# MEEG336 Fluid Mechanics II

### Spring 1998 DuPont 140; TR 11:00-12:15

INSTRUCTOR
Dr. Ajay K. Prasad; 228 Spencer Lab; Office Hours: TR 3:00-4:00 pm

TA
Mohan Kompella; 332 Spencer Lab; Office Hours: MW 1:00-3:00 pm

### Home-Work 1

Assigned: Tuesday 2/17/98
Due date: Tuesday 2/24/98
Solve the following eight problems from Chapter 6:

6.7, 6.8, 6.9, 6.21, 6.49, 6.53, 6.54, 6.61

Neglect minor losses in these problems (entrance, bends, etc.)

6.7: (a) 26 sec; (b) T = 66 deg C

6.8: (a) 64 cm; (b) 62 cm

6.9: (a) 217 Pa; (b) 163 Pa

6.21: For H=50cm, Q = 0.00667m^3/hr; Flow will go turbulent for H ~ 0.87 m

6.49: epsilon = 0.012 mm;

6.53: Use iterative method to converge to Q = 27 m^3/hr

6.54: Use iterative method to converge to D = 0.118 m

6.61: Q = 87 m^3/hr

What effect does the length of the pipe have in this problem?

### Home-Work 2

Assigned: Tuesday 2/24/98
Due date: Tuesday 3/3/98
Solve the following six problems from Chapter 7:

7.5, 7.8, 7.9, 7.16, 7.25, 7.33

7.5:

• From flat plate considerations: 6.55 ft
• From Entry length correlation (is it laminar or turbulent flow): 39 ft
• Why is the flat plate estimate too low? What is different about the two cases? (Hint: Think about the U_infinity.)
7.8:
• a: C_f = 0.577/Re_x**0.5
• b: theta/x = 0.577/Re_x**0.5
• c: delta_*/x = 1.732/Re_x**0.5
• d: H = 3.0
7.9:
• a: C_f = 0.655/Re_x**0.5
• b: theta/x = 0.655/Re_x**0.5
• c: delta_*/x = 1.743/Re_x**0.5
• d: H = 2.66
7.16: (a) laminar or turbulent? F = 181 N; (b) F = 256 N

7.25: You will need to use Table 7.1 for this problem: (a) L = 62 cm; (b) h_2 = 4.6 mm

### Home-Work 3

Assigned: Tuesday 3/3/98
Due date: Tuesday 3/10/98
Solve the following six problems from Chapter 7:

7.56, 7.63, 7.64, 7.80, 7.87, 7.89

7.56:

• (a) Laminar or turbulent? F = 10 N
• (b) F_normal = 80 N; F_tangential = 10 N (assume that the two forces may be superposed; not strictly true.)
7.63:
Use Bernoulli's to convert velocity to pressure. Integrate the pressure distribution from theta = 0 to pi/2 to get the total frontal pressure. Back pressure is given.

(This theoretical answer is too high--- check with Fig. 7.13 --- because the assumed p_rear is much too low.)

7.64:

Assume that the acceleration is very small from 2000 m to 1000 m (i.e. velocity changes very little). Use Table A-6 for air properties at different altitudes.

7.80: theta = 72 deg.

Use Fig. 7.16 to determine C_D. You may also consult Table 7.3.

First you have to determine Re_D. From Fig. 7.16 it appears that the boundary layer over a sphere goes *turbulent before separation* for Re > 3e5. For Re < 3e5, the boundary layer stays *laminar until separation*.

For the latter case (Re < 3e5) C_D is higher. Why? (Answer: separation is earlier, turbulent wake is larger, form drag is higher).

7.87: (a) 188 hp; 172 hp (b) 416 hp; 376 hp

7.89: 1e6 N-m.

### Home-Work 4

Assigned: Tuesday 3/10/98
Due date: Tuesday 3/17/98
Solve the following FIVE problems from Chapter 7:

7.72, 7.79, 7.101, 7.103, 7.112 (20 points)

7.72:

• (a) 6.3 m;
Use Fig. 7.16(b) to iterate to the correct C_D. i.e., guess C_D, determine settling velocity, then Re, etc.

• (b) 120 m; is this Stoke's flow regime?
7.79: (a) 1710 days; (b) 17 days 7.101:
You will need to solve a quadratic equation for V_berg.

• V_berg = U_wind ([(sqrt{alpha} -1)/(alpha -1)])

where alpha = (7 rho_w C_Dw)/(rho_a C_Da)

But alpha >>1 (It is Order(5000))

• So, V_berg = U_wind/sqrt(alpha)
7.103: theta_Pode = 44 deg.

(Take moments about A; tangential drag does not contribute!).

7.112

As the ball travels on its path, it slows down due to drag, and drops due to the action of lift and gravity.

