# Solutions of Mid Semester Examination Mid Semester Examination Autumn Semester, 2013-14 CH-205:...

date post

08-May-2020Category

## Documents

view

5download

0

Embed Size (px)

### Transcript of Solutions of Mid Semester Examination Mid Semester Examination Autumn Semester, 2013-14 CH-205:...

Mid Semester Examination Autumn Semester, 2013-14

CH-205: Fluid Dynamics 2nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering)

Solutions of Mid Semester Examination

Data Given: Density of water, ρ = 1000 kg/m3, gravitational acceleration, g = 9.81 m/s2

Question # 1: [4+4+2]

Consider the flow of water through a clear tube (cross-section area A1). It is sometimes possible to observe

cavitation in the throat created by pinching off the tube to a very small diameter of cross-section area

A2. Let the average velocities and pressures at points 1 (inlet) and 2 (throat) are (V1, P1) and (V2,

P2), respectively. Both the maximum velocity and minimum pressure occur at throat. Consider that the

throat diameter is 1/20th of the inlet diameter. (a) estimate the minimum average inlet velocity at which

cavitation is likely to occur for water entering at pressure of 20.803 kPa and temperature of (a) 20oC

and (b) 50oC, respectively. Explain why the required velocity in part (b) is higher or lower than that

part (a). Under the incompressible flow with negligible gravitational effects and negligible irreversibilities

conditions, both (V A) and [P+ρ(V 2/2)] may be taken as constant along the flow. The saturation pressure

of water may be taken as 2.34 and 12.35 kPa at 20oC and 50oC, respectively.

Solution # 1

Given that the flow is incompressible with negligible gravitational effects and negligible irreversibilities, V A

and [P + ρ(V 2/2)] may be taken as constant along the flow. Therefore, applying the prescribed balances

in between inlet (point 1) and throat (point 2), we get

V1A1 = V2A2 or V2 = V1(A1/A2) or V2 = V1(D1/D2)2 (1)

and

P1 + ρ V 21 2

= P2 + ρ V 22 2

(2)

On substituting Eq. (1) in Eq. (2), we get

(P1 − P2) = ρ V 21 2

( D41 D42 − 1 )

or V1 =

√ 2(P1 − P2)

ρ

( D41 D42 − 1 )−1

(3)

Given P1 = 20.803 kPa, D1/D2 = 20, ρ = 1000 kg/m3

We understand that the the pressure (P ) anywhere in flow should not be allowed to drop below the vapor

pressure (Pv) at the given temperature to avoid the cavitation. For a pure substance, the vapor pressure

(Pv) equals to the saturation pressure (Psat). Therefore, the cavitation is likely to occur at the throat

when P2 ≤ Psat. (a) At 20oC, P2 = Psat = 2.34 kPa

After substituting the numerical values in Eq. (3), we get

V1 =

√ 2(20.803− 2.34)× 103

1000 (204 − 1)−1 = 0.015192 m/s (4)

(b) At 50oC, P2 = Psat = 12.35 kPa

After substituting the numerical values in Eq. (3), we get

V1 =

√ 2(20.803− 12.35)× 103

1000 (204 − 1)−1 = 0.010279 m/s (5)

Instructors: RPB & SC Page 1 of 6 Sept 11, 2013 (10 a.m. - 11:30 a.m.)

Mid Semester Examination Autumn Semester, 2013-14

CH-205: Fluid Dynamics 2nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering)

The results show that at 50oC, cavitation may occur at lower value of the inlet average velocity compared

to that at 20oC. It is simply because for the given decrease in the duct cross-sectional area, the velocity

increases and the pressure decreases according to given flow continuity and mechanical energy balances.

We also know that the vapor pressure increases with increasing value of temperature. Therefore, the pres-

sure difference (P1 − P2) to be maintained to avoid the cavitation decreases with increasing temperature, In first case, the pressure (P1 − P2) is very large and, thus, the corresponding inflow velocity is larger compared to the second case.

Question # 2: [2+2+1+1+3+1]

A metal cylinder (diameter D = 3 m and height L = 3 m) of specific weight of 5886 N/m3 is required

to float in water with its axis vertical. Determine the (i) depth of immersion h of cylinder in water, (ii)

distance of center of buoyancy B and center of gravity G from the bottom point of cylinder O (iii) distance

of center of gravity G from the center of buoyancy B (iv) volume of the cylinder submerged in the water

Vsub, (v) metacentric height (GM). Based on the above results, state whether the cylinder is in stable

equilibrium.

