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sergey [27]
3 years ago
13

While discussing PCV valve operation: Technician A says that the PCV valve opening is decreased at part-throttle operation compa

red to idle operation. Technician B says that the PCV valve opening is decreased at wide-open throttle compared to part throttle. Who is correct?
Engineering
1 answer:
Levart [38]3 years ago
5 0

Answer:

Both are incorrect.

Explanation:

PCV valve opening is dependent on amount of manifold vacuum value. The opening cannot decrease due to part throttle operation. The PCV system is a tampered valve whose opening depend upon intake manifold vacuum. PCV valve is supported by a spring and is initially in a closed position.

You might be interested in
Why do electricians require critical thinking skills? In order to logically identify alternative solutions to problems in order
Rina8888 [55]

Answer:

In order to logically identify alternative solutions to problems

Explanation:

Electricians are specialized in electrical wiring of buildings, transmission lines, stationary machines, and related equipment. They are either employed in the installations of new electrical components, or to maintain an already installed component. The job of an electrician can be mentally tasking, especially in troubleshooting for fault, and methods of fixing of faults. Some problems might require an out-of-norm approach to solve, and the electrician has to be able to logically identify alternative solutions to problems.

8 0
3 years ago
In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)
docker41 [41]

Answer:

(b) Constant (minimum) volume

Explanation:

In the idealized Otto cycle there are 4 process that are

  • Reversible adiabatic compression  
  • Addition of heat at constant volume
  • Reversible adiabatic expansion
  • Rejection of constant volume

So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

3 0
3 years ago
The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published
Iteru [2.4K]

Answer:

a

The rate of radiation of the energy is  E_r = 1.523747635*10^9 W/m^2

b

The irradiation is  G =46.177\ kW/m^2

c

The amount of energy absorbed is E_B = 461.772 KJ

d

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

  From the question we are told that

        The  temperature is  T =  7200K

        The diameter of the ball is  d = 1.5 km = 1.5 *1000 = 1500m

       Hence the radius  == \frac{1500}{2} = 750m

 The total energy radiated can be mathematically represented as

                         E = \sigma A T^4

Where \sigma is the Stefan-Boltzmann constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}

            A is the area of a sphere  = \pi d^2  = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2

 Substituting values we have

                    E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4

                        =1.077*10^{15} W

Now the state of the energy is mathematically represented as

                           Rate  \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4

                                                            = 5.67*10^{-8} * 7200^2

                                                            = 1.523747635*10^9 W/m^2

A sketch illustrating the b part of the question is shown on the first uploaded image

     looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

      from the diagram x can be calculated as follows

                      x = \sqrt{40^2 + 16^2}

                        = 43.0813 Km

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

                              G = \frac{E}{4 \pi r^2}

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

                            G= \frac{1.077*10^15}{4 \pi (431081.3^2)}

                                G =46.177\ kW/m^2

Generally the amount of energy absorbed can be mathematically represented as

                            Amount \ of \ energy \ absorbed \ (E_B ) = G * t

Where t is the time taken

       Therefore     E_B = 46.177 *10 = 461.77 KJ

       

                         

                       

             

   

6 0
3 years ago
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as
Llana [10]

Answer:

Explanation:

a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2

We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX

If AB = 00 select I_o

If AB = 01 select I_1

If AB = 1_(B is don't care), select I_2

However, the truth table is attached and shown in the first file below.

Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.

b) We are show how  two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.

The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.

Where ; I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7 are the inputs of the multiplexer and Z is the output.

c)  Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.

For  four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.

Where ; I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7 are the inputs of the multiplexer and Z is the output

5 0
4 years ago
A reversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP?
jonny [76]

Answer:

Both heat pump has same COP

Explanation:

Given that a reversible heat pump and a Carnot heat pump which is operating between the same temperature limits will have the same COP because the temperature limits are same, that is the source temperature and the sink temperature. Also because we know that a Carnot cycle is a totally reversible cycle. Therefore for a perfectly reversible Carnot cycle, the COP will be same for a reversible heat pump and a Carnot heat pump operating between two same thermal reservoirs.

Thus, both heat pump will have same COP.

4 0
4 years ago
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