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kow [346]
1 year ago
11

There are signs of oil spray on the compressor clutch hub and nearby underhood areas. Technician A says that a faulty compressor

clutch bearing could be the cause. Technician B says that a leaking compressor shaft seal could be the cause. Who is correct
Engineering
1 answer:
Nina [5.8K]1 year ago
8 0

The statements regarding the compressor as expressed by the technicians are correct. Hence, Both the technicians are correct.

<h3>What is a compressor?</h3>

A compressor is a mechanical tool that will increase the pressure of a gas with the aid of using decreasing its volume. An air compressor is a selected form of gas compressor.

Hence, The statements regarding the compressor as expressed by the technicians are correct. Hence, Both the technicians are correct.

learn more about compressors:

brainly.com/question/10174214

#SPJ1

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What are the main microsoft ware packages widely used today​
RSB [31]

Answer:

» Microsoft word ( word processing )

» Microsoft powerpoint ( presentation )

» Microsoft access ( database mamagement )

» Microsoft excel ( spread sheets )

Explanation:

.

7 0
2 years ago
Read 2 more answers
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
2 years ago
What is a motor cycle motor made out of
Natali5045456 [20]

Explanation:

a motorcycle Motor is made out of iron

4 0
3 years ago
Read 2 more answers
If your accelerator pedal gets stuck, what is the first thing you should do?
Anna35 [415]

If your accelerator gets stuck down, do the following: Shift to neutral. Apply the brakes. Keep your eyes on the road and look for a way out.If your accelerator gets stuck down, do the following:

Shift to neutral.

Apply the brakes.

Keep your eyes on the road and look for a way out.

Warn other drivers by blinking and flashing your hazard lights.

Try to drive the car safely off the road.

Turn off the ignition when you no longer need to change direction.

8 0
2 years ago
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
2 years ago
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