MECH3700 Tutorial 2

1. The basic component of a pressure gage tester consists of a piston-cylinder apparatus as shown in Fig. 1. The piston, 6 mm in diameter, is loaded to develop a pressure of known magnitude. The piston length is 25 mm. The vertical movement of the piston due to leakage is 1 mm/min. The specific gravity of the fluid is 0.92. (a) Calculate the mass, M, required to produce 1.5 MPa (gauge) in the cylinder. (b) Determine the leakage flow rate as a function of radial clearance, a, for this load if the dynamic viscosity of the liquid 0.42 N-s/m2. (c) Specify the maximum allowable radial clearance.

Fig.1

(a) ( )∑ =−−= 04

2

MgppDF atmyπ , kg32.4=M

(b) From Example 8.1, Lpa

WQ

µ12

3Δ= , where DW π= .

(c) The piston moves downwards at speed U to displace the fluid at volume flow rate Q. By conservation of volume of fluid,

/sm1071.4601000/1

6)1000/6(

4310

22−×===

ππ UDQ , m1028.112 53/1

−×=⎟⎟⎠

⎞⎜⎜⎝

⎛

Δ=

pDLQa

πµ

Check if the flow is laminar:

( ) 2722

522 m102.1100061028.1

10006

4][

4−− ×=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡ ×+=−+=ππ DaDA

m/s1089.3 3−×==AQV

453

1009.142.0

)1028.1)(1089.3)(1000(92.0Re −−−

×=××

==µρ aV (laminar)

2. Consider fully developed laminar flow between two infinite parallel plates in Fig. 2 separated by gap width d = 10 mm. The upper plate moves to the right with speed U1 = 0.5 m/s; the lower plate moves to the left with speed U2 = 0.25 m/s. The pressure gradient in the direction of flow is zero. Find the velocity distribution in the gap and the volume flow rate if the depth is b.

Fig. 2 From the Navier-Stokes equations as in section 8.2,

it is concluded that2

2

0dyud

xp

µ+∂

∂−=

because the flow is fully developed, )(yuu = .

( ) 212

21 cycy

xpyu ++⎟⎠

⎞⎜⎝

⎛∂

∂=

µµ, 0=∂

∂

xp , ( ) 2

1 cycyu ++=µ

Boundary conditions: 2)0( Uyu −== , 1)( Udyu ==

( ) 121)( UdyUUyu −+= ,

⎟⎠

⎞⎜⎝

⎛ −⎟⎠

⎞⎜⎝

⎛=−

== ∫= 2

25.05.0100010

2)( 12

0

bUUbdbdyuQd

y

00125.0=bQ m3/s/m

3. A viscous-shear pump in Fig. 3 is made from a stationary housing with a close-fitting rotating drum inside. The clearance is small compared with the diameter of the drum, so flow in the annular space may be treated as flow between parallel plates. Fluid is dragged around the annulus by viscous forces. Evaluate the pressure differential and input power as functions of volume flow rate. Assume that the depth normal to the diagram is W.

Fig. 3

From section 8.2b, ( ) 212

21 cycy

xpyu ++⎟⎠

⎞⎜⎝

⎛∂

∂=

µµ

Boundary conditions: 0)0( ==yu , Uayu == )( .

02 =c ,

acaxpU

µµ12

21

+⎟⎠

⎞⎜⎝

⎛∂

∂=

aa

y

ycyxpWdyuWQ

0

21

3

0 2321

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛∂∂

== ∫= µµ

, 3

121

2a

xpUa

WQ

⎟⎠

⎞⎜⎝

⎛∂

∂−=

µ

From ωRU = and

Lp

xp Δ

−=∂

∂ , ⎟⎠⎞

⎜⎝⎛ −=Δ−

WQRa

aLp

212

3ωµ .

