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valentina_108 [34]
2 years ago
13

Technician A says that to depressurize high-pressure components of the electronic brake control (EBC) system, research the proce

dure for depressurizing the accumulator in the service information. Technician B says to remove the ABS fuse from the fuse box and apply the brake firmly at least 40 times when depressurizing the components of the EBC system. Who is correct
Engineering
1 answer:
Naddik [55]2 years ago
7 0

Answer:

Both Technician A and Technician B

Explanation:

Information regarding the depressurization of the of the electronic brake control (EBC) is contained in the service manual. Information regarding the depressurization of the accumulator should be researched and information based on the manufacturers information should be followed

Other steps required in depressurization are;

The ignition is placed in the off position

The ABS fuse should be removed from the fuse box

Firmly apply the brakes for at least 40 times

The brakes should be applied to verify that there is no power assist remaining. The brakes should be further pumped for an additional 10 times if there is still power assist.

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2 years ago
Thermosets burn upon heating. a)-True b)- false?
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Answer:

true

Explanation:

True, there are several types of polymers, thermoplastics, thermosets and elastomers.

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3 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0
3 years ago
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The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.

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The efficiency of the incandescent bulb will be:

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In this case, the LED bulb is more efficient than the incandescent bulb.

Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient

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4 0
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