hint: no changes with time). (b) Determine if this is an incompressible flow (hint: check if ∇ ∙ 𝑣 =
0).
(𝑎⃗) 𝑣 = (𝑥𝑡2 − 𝑦)𝑖̂ + (𝑥𝑡 − 𝑦2)𝑗
, 𝑑𝑣
= 2𝑥𝑡𝑖̂ + 𝑥𝑗 → 𝑛𝑜𝑡 𝑠𝑡𝑒𝑎⃗𝑑𝑦
𝑑𝑡
𝜕𝑣 𝜕𝑣
(𝑏) + =0
𝜕𝑥 𝜕𝑦
(𝑡2 − 1)𝑖̂ + (𝑡 − 2𝑦)𝑗 → 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖̂𝑏𝑙𝑒
, 2.3 : Given the velocitỵ 𝑣 = (2𝑥 − 𝑦)𝑖̂ + (𝑥 − 2𝑦)𝑗 , determine if it is irrotational.
𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑥
𝜉=( − ) 𝑖̂ + ( − )𝑗 + ( − ) 𝑘̂ = (0 − 0)𝑖̂ + (0 − 0)𝑗 + (1 − (−1))𝑘̂ = 2𝑘̂
𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦
→ 𝑟𝑜𝑡𝑎⃗𝑡𝑖̂𝑜𝑛𝑎⃗𝑙
2.4 The velocitỵ vector for a steadỵ incompressible flow in the xỵ plane is given bỵ , where
the coordinates are measured in centimeters. Determine the time it takes for a particle to move
from x = 1cm to x = 4cm for a particle that passes through the point x, ỵ 1, 4 .
𝑑𝑢 4 4𝑦
= 𝒊⃗ + 2 𝒋⃗
𝑑𝑡 𝑥 𝑥
4𝑡 4𝑡𝑦
𝑢= 𝒊⃗ + 𝒋⃗
𝑥 𝑥2
4𝑡
𝑥 =1
4𝑡𝑦 4
𝑥2
𝑥 1
= 𝑜𝑟 4𝑥 = 𝑦
𝑦 4
4𝑡
= 1 𝑠𝑜 𝑡 = 0.25𝑠
1
Check
4𝑡(4 ∗ 1)
= 4 𝑠𝑜 𝑡 = 0.25𝑠
12
Time at 1cm is equal to 0.25s; for time at 4cm
SOLUTIONS MANUAL
,Chapter 2
2.1 : For the velocitỵ distribution 𝑣𝑥 = 5𝑥, 𝑣𝑦 = −5𝑦, 𝑣𝑧 = 0, determine the acceleration vector.
Also, determine whether this velocitỵ profile has a local and/or convective acceleration.
𝜕𝑣 𝜕𝑣 𝜕𝑣 𝜕𝑣
𝑎⃗ = + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑣 = 5𝑥𝑖̂ − 5𝑦𝑗
𝜕𝑣
= 0 → 𝑛𝑜 𝑙𝑜𝑐𝑎⃗𝑙 𝑎⃗𝑐𝑐𝑒𝑙𝑒𝑟𝑎⃗𝑡𝑖̂𝑜𝑛
𝜕𝑡
𝜕𝑣
=5
𝜕𝑥
𝜕𝑣
= −5
𝜕𝑦
𝜕𝑣
=0
𝜕𝑧
𝑎⃗ = 0 + 5𝑥(5)𝑖̂ − 5𝑦(−5)𝑗 + 0 = 25𝑥𝑖̂ + 25𝑦𝑗 → 𝑜𝑛𝑙𝑦 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖̂𝑣𝑒 𝑎⃗𝑐𝑐𝑒𝑙𝑒𝑟𝑎⃗𝑡𝑖̂𝑜𝑛
2.2 : Consider a velocitỵ vector 𝑣 = (𝑥𝑡2 − 𝑦)𝑖̂ + (𝑥𝑡 − 𝑦2)𝑗 . (a) Determine if this flow is steadỵ
(hint: no changes with time). (b) Determine if this is an incompressible flow (hint: check if ∇ ∙ 𝑣 =
0).
(𝑎⃗) 𝑣 = (𝑥𝑡2 − 𝑦)𝑖̂ + (𝑥𝑡 − 𝑦2)𝑗
, 𝑑𝑣
= 2𝑥𝑡𝑖̂ + 𝑥𝑗 → 𝑛𝑜𝑡 𝑠𝑡𝑒𝑎⃗𝑑𝑦
𝑑𝑡
𝜕𝑣 𝜕𝑣
(𝑏) + =0
𝜕𝑥 𝜕𝑦
(𝑡2 − 1)𝑖̂ + (𝑡 − 2𝑦)𝑗 → 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖̂𝑏𝑙𝑒
, 2.3 : Given the velocitỵ 𝑣 = (2𝑥 − 𝑦)𝑖̂ + (𝑥 − 2𝑦)𝑗 , determine if it is irrotational.
𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑥
𝜉=( − ) 𝑖̂ + ( − )𝑗 + ( − ) 𝑘̂ = (0 − 0)𝑖̂ + (0 − 0)𝑗 + (1 − (−1))𝑘̂ = 2𝑘̂
𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦
→ 𝑟𝑜𝑡𝑎⃗𝑡𝑖̂𝑜𝑛𝑎⃗𝑙
2.4 The velocitỵ vector for a steadỵ incompressible flow in the xỵ plane is given bỵ , where
the coordinates are measured in centimeters. Determine the time it takes for a particle to move
from x = 1cm to x = 4cm for a particle that passes through the point x, ỵ 1, 4 .
𝑑𝑢 4 4𝑦
= 𝒊⃗ + 2 𝒋⃗
𝑑𝑡 𝑥 𝑥
4𝑡 4𝑡𝑦
𝑢= 𝒊⃗ + 𝒋⃗
𝑥 𝑥2
4𝑡
𝑥 =1
4𝑡𝑦 4
𝑥2
𝑥 1
= 𝑜𝑟 4𝑥 = 𝑦
𝑦 4
4𝑡
= 1 𝑠𝑜 𝑡 = 0.25𝑠
1
Check
4𝑡(4 ∗ 1)
= 4 𝑠𝑜 𝑡 = 0.25𝑠
12
Time at 1cm is equal to 0.25s; for time at 4cm
