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ohaa [14]
2 years ago
9

"Vehicles begin to arrive at a parking lot at 6:00 A.M. at a rate of eight per minute. Due to an accident on the access highway,

no vehicles arrive from 6:20 to 6:30 A.M. From 6:30 A.M. on, vehicles arrive at a rate of two per minute. The parking-lot attendant processes incoming vehicles (collects parking fees) at a rate of four per minute throughout the day. Assuming D/D/1 queuing, determine total vehicle delay."

Engineering
1 answer:
marin [14]2 years ago
8 0

Answer:

The solution and explanation is attached.

Explanation:

You might be interested in
Toeboards are usually ___ inches high and used on landings and balconies.
Ksju [112]

Answer:4 inches

Explanation:

Google told me

6 0
2 years ago
Chapter 19: Diesel Engine Operation and Diagnosis -Chapter Quiz
Llana [10]

Answer: See explanation

Explanation:

1. How is diesel fuel ignited in a warm diesel engine?

B. Heat compression

2. Which type of diesel injection produces less noise?

A. Indirect injection (IDI)

3. Which diesel injection system requires the use of a glow plug?

A. Indirect injection (IDI)

4. The three phases of diesel ignition include:

C. Ignition delay, repaid combustion, controlled combustion.

5. What fuel system component is used in a vehicle equipped with a diesel engine that is seldom used on the same vehicle when it is equipped with a gasoline engine?

D. Water-fuel separator

6. The diesel injection pump is usually driven by a _________________.

A. Gear off the camshaft

7. Which diesel system supplies high-pressure diesel fuel to all the injectors all of the time?

C. High-pressure common rail

8. Glow plugs should have high resistance when _____________and lower resistance when __________________.

B. Warm/cold

9. Technician A says that glow plugs are used to help start a diesel engine and are shut off as soon as the engine starts. Technician B says that the glow plugs are turned off as soon as a flame is detected in the combustion chamber. Which Technician is correct?

D. Neither Technicians A NOR B

10. What part should be removed to test cylinder compression on a diesel engine?

D. A glow plug

6 0
2 years ago
Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer
Andrei [34K]

Answer:

(a) Work out put=692.83\frac{KJ}{Kg}

(b) Change in specific entropy=0.0044\frac{KJ}{Kg-K}

Explanation:

Properties of steam at 1 MPa and 200°C

        h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

      h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}

 s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}

h_2=2134.57\frac{KJ}{Kg}

s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}

s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}

s_2=6.6944\frac{KJ}{Kg-K}

(a)

Work out put =h_1-h_2

                      =2827.4-2134.57 \frac{KJ}{Kg}

Work out put =692.83 \frac{KJ}{Kg}

(b) Change in specific entropy

     s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}

Change in specific entropy =0.0044\frac{KJ}{Kg-K}

3 0
2 years ago
Write multiple if statements: If carYear is before 1967, print "Probably has few safety features." (without quotes). If after 19
Free_Kalibri [48]

Answer:

The solution code is written in Python 3.

  1. carYear = 1995
  2. if(carYear < 1967):
  3.    print("Probably has few safety features.\n")
  4. if(carYear > 1970):
  5.    print("Probably has head rests. \n")
  6. if(carYear > 1991):
  7.    print("Probably has electronic stability control.\n")
  8. if(carYear > 2002):
  9.    print("Probably has airbags. \n")

Explanation:

Firstly, create a variable, <em>carYear</em> to hold the value of year of the car make. (Line 1)

Next, create multiple if statements as required by the question (Line 3-13). The operator "<" denotes "smaller" and therefore <em>carYear < 1967</em> means any year before 1967. On another hand, the operator ">" denotes "bigger" and therefore <em>carYear > 1970 </em>means any year after 1970.

The print statement in each of the if statements is done using the Python built-in function <em>print()</em>. The "\n" is an escape sequence that create a new line at the end of each printed phrase.

5 0
2 years ago
A three-phase line has an impedance of 0.4 j2.7 ohms per phase. The line feeds two balanced three-phase loads that are connected
Viktor [21]

Answer:

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

Complex power of second load:

S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload)   ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

Ploss = 12000 W

Ploss = 12 kW

The 3-phase reactive power loss is given by

Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar

7 0
2 years ago
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