The deceleration in the x-direction comes partially from Drag force and partially from the lift force.

The acceleration in the y-direction comes partially from drag, partially from lift, and gravity.

DRAW A CORRECT FREE BODY DIAGRAM TO SHOW THIS.

Write the differential equations for the horizontal and vertical deceleration/acceleration of the ball and solve numerically for the position of the ball in x-y space, using Runge-Kutta or similar solver.

• Answer: Ball will never reach end of table for top-spin. Even for large underspin (C_L = -0.2), ball will only travel about 2.25 m.

### Home-Work 5

Assigned: Tuesday 3/24/98
Due date: Tuesday 3/31/98
Solve the following six problems from Chapter 4:

47, 51, 60, 62, 66, 67

NOTE: Many of these problems require you to plot streamlines. Use a computer to do the plotting.

47: psi = y**2 - 3/2*x**2 + constant.

• Assume constant = 0, and plot for various values of psi (say -2, -1, 0, 1, 2).
• The streamlines represent flow near unequal corners (51 deg or 39 deg).

51: Plot streamlines in all but the bottom right quadrant.

You may first wish to convert from (r,theta) to (x,y). It represents inviscid flow around a 90 deg. bend.

60: Answer: Irrotational (is this counter-intuitive?)

Use Bernoulli... z_o = H - omega**2*R**2/(2g)

62: psi = V*y**2/2h + constant.

Is this velocity field irrotational?

66: psi = - Ksin(theta)/r

Plot the lines using a computer. The velocity field corresponds to a "doublet".

67: phi = C ln(r) + K*theta

Plot streanmlines using NEGATIVE values of C (i.e., a line sink). What familiar flow pattern does this represent?

### Home-Work 6

Assigned: Friday 4/3/98
Due date: Thursday 4/16/98 (Thursday after Spring Break)
Solve the following five problems from Chapter 8:

5, 25, 26, 47, 109

5:

psi = -B/2(x**2 + y**2) + constant; phi does not exist (why?)

what flow does this represent? Convert psi to polar coordinates to help visualize the flow.

25:
m = 1980 m^2/s ; beta = 55.6 deg.

p - p_infinity = -15700 Pa; or p = 85 kPa (abs).

26: V = 9.4 m/s; at angle theta = -46.8 deg.

47:

Note: theta is measured from the backward stagnation point, so the forward stag. point is located at theta = pi

U_infinity = 1./sin(105 deg)*sqrt((p_a-p_b)/2 rho)

This works well as a velocimeter, provided point 'a' coincides exactly with the forward stagnation point (alignment is critical).

109:
The stream function value at the lower wall is 0, and at the upper wall is 100 m^2/s (why 100?).

Set up the Fortran (or whatever) program to solve for the interior stream function values.

To find pressure distributions (C_p) along the lower and upper wall, determine the total velocity (= u**2+v**2) there, and use Bernoulli's to determine the pressure. Plot C_p along both walls. Discuss the plots.

### Home-Work 7

Assigned: Thursday 4/30/98
Due date: Thursday 5/07/98
Solve the following 7 problems from Chapter 9:

19, 21, 22, 30, 41, 43, 63

19: T_nose = 173 C; Ma = 3.42.

21:

T_2 = 328 K
a_2 = 0.895
T_o1 = 381 K
p_o1 = p_o2 = 135 kPa
V_1 = 126 m/s
M_1 = 0.33

22:
(a) 267 m/s
(b) 286 m/s

30: Write the equations for stagnation temp. and mass flux and solve simultaneously:
V = 262 m/s
M = 0.563
rho_o = 0.905 kg/m^3.

41: There is a high pressure subsonic entrance region and a low pressure supersonic entrance region. Both are bounded by a critical (sonic) throat.

You may need to write a small program to convert implicitly from area ratio to Mach number.

43: Use iteration (Table B-1).

M_1 = 0.729; M_2 = 1.32
m_dot = 0.671 kg/s.
A* = 23.3 cm^2

63:
• p_3 = 59200 Pa
• m_dot = 0.241 kg/s

### Home-Work 8

Assigned: Tuesday 5/12/98
Due date: Tuesday 5/19/98
Solve the following 6 problems from Chapter 9:

36, 70, 91, 92, 93, 102

36:

• Need to solve a first order o.d.e. and obtain an exponential solution for p_o(t)
• (a) 39 seconds; (b) 150 seconds

70: (a) 388 kPa; (b) 19 kPa

91: (a) M_1 = 0.30; (b) 1030 kPa

92: (a) M_2 0.45; (b) mdot = 2.04 kg/sec

93: mdot = 2.27 kg/s

102:

• mdot = 0.69 kg/sec (choked)
• Assume that the flow is choked, and take M_e = 1.0. Work backward, assuming an average viscosity of mu=2.2e-5 kg/m-s