Solutions of Mid Semester Examination · Mid Semester Examination Autumn Semester, 2013-14 CH-205:...
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 Dynamics2nd 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 + ρV 21
2= P2 + ρ
V 22
2(2)
On substituting Eq. (1) in Eq. (2), we get
(P1 − P2) = ρV 21
2
(D4
1
D42
− 1
)or V1 =
√2(P1 − P2)
ρ
(D4
1
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 Dynamics2nd 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
4D2L× wc =
π
4D2h× 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
4D2h =
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 Dynamics2nd 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,cVsub
For the circular cross-sections, the second moment of inertia,
Ixx,c =πD4
64⇒ BM =
Ixx,cVsub
=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 Dynamics2nd 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 cosαg − 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 (6)
P = P0 −ρ
2
[2U0bx+ b2(x2 + y2)
](7)
where U0 and P0 are the horizontal velocity and pressure, respectively, at x = 0. Note that this equation
ignores viscous effects along the walls but is a reasonable approximation throughout the majority of the
flow field. Calculate the material acceleration for fluid particles passing through this duct. Further, give
your answer in two ways: (i) as acceleration components ax and ay, and (ii) as acceleration vector #»a .
Also, obtain an expression for the rate of change of pressure following a fluid particle.
Instructors: RPB & SC Page 4 of 6 Sept 11, 2013 (10 a.m. - 11:30 a.m.)
Mid Semester Examination Autumn Semester, 2013-14
CH-205: Fluid Dynamics2nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering)
Solution # 4(A)
The velocity and pressure fields for steady, incompressible, two-dimensional flow through a converging duct
is given as,#»
V = (u, v) = (U0 + bx)#»i − by #»
j
P = P0 −ρ
2
[2U0bx+ b2(x2 + y2)
]The acceleration field components are obtained from its definition (the material acceleration) in Cartesian
coordinates,D
#»
V
Dt=∂
#»
V
∂t+
#»
V .∇ #»
V
The material acceleration components
ax =Du
Dt=����∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+����
w∂u
∂z= (U0 + bx)b+ (−by)0 = (U0 + bx)b
ay =Dv
Dt=����∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+����
w∂v
∂z= (U0 + bx)0 + (−by)(−b) = b2y
In terms of the vector notations, material acceleration vector
#»a = ax#»i + ay
#»j = b(U0 + bx)
#»i + b2y
#»j
By definition, the material derivative, when applied to pressure, produces the rate of change of pressure
following a fluid particle. Therefore, the rate of change of pressure following a fluid particle
DP
Dt=���7∂P
∂t+u
∂P
∂x+v
∂P
∂y+����
w∂P
∂z= (U0+bx)(−ρU0b−ρb2x)+(−by)(−ρb2y) = ρ
[−U2
0 b− 2U0b2x+ b3(y2 − x2)
]
Question # 4(B): [2+2]
Consider the following steady, two-dimensional velocity field
#»
V = (u, v) = [a2 − (b− cx)2] #»i + (2c2xy − 2cby)
#»j (8)
Is there a stagnation point in this flow field? If so, calculate the location of stagnation point.
Solution # 4(B)
The given velocity field,
#»
V = (u, v) = [a2 − (b− cx)2] #»i + (2c2xy − 2cby)
#»j
At a stagnation point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity
components u and v are given as
u = a2 − (b− cx)2 v = c2xy − 2cby
Instructors: RPB & SC Page 5 of 6 Sept 11, 2013 (10 a.m. - 11:30 a.m.)
Mid Semester Examination Autumn Semester, 2013-14
CH-205: Fluid Dynamics2nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering)
Setting these to zero and solving simultaneously for x and y yields the stagnation point
u = 0 = a2 − (b− cx)2 ⇒ x =b− ac
v = 0 = c2xy − 2cby ⇒ y = 0
So, yes there is a stagnation point; its location is x = (b - a)/c, y = 0.
If the flow were three-dimensional, then set w = 0 as well to determine the location of the stagnation point.
Some flow fields may have more than one stagnation point.
Instructors: RPB & SC Page 6 of 6 Sept 11, 2013 (10 a.m. - 11:30 a.m.)