Solution Manual for A Student's Guide to the Navier–Stokes Equations 1st Edition By Justin Garvin

Solution Manual for A Student's Guide to the Navier–Stokes Equations 1st Edition by Justin Garvin

,(A Student's Guide to the Navier–Stokes Equations, 1e Justin Garvin)

(Solution Manual all Chapters)



1
Chapter 1 Solutions




Problems
1.1 Find the mass flow rate of a fluid at a constant density of 1.2 kg/m3
passing through a surface whose area is 1 m2 and whose normal is ⃗n =
√ ˆi + 2 ˆj + 3kˆ . The velocity of the flow is: V
⃗ = 2ˆi + 3 ˆj + 0kˆ m/s.
1
14

Solution: The mass passing through a surface is defined as:


ṁ pass = ρ V⃗ · ⃗ndA
A

We can plug our numbers in:


ṁ pass = ρ V⃗ · ⃗ndA
A
!
kg ˆ ˆ 1 ˆ ˆ ˆ
= 1.2 2 i + 3 j m/s · √ i + 2 j + 3k dA
m 1 14
A 3
kg
= 1.2 (2 + 6 + 0) m/sdA
√
A m 3 14
1 m2
kg 8
= 1.2 2 √ A
sm 14 0
9.6
= kg/s
√
14
1.2 Approximate how long will it take, in minutes, to fill up a bathtub with
dimensions of 1.3 m x 0.6 m x 0.4 m if water is coming out of a 40 mm
diameter faucet at a speed of 1.2 m/s. Assume the density of water is
1000 kg/m3.

3

, 4 Chapter 1 Solutions


Solution: This is a basic mass conservation problem. In words, we have:

the change
time rate of time rate of
of mass of water
mass of water mass of water
in the bathtub = –
entering the leaving the
in a given
bathtub bathtub
unit of time

There is no mass coming out, which leaves us with:

the change
time rate of
of mass of water
mass of water
in the bathtub =
entering the
in a given
bathtub
unit of time

So, the change of the mass in a given unit of time in the tub is just
equal to the mass flow rate into the tube:

simplifies to
ṁin = ρ V⃗ · ⃗ndA −−−−−−−−→ ρVA
A


It simplifies to ρVA because ρ and V⃗ are constant and can be pulled
out of the area integral. In addition, it is assumed that the velocity and
the normal are in the same direction. Plugging in numbers:

π
ṁin = ρVA = (1000) (1.2) 0.042 = 0.302kg/s
4
Thus the tube fills at a rate of 0.302 kg/s. The total mass of water that
the tube (masstotal) can hold is the volume of the tub multiplied by the
density of water:


masstube = (1000) (1.3 × 0.6 × 0.4) = 312kg

The length of times to fill is:

masstotal 312kg
time = = = 1033 s = 17 minutes
ṁin 0.302kg/s




Solution Manual for A Student's Guide to the Navier–Stokes Equations 1st Edition by Justin Garvin

, Problems 5

1.3 The divergence of the velocity vector in spherical coordinates can be
written as:

2
1 ∂ r Vr 1 ∂ 1 ∂Vϕ
⃗ ·V
∇ ⃗ = + (Vθ sin (θ)) +
r2 ∂r r sin (θ) ∂θ r sin (θ) ∂ϕ
where Vr, Vθ, and Vϕ are the velocity coordinates in the r−, θ−, and ϕ−
direction, respectively. Determine if a flow with the following flow field
velocity is incompressible:

!
3R R3
Vr = −U cos (θ) 1 – +
2r 2r3
!
3R R3
Vθ = U sin (θ) 1 – −
4r 4r3

Vϕ = 0

where R and U are constants (note, R is not the gas constant in this prob-
lem).

Solution: This one is simple enough. We just have to plug in the Vr,
Vtheta, and Vz expression into the divergence of velocity equation given
for spherical coordinates and see if it is equal to zero, since the diver-
gence of velocity is zero for an incompressible flow. The easiest term in
the velocity divergence term is the last one since Vϕ is zero:

1 ∂Vphi 1 ∂0
= =0
r sin θ ∂ϕ r sin θ ∂ϕ
The next term we will look is the first term of the velocity divergence.
Before we do anything, let’s break out the derivative using the product
rule:


1 ∂ r2Vr 1 2 ∂Vr 1 ∂r2
= 2r + 2 Vr (product rule)
r 2 ∂r r ∂r r ∂r
x˛zx x˛zx (1.1)
=1 =2r
∂Vr 2
= + Vr
∂r r
We can insert Vr into Equation 1.1 to get: