Answer:
Assume Base free flow speed (BFFS) = 70 mph
Lane width = 10 ft
Reduction in speed corresponding to lane width, fLW = 6.6 mph
Lateral Clearance = 4 ft
Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph
Interchanges/Ramps = 9/ 6 miles = 1.5 /mile
Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph
No. of lanes = 3
Reduction in speed corresponding to number of lanes, fN = 3 mph
Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph
Peak Flow, V = 2000 veh/hr
Peak 15-min flow = 600 veh
Peak-hour factor = 2000/ (4*600) = 0.83
Trucks and Buses = 12 %
RVs = 6 %
Rolling Terrain
fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806
fP = 1.0
Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln
Vp < (3400 – 30 FFS)
S = FFS
S = 54.6 mph
Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln
Answer:
If i am correct It should be 1/4 of an inch
Explanation:
Sorry but i can't quite explain
Answer:
,
, ![\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdx%7D%20%3D%20-v_%7Bin%7D%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%20%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%20%5Ccdot%20%5Cleft%5B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D%20-1%20%5Cright%29%20%5Ccdot%20x%20%5Cright%5D%5E%7B-2%7D)
Explanation:
Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:
![\dot m_{in} - \dot m_{out} = 0](https://tex.z-dn.net/?f=%5Cdot%20m_%7Bin%7D%20-%20%5Cdot%20m_%7Bout%7D%20%3D%200)
![\dot m_{in} = \dot m_{out}](https://tex.z-dn.net/?f=%5Cdot%20m_%7Bin%7D%20%3D%20%5Cdot%20m_%7Bout%7D)
![\dot V_{in} = \dot V_{out}](https://tex.z-dn.net/?f=%5Cdot%20V_%7Bin%7D%20%3D%20%5Cdot%20V_%7Bout%7D)
![v_{in} \cdot A_{in} = v_{out}\cdot A_{out}](https://tex.z-dn.net/?f=v_%7Bin%7D%20%5Ccdot%20A_%7Bin%7D%20%3D%20v_%7Bout%7D%5Ccdot%20A_%7Bout%7D)
The following relation are found:
![\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bout%7D%7D%7Bv_%7Bin%7D%7D%20%3D%20%5Cfrac%7BA_%7Bin%7D%7D%7BA_%7Bout%7D%7D)
The new relationship is determined by means of linear interpolation:
![A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x](https://tex.z-dn.net/?f=A%20%28x%29%20%3D%20A_%7Bin%7D%20%2B%5Cfrac%7BA_%7Bout%7D-A_%7Bin%7D%7D%7BL%7D%5Ccdot%20x)
![\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x](https://tex.z-dn.net/?f=%5Cfrac%7BA%28x%29%7D%7BA_%7Bin%7D%7D%20%3D%201%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%20%5Cfrac%7BA_%7Bout%7D%7D%7BA_%7Bin%7D%7D-1%5Cright%29%5Ccdot%20x)
After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:
![\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bin%7D%7D%7Bv%28x%29%7D%20%3D%201%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%5Cright%29%20%5Ccdot%20x)
![v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}](https://tex.z-dn.net/?f=v%28x%29%20%3D%20%5Cfrac%7Bv_%7Bin%7D%7D%7B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%5Ccdot%20x%7D)
![v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}](https://tex.z-dn.net/?f=v%20%28x%29%20%3D%20v_%7Bin%7D%5Ccdot%20%5Cleft%5B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%5Ccdot%20x%20%5Cright%5D%5E%7B-1%7D)
The acceleration can be calculated by using the following derivative:
![a = v\cdot \frac{dv}{dx}](https://tex.z-dn.net/?f=a%20%3D%20v%5Ccdot%20%5Cfrac%7Bdv%7D%7Bdx%7D)
The derivative of the velocity in terms of position is:
![\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdx%7D%20%3D%20-v_%7Bin%7D%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%20%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D-1%20%20%5Cright%29%20%5Ccdot%20%5Cleft%5B1%20%2B%20%5Cleft%28%5Cfrac%7B1%7D%7BL%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7Bv_%7Bin%7D%7D%7Bv_%7Bout%7D%7D%20-1%20%5Cright%29%20%5Ccdot%20x%20%5Cright%5D%5E%7B-2%7D)
The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ
= W / A
we substitute
σ
= 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ
- σ
)A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb