g

Vmax

b h

R

SGb

SGoilΥoil

Z

Figure 1: Schematic of the system

ENG 103 -MIDTERM/SOLUTION 4/28/10

1. Viscous Flow in Gap (10 points)

A cylindrical body is falling under gravity g in a tube �lled with oil. Let h be the small gap between the body andthe cylindrical tube of radius R and b the length of the body. The speci�c gravity of the oil is SGoil and that of thebody SGb. The kinematic viscosity of the oil is νoil. After a short time, the velocity of the falling object remainsconstant and equal to Vmax. See Fig. 1.

1.1 Description of the Forces Acting on the Body

The forces acting on the body are:- the buoyancy force Fb = ρoilϑbg, where ϑb = πR2b is the volume of the body. The force is up,- the friction force that opposes the motion Fτ = τA, where τ = µoilV/h is the shear stress and A = 2πRb is thewetted area. The force is up,- the weight W = ρbϑbg. The force is down.

1.2 Condition for Zero Acceleration

The condition for the body to fall at constant speed (terminal velocity) is∑

forces = 0, i.e.

Fb + Fτ − W = 0

1.3 Terminal Velocity

It is given by

ρoilπR2bg + 2πRbµoilV

h= ρbπR2bg

1

Dividing by ρwater brings the speci�c gravities

SGoilRg + 2µoil

ρwater

V

h= SGbRg

Butµoil

ρwater=

µoil

ρoil

ρoil

ρwater= νoilSGoil

Finally one obtains the formula for the terminal velocity V = Vmax

Vmax =SGb − SGoil

SGoil

hRg

2νoil

The terminal velocity does not depend on b because each of the forces is proportional to b, hence the result isindependent of b.If SGb = SGoil the body �oats in the �uid and is neutrally buoyant. Vmax = 0.The homogeneity of the expression for Vmax is satis�ed since we have[

SGb − SGoil

SGoil

hRg

2νoil

]=

[hRg

νoil

]=

LLLT−2

L2T−1= LT−1

which has unit of a velocity.Application: if g = 10 m/s2, νoil = 3.0 10−4 m2/s, R = 0.1 m, h = 10−4 m, SGoil = 0.9 and SGb = 2.9, then

Vmax = 0.37 m/s

2. Reservoir and Turbine (10 points)

A reservoir feeds a turbine located at H = 20 m below the water level. The inlet pipe to the turbine has a diameterD = 1 m. the mass �ow rate is m = 103 kg/s and the jet velocity is V = 10 m/s. Take g = 10 m/s2 andρ = 103 kg/m3. See Fig. 2.

2.1 Power Generated

Neglecting the losses we calculate the power generated by the turbine using the Steady Energy Equation (SEE)(p

ρg+

V 2

2g+ z

)1

=(

p

ρg+

V 2

2g+ z

)2

+ hf − hs − hq

Here hf = hq = 0. Furthermore p1 = p2 = pa, V1 = 0 and z1 − z2 = H. One �nds

hs = −H +V 2

2g=

Ws

mg, ⇒ Ws = −mg

(H − V 2

2g

)One �nds Ws = −150 kW . The minus sign corresponds to a turbine.

2.2 Force on the Assembly

To calculate the horizontal force Fx in (N) due to the �ow of water on the assembly at point 3, one uses thex-momentum applied to the control volume shown in Fig. 2.The theorem reads

d

dt

∫CV

ρudϑ +∫

CS

ρu(~V .~n)dA = −∫

CS

(p − pa)nxdA +∫

CS

τxdA +∫

CV

ρgxdϑ − Fx

Note the minus sign in front of the exterior force: this is because Fx represents the force exerted by the �uid onthe assembly (and −Fx is the force applied to the control surface that contains the �uid system). This will simplifysince the �ow is steady ( d

dt = 0) and inviscid (~τ = ~0), gravity is acting perpendicular to the x-axis (gx = 0). Theonly pressure contribution is on the inlet 3.

ρV3(−V3)A3 + mV = −(p3 − pa)(−1)A3 − Fx

Solving for the forceFx = (p3 − pa)A3 − m(V − V3)

We need to evaluate p3 and V3. The latter is given by conservation of mass �ow rate as

2

g

H

TV

´1

´

2

´3D

pa

pa

Fx

Ñ

n2n3

n4

n5

X

Z

Figure 2: Sketch of reservoir, turbine and control volume

m = ρV3A3, ⇒ V3 =4m

ρπD2

Using Bernoulli between 1 and 3 gives

p3 − pa = ρgH − 12ρV 2

3

Substitution in the equation for the force gives

Fx = (ρgH − 12ρV 2

3 )A3 − m(V − V3) = ρgHA3 −12mV3 − m(V − V3) = ρgHA3 − m(V − V3

2)

There is no further need of replacing A3 and V3.Application: one �nds A3 = 0.785 m2 and V3 = 1.27 m/s, ⇒Fx = 147, 635 N = 148 kN .

3. Tank and Pipe (10 points+2 bonus points)

A tank of height H = 2l is connected in its mid-height to a constant cross section pipe having three sections oflength l each: section 1 is horizontal, section 2 is vertical and section 3 is at the level of the tank base. See Fig. 3.Assume steady, inviscid (reversible) �ow.

3.1 Velocity

The velocity V of the jet is given by the classic formula from Bernoulli: V =√

2gH.

3.2 Pressure in the Pipe

The velocity in the pipe is constant and equal to V , since it is a constant cross section pipe, the pressure will behydrostatic everywhere as a consequence of the Bernoulli equation

p − pa

ρg+

V 2

2g+ z = const = H, ⇒ p − pa

ρg+ z = H − V 2

2g= 0

In the �rst section the pressure is p1(x) = p1 = pa − ρgl.In the second section, the pressure will be p2(z) = pa − ρgz.In the third section of the pipe, p3(x) = p3 = pa.

3

g

H=2l

V

l

l

l

´1

´

2

O

water pa

pa

ÑEGLHinv.L

HGLHinv.L

EGLHvis.L

HGLHvis.L

X

Z

Figure 3: Tank and pipe system

3.3 Energy and Hydraulic Grade Lines

The Energy Grade Line (EGL) is horizontal in inviscid �ow and stands at zEGL = H. Given that

zEGL =p − pa

ρg+

V 2

2g+ z, zHGL =

p − pa

ρg+ z

the Hydraulic Grade Line (HGL) is located below the EGL such that

zHGL = zEGL − V 2

2g= zEGL − H = 0

This is shown in Fig. 3.

3.4 Bonus Question

Let the total head loss due to friction in the pipe be hf < H, say hf = l, then the EGL and HGL in this casewould be as shown in Fig. 3. One third of the loss occurs in each pipe.

4