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Crazy boy [7]
3 years ago
9

Vehicles arrive at a single toll booth beginning at 8:00 A.M. They arrive and depart according to a uniform deterministic distri

bution. However, the toll booth does not open until 8:10 A.M. The average arrival rate is 8 veh/min, and the average departure rate is 10 veh/min. Assuming D/D/1 queuing, when does the initial queue clear and what are the total delay, the average delay per vehicle, longest queue length (in vehicles), and the wait time of the 100th vehicle to arrive (assuming first-in-first-out)?

Engineering
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

Explanation:

answers attached bellow

You might be interested in
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter
slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg ,  1980.39 MJ/Kmol

Explanation:

molar mass of  ethanol (C2H5OH) = 46 g/mol

molar mass of   octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol =  0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value  ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg

HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0
2 years ago
8. The operation of a TXV is controlled by the
Katena32 [7]
Answer should be C hopefully
4 0
3 years ago
You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240V. If you use the lightbulb in the United Stat
givi [52]

Answer:

The bulb will be \frac{1}{4} times as bright as it is in Europe.

Explanation:

Data provided in the question:

Power of bulb in Europe = 75 W

Voltage provided in Europe = 240 V

Voltage provided in United Stated = 120 V

Now,

We know,

Power = Voltage² ÷ Resistance

Therefore,

75 = 240² ÷ Resistance

or

Resistance = 240² ÷ 75

or

Resistance = 768 Ω

Therefore,

Power in United States = Voltage² ÷ Resistance

= 120² ÷ 768

= 18.75 W

Therefore,

Ratio of powers

\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}

= \frac{1}{4}

Hence,

The bulb will be \frac{1}{4} times as bright as it is in Europe.

7 0
3 years ago
Consider a junction that connects three pipes A, B and C. What can we say about the mass flow rates in each pipe for steady flow
Elis [28]

Answer:

The statement regarding the mass rate of flow is mathematically represented as follows \Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}

Explanation:

A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure

As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.

The above statement can be mathematically written as

Mass_{Junction}=Constant\\\\\Rightarrow Mass_{in}=Mass_{out}

this is known as equation of conservation of mass / Equation of continuity.

Now we know that in a time 't' the volume that enter's the Junction 'O' is

1) From pipe 1 = V_{1}=Q_{1}\times t

1) From pipe 2 = V_{2}=Q_{2}\times t

Mass leaving the junction 'O' in the same time equals

From pipe 3 = V_{3}=Q_{3}\times t

From the basic relation of density, volume and mass we have

\rho =\frac{mass}{Volume}

Using the above relations in our basic equation of continuity we obtain

\rho \times V_{3}=\rho \times V_{1}+\rho \times V_{2}\\\\Q_{3}\times t=Q_{1}\times t+Q_{2}\times t\\\\\Rightarrow Q_{3}=Q_{1}+Q_{2}

Thus the mass flow rate equation becomes \Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}

6 0
3 years ago
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