Solution Manual for A Student’s Guide to the Navier–Stokes Equations (New Edition, 2024) by Chris Garvin

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SOLUTIONS + EXTRA

,Solutions to A Student’s Guide to
the Navier-Stokes Equations

Justin W. Garvin
The University of
Iowa

,
, 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 =
î + 2 ĵ + 3k̂ . The velocity of the flow is: ⃗V = 2 î + 3 ĵ + 0k̂ m/s.
1
√
14
Solution: The mass passing through a surface is defined as:

ṁpass = ρV⃗ · ⃗ndA
A

We can plug our numbers in:

ρV⃗ · ⃗ndA
ṁpass = A
kg !
= 1.2 2 î + 3 ĵ m/s
√1 î + 2 ĵ + 3k̂ dA
· 14
A kgm3 1 m/sdA
= 1.2 (2 + 6 + 0)
√
A m3 14
2



kg 1m
= 1.2 8
√ A
sm2 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
of mass of time rate of mass of
water mass of water
in the bathtub = water –
leaving the
in a given entering the bathtub
unit of bathtub
time

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

the change
of mass of time rate of
water mass of
in the bathtub = water
in a given entering the
unit of bathtub
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 = ρV⃗ · ⃗ndA −−−−−−−→ ρV A
A
It simplifies to ρVA because ρ and V ⃗ 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:

ṁin = ρVA = (1000) (1.2) π0.042 = 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

, Problems 5
1.3 The divergence of the velocity vector in spherical coordinates can
be written as:

∂ r2V 1 1 ∂Vϕ
⃗⃗ ∂
1 +
r
∇·V = 2 (Vθ sin (θ)) + r sin (θ) ∂ϕ
r sin (θ) ∂θ
r ∂r
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 ∂ r Vr
2
1 2 ∂Vr 1 ∂r 2
= (product
rule)
r2 ∂r r ∂r
r2 + 2 Vr (1.1)
|{ z} ∂r r |{ z}
∂Vr 2
=1 =2r
= + Vr
∂r r
We can insert Vr into Equation 1.1 to get:

,6 Chapter 1 Solutions


1 ∂ r Vr
2
∂Vr 2
= + V r
r2 ∂r ∂r r 3 !! 3 !!
3R R 2 3R R
∂ 1− + 3 1− + 3
= −U cos (θ) 2r 2r + −U cos (θ) 2r 2r
∂r !! r !!
3R R3 2 3R R 3

= −U cos θ +
2 − 3
4 – U cos θ r − r2 + r4
2r 2r !
2 3 1 R3
−
= −U cos θ r 2r2 2 r4 −

Now onto the second term. We can do the same thing for the
second term as we did for the first term. That is, we can expand out
the derivative using the product rule:

1 ∂ 1 1 ∂sin θ
(Vθ sin (θ)) = sin θ ∂Vθ Vθ
rsinθ∂θ r sin θ ∂θ + r sin θ ∂θ
| {z =
} |{ z }
1r =cos θ
So now we can plug in Vθ into the above equation to get:

1 ∂ cos θ
1 ∂Vθ Vθ
(Vθ sin (θ)) = r ∂θ3 + r sin θ
r sin (θ) ∂θ 3R !! 3!
1 ∂ U sin θ R cos θ U sin θ 1 − 3R R
= r ∂θ 1− − 4r3 + r sin θ 3
4r 4r3
4r 3 −
1 3R U !
= U cos θ 1 ! 3R R
R cos θ 1 −
− − 3 + − 3
r 4r 4r r 4r 4r
3R
= U cos θ 2 !
r − 2 2r R3
4
r
− (1.2)
As you can see, Equation 1.3 and 1.2 are negatives of each
other. Therefore, adding them up gives us zero and, hence,
constitutes an in- compressible flow.
1.4 An incompressible fluid with density ρ = 1000 kg/m3 travels through
a channel with a rectangular cross-section of dimensions 25 mm by
30 mm. The average velocity of the flow in this portion of the
channel is 1 m/s. If the flow is at a steady state, what will the
velocity be if the rectangular cross-section increases to 50 mm by
50 mm? Start from the integral form of the continuity equation.

, Problems 7
Solution: Starting from the integral form of the continuity equation
we have:

$ ρV⃗ · ⃗ndA = 0
∂t ∂ρ +A
V
The slow is at steady state, so the time derivative goes away and we
are left with:

⃗ ·⃗ndA = 0
ρV
A
We essentially have two sections (areas) that have flow: an inlet
and an exit. So, we can break up the area integral into two pieces:
" " ⃗ ·⃗ndA = 0
ρV
⃗ ·⃗ndA +
ρV
outle
inlet
t
We are going to assume the density and the velocity do not
change with the cross-sectional area. In addition, we will assume
the inlet nor-
ˆ ˆ
mal is −i and the outlet normal is i, giving us:

⃗ inlet · −iAˆinlet + ρoutlet⃗V
ρinletV · îA outlet = 0
outle
t
If the velocity is only in the x−direction, then we have:

– ρinlet uinlet Ainlet + ρoutlet uoutlet · îAoutlet = 0
– uinlet Ainlet + uoutlet Aoutlet = 0
solve for uoutlet Ainlet
– −−−−−−−→ uoutlet = uinlet
Aoutlet
Plugging in numbers leads
to: (30) (25)
uoutlet = 1m/s = 0.3m/s
(50) (50)
1.5 Air in a pipe with a diameter of 10 cm starts out at a temperature of
700 K and pressure of 4 × 105 Pa. The initial flow velocity is 10 meters
per second. If the pipe diameter contracts to 5 cm with the air
speeding up to 115 m/s and the temperature decreases to 500 K,
what is the pressure af- ter the contraction? You can assume the
molecular weight of air is 28.97 kg/kmol and a steady state. Start
from the integral form of the conti- nuity equation.

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