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3 years ago

What do auto mechanics do when they take out the spark plugs

1 answer:

7 0

Check them to see if the cylinder is running rich or lean. This can be determined by looking at the electrodes on the plugs
Hope this helps.

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The question belongs to Electrical Engineering (Linear System).

-Dominant- [34]

I’m crying looking at that.

5 0

3 years ago

Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press

Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0

3 years ago

An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied

lyudmila [28]

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

4 0

3 years ago

The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r

Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

$K = \sigma Y \sqrt{\pi a}$

where;

fracture toughness K = 137 MPa $m^{1/2}$

geometry factor Y = 1

applied stress $\sigma$ = ???

crack length a = 2mm = 0.002

∴

$137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }$

$137 =\sigma \times 0.07926$

$\dfrac{137}{0.07926} =\sigma$

$\sigma = 1728.489 MPa$

Now, the tensile impact obtained is:

$\sigma = \dfrac{P}{A}$

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0

3 years ago

Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3t

IrinaK [193]

Answer:

V=L(di/dt) where i is current, V=0.208

Explanation:

using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)

V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208

3 0

3 years ago