Solution # 2

Given, the length of cylinder, L = 3 m, diameter of the cylinder D = 3m, specific weight of cylinder

wc = 5886 N/m3 Consider that h height of the cylinder is immersed in the liquid. Taking the force bal-

ance, Weight of cylinder = weight of water displaced

π

4 D2L× wc =

π

4 D2h× w

where w is the specific weight of water.

(i) the depth of immersion (h) of cylinder in water

h = L× wc w

= 3× 5886 9810

= 1.8 m

(ii) Distance of center of buoyancy (B) from the bottom center (O)

OB = h

2 = 0.9 m

Distance of center of gravity (G) from the bottom center (O)

OG = L

2 = 1.5 m

(iii) Distance of center of gravity (G) from center of buoyancy (B)

BG = OG−OB = 0.6 m

(iv) volume of cylinder submerged in water

Vsub = π

4 D2h =

π

4 × 32 × 1.8 = 12.723 m2

Instructors: RPB & SC Page 2 of 6 Sept 11, 2013 (10 a.m. - 11:30 a.m.)

Mid Semester Examination Autumn Semester, 2013-14

CH-205: Fluid Dynamics 2nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering)

(v) metacentric height, GM is a measure of stability for the floating bodies. It is the distance between the

center of gravity (G) and the metacenter (M, the intersection point of the lines of action of buoyancy force

through the body before and after rotations).

GM = BM −BG

where BM is metacentric radius.

The second moment of inertia (Ixx,c) would be related to the buoyant force for the the small angle of

rotations due to displacement as follow

w × Ixx,c = BM × FB ⇒ w × Ixx,c = BM × w × Vsub ⇒ BM = Ixx,c Vsub

For the circular cross-sections, the second moment of inertia,

Ixx,c = πD4

64 ⇒ BM = Ixx,c

Vsub =

D2

16h = 0.3125 m

GM = 0.3125− 0.6 = −0.2875 m

The negative value of the GM indicates that the metacentre is below the center of gravity. Thus, the

cylinder is in unstable equilibrium.

Question # 3(A): [3]

Consider two identical glasses of water, one stationary and other moving on a horizontal plane with con-

stant acceleration. Assuming no splashing or spilling occurs, give your to-the-point explanation that which

glass will have a higher pressure at the (i) front (ii) midpoint and (iii) back of the bottom surface?

Solution # 3(A)

We know that the pressure in all cases is the hydrostatic pressure, which is directly proportional to the

fluid height. The pressure at the bottom surface is constant when the glass is stationary. For a glass

moving on a horizontal plane with constant acceleration, water will collect at the back but the water depth

will remain constant at the center. Therefore, the pressure at the midpoint will be the same for both

glasses. But the bottom pressure will be low at the front relative to the stationary glass, and high at the

back (again relative to the stationary glass).

Question # 3(B): [3+2+2]

A water tank is being towed on an uphill road (inclined by 20o with the horizontal) with constant accel-

eration of 5 m/s2 in the direction of motion. Determine the angle the free surface of water makes with

horizontal. What would your answer be if the direction of motion were downward on the same road with

the same acceleration? If the water is replaced by oil (specific gravity 0.8) then what would be the new

values of angle in both the cases?

Solution # 3(B)

The effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be negligible and

the acceleration remains constant.

Instructors: RPB & SC Page 3 of 6 Sept 11, 2013 (10 a.m. - 11:30 a.m.)

Mid Semester Examination Autumn Semester, 2013-14

CH-205: Fluid Dynamics 2nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering)

From geometrical considerations, the horizontal and vertical components of acceleration are

ax = a cosα az = a sinα

The tangent of the angle the free surface makes with the horizontal is

tan θ = ax

g + az =

a cosα

g + a sinα =

5 cos 20o

9.81 + 5 sin 20o = 0.4078 ⇒ θ = 22.20o

When the direction of motion is reversed, both ax and az are in negative x- and z-directions, respectively,

and thus become negative quantities,

ax = −a cosα az = −a sinα

Then the tangent of the angle the free surface makes with the horizontal becomes

tan θ = ax

g + az = −a cosα g − a sinα

= −5 cos 20o

9.81− 5 sin 20o = −0.5801 ⇒ θ = −30.1o

The analysis is valid for any fluid with constant density, not just water. Since we used no information that

pertains to water in the solution and so there will not be any change in the results after changing the fluid

from water to oil.

Question # 4(A): [2+2+2]

Consider steady, incompressible, two-dimensional flow through a converging duct. A simple approximate

velocity and pressure field for this flow are given as

#»

V = (u, v) = (U0 + bx) #» i − by #»j

Recommended

*View more*