CE111L: Air and Water Quality Laboratory
Combustion Stoichiometry and CMFR Dynamics
Objective and activity in brief:
This laboratory demonstrates fundamentals of combustion stoichiometry and dynamics of a continuously
mixed flow reactor (CMFR). Participan
DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING
UNIVERSITY OF CALIFORNIA, BERKELEY
CE 111, FALL 2016
HOMEWORK ONE
DUE FRI, SEPTEMBER 2
1. San Pablo Reservoir. The San Pablo Reservoir is located about 5 km east of campus,
near the Town of El Sobrante. Wh
1. Calculate the basic values In the design of a road, a spiral is
interposed between a straight-line segment and a circular curve to
effect a gradual transition from rectilinear to circular motion, and vice
versa. The type of spiral most frequently used
main tangent, with transit at TS. Although extensive tables of spiral
values have been compiled, this example is solved without recourse to
these tables in order to illuminate the relatively simple mathematical
relationships that inhere in the clothoid. C
Method 2: The prismoidal method postulates that the earthwork
between the stations is a prismoid (a polyhedron having its vertices in
two parallel planes). The volume of a prismoid is L(A1 + 4Am + A2)
y=_l 27 (15) 6 FIGURE 10 where Am = area of center sec
= 0.000 Ix2 0.046*. Record the calculations for y in tabular form.
The results, as shown, agree with those obtained by the averagegrade
method. Tangent-Offset Method 0.0001 Jc2 , ft Station x, ft (m) (m)
0.046*, ft (m) y, ft (m) 52 + 60 O (O) O (O) O (O)
kN-m). The concrete section is therefore excessive, but a 6-in (152.4mm) slab would be inadequate. The steel is stressed to capacity at
design load. Or, As = 40,500/(20,0OO x 0.875 x 4.37) = 0.53 in2 (3.4
cm2 ). Use No. 6 bars 10 in (254 mm) on centers, t
cm2 ); / - 9012 in4 (37.511 dm4 ); d = 35.84 in (910.336 mm); S = 503
in3 (8244.2 cm3 ); flange thickness = 1 in (25.4 mm), approximately.
2. Compute the section moduli of the noncomposite section where the
cover plate is present To do this, compute the s
932(15,275.5) Composite, moving loads 9,840 (1,111.7)
1,179(19,323.8) 2,936(48,121.0) 1,826(29,928.1) 8. Compute the
critical stresses in the member To simplify the calculations, consider
the sections of maximum live-load and dead-load stresses to be
coin
of the earth causes a star to appear to describe a circle on the celestial
sphere centered at the celestial axis. The star is said to be at
culmination or transit when it appears to cross the observer's
meridian. In Fig. 12, P and M represent the position
curvature at given point on spiral; Ts , = length of main tangent from
TS to PI; E3 = external distance, i.e., distance from PI to midpoint of
circular curve. In addition, there is a long tangent (LT), short tangent
(ST), and long chord (LC), as indicated
curved highway route AB. Distances on the route are measured along
the arc. Applying the recorded data, determine the station of the
intersection point P. Calculation Procedure: 1. Apply trigonometric
relationships to determine three elements in triangle
will be taken as 2.5 times the bar diameter. Then, d = 5.75 + 33.5 - 2 0.5 - 1.375(0.5 + 2.5) = 32.62 in (828.548 mm); (b) Loading for
maximum shear FIGURE 38 (a) Loading for maximum moment v =
Vlb'jd = 66,2807(14 x 0.875 x 32.62) = 166 < 225 lb/in2 (1144
civil time of upper culmination at site 3* 19W115 Correction to
standard meridian 24"1OO5 EST of upper culmination at site 3*43m
1 Pa.m. PLOTTING A CIRCULAR CURVE A horizontal circular
curve having an intersection angle of 28 is to have a radius of 1200 f
11. By the law of sines, A'A = 1450.7 sin 1 l/sin 117 = 310.7 ft
(94.70 m). This is acceptable. CHARACTERISTICS OFA
COMPOUND CURVE The tangents to a horizontal curve intersect at
an angle of 6822'. To fit the curve to the terrain, it is necessary to use
a
Analyze the specific speed and suction specific speed at each of the
various operating speeds, using the data in Tables 1 and 2. These
tables show that at 870 and 1160 r/min, the suction specific-speed
rating is poor. At 1750 r/min, the suction specific-s
member. Thus, the steel alone supports the concrete slab, and the steel
and concrete jointly support the wearing surface. Plastic flow of the
concrete under sustained load generates a transfer of compressive
stress from the concrete to the steel. Conseque
H, ft(m) ^,ft(m) Elevation, ft (m) 1 544.8(166.06) 25.4 (7.74)
505.6(154.11) 2 622.0(189.59) 34.8 (10.61) 520.0(158.50) 3
482.5(147.07) -15.7 (-4.79) 468.5(142.80) VOLUME OF
EARTHWORK Figure 10 and b represent two highway cross sections
100 ft (30.5 m) ap
sphere is of infinite radius and has the earth as its center. The celestial
equator, or equinoctial, is the great circle along which the earth's
equatorial plane intersects the celestial sphere. The celestial axis is the
prolongation of the earth's axis o
readings were taken with the instrument at a station of elevation 483.2
ft (147.28 m), the height of instrument being 5 ft (1.5 m). The stadia
interval factor is 100, and the value of C is negligible. Compute the
horizontal distances and elevations. Point
percent; Gb = +1.8 percent; FIGURE 19. Parabolic arc. r = rate of
change in grade = 0.02 percent per foot; L = (Gb - G0)Ir = [1.8 - (4.6)]/0.02 = 320 ft (97.5 m). 2. Locate the PC and PT The station of
the PC = station of the PI - 'L/2 = (54 + 20) - (1 +
and submerged, free, and varying discharge head. Prepare similar
sketches for the same pump with static suction head. Label the various
heads. Compute the total head on each pump if the elevations are as
shown in Fig. 4 and the pump discharges a maximum o
the spiral into an integral number of arcs, these equations may be
converted to these more suitable forms: 8h = (2np + nb)(np-np) -J^
(38a) 8r (2np + nf)(nf -np) -^ (396) where n denotes the number of
arcs to the designated point. Tangent through P Main t
From the geometric relationships, T = R tan 1M (19) T = 1200(0.2493)
= 299.2 ft (91.20 m) Also C = 21 ISm1M (20) C = 2(1200)(0.2419) =
580.6 ft (176.97 m) FIGURE 13. Circular curve. And, M = ^(ICOs1X2A) (21) M = 1200(1 - 0.9703) = 35.6 ft (10.85 m) Lastly
= arcsin 0.04167; 1X2Z) = 223.3'; Z) = 446.6'. 4. Determine the
station at the PT Number of stations on the curve = 28/446.6' =
5.862; station of PT = (82 + 30) + (5+ 86.2)-88+ 16.2. 5. Calculate the
deflection angle of station 83 and the difference betwe
0.183262 /42) = 21.33 ft (6.501 m); k =348.83 -954.93 sin 1030' =
174.80 ft (53.279m); p = 21.33 - 954.93(1 - cos 1030') = 5.34 ft (1.628
m). 2. Locate the TS and SC Thus, T5 = (954.93 + 5.34) tan 2624' +
174.80 = 651.47; station of TS = (34 + 93.81) -(6
is necessary to consider the possibility that several trucks will be
present simultaneously. To approximate this condition, the AASHTO
Specification offers various lane loadings, and it requires that the
bridge be designed for the lane loading if this yie