⎥⎦

⎤⎢⎣

⎡+⎟⎠

⎞⎜⎝

⎛ Δ−==

µµµµτ 1c

Lpy

dydu

At ay = , ⎥⎦

⎤⎢⎣

⎡+⎟⎠

⎞⎜⎝

⎛ Δ−=

µµµτ 1c

Lpa

Torque ( )RRWπτ 2=Γ . Power ωΓ=P .

4. Water is pumped at the rate of 0.075 m3/s from a reservoir 20 m above a pump to a free discharge 35 m above the pump, as shown in Fig. 4. The pressure on the intake side of the pump is 150 kPa (gauge) and the pressure on the discharge side is 450 kPa (gauge). All pipes are commercial steel of 15 cm diameter. Determine (a) the head supplied by the pump and (b) the total head loss between the pump and point of free discharge.

Fig. 4

(a) 3

233

2

222

22z

gV

gp

HzgV

gp

p ++=+++ρρ

32 zz = ,

32 VV = ,

m6.3023 =−

=gpp

H p ρ

(b) LhzgV

gpz

gV

gp

+++=++ 4

244

3

233

22 ρρ

43 VV =

m9.104343 =−+

−= zz

gpp

hL ρ

5. Two galvanized iron pipes (e = 0.15 mm) of diameter D are connected to a large water reservoir ( 6101 −×=ν m2/s) as shown in Fig. 5. Pipe A has length L and pipe B has length 2L. Both pipes discharge to atmosphere. Which pipe will have the larger flow rate? Compute the flow rates, if H = l0 m, D = 50 mm and L = 61 m.

Fig. 5 From 1 to 2,

2222

22

22

2

2

1

2 VKVDLfgzVpgzVp

+=⎟⎟⎠

⎞⎜⎜⎝

⎛++−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

ρρ

2)5.01(

2

22

22 VV

DLfgH ++= … (1)

From 1 to 3,

222

22

23

23

3

2

1

2 VK

VDLfgzVpgzVp

+=⎟⎟⎠

⎞⎜⎜⎝

⎛++−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

ρρ

2)5.01(

22 2

323 VV

DLfgH ++= … (2)

Since LHS of (1) is the same as LHS of (2), 32 VV > and 32 QQ > .

Starting with fully turbulent flow where 0Re1≈ ,

66.22 =V m/s, 52 1033.1/Re ×== νDV

62.22 =V m/s, 32 1014.5 −×== AVQ m3/s

0263.027.0ln

325.12 =

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

De

f

0271.0

Re174.527.0ln

325.129.0=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛+

=

De

f

6.(a)Water at 65°C (density 980 kg/m3, kinematic viscosity 4.4×10 −7 m2/s) flows through a 75-mm diameter orifice installed in a 150-mn inner-diameter pipe. The flow rate is 20 L/s. Determine the pressure difference between the corner taps.

( ) pKAppKAm ttactual Δ=−= ρρ 22 21

⎟⎟⎠

⎞⎜⎜⎝

⎛=

11,ReDD

fK tD

511 1085.3Re ×==

νVD

D , m/s13.1421

===DQ

AQV

π. 5.0

1

==DDtβ

From Fig. 8.21, 624.0=K .

pKAQm tactual Δ== ρρ 2 , kPa8.25=Δp

(b) Consider a horizontal Ventuli with D1 = 50 mm and Dt = 25 mm. For a differential pressure of 150 kPa, calculate the flow rate of water (kinematic viscosity 1.01×10 −

6 m2/s).

( ) pCA

ppCA

m ttactual Δ

−=−

−= ρ

βρ

β2

12

1 4214

For 511 102Re ×>=

νVD

D , 995.098.0 << C .

Assume 99.0=C .

5.01

==DDtβ

/sm1069.821

33

4

−×=Δ−

== pCAm

Q tactual ρβρρ

Check the Reynolds number:

m/s43.4421

===DQ

AQV

π

51

1 1019.2Re ×==νVD

D

As 511 102Re ×>=

νVD

D , the assumption is valid.