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Bess [88]
2 years ago
14

A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turne

d out to be 0.78 mile to the overpass from the point at which she started her vehicle count. The vehicles were moving at approximately 52 mph at the time. Traffic was distributed evenly over the three EB lanes.
Required:
a. Calculate the vehicle density per lane (excluding the driver's rental vehicle) on that stretch of EB I-80.
b. What was the flow rate for that section of I-80?
Engineering
1 answer:
SVEN [57.7K]2 years ago
6 0

Answer:

vehicle density =  28.205 veh/mile

flow rate =  0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance    ...............1

vehicle density = ( 21 + 1 )  ÷ 0.78

vehicle density  k =  28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs    .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate =  0.909 veh/hr

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Answer:

(d) Spheroidizing

Explanation:

Spheroidizing

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In this process a hard and brittle materials convert into soft and ductile after this it improve the machine ability as well as improve the tool life.

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3 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
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Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

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3 years ago
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<u></u>

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