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zzz [600]
3 years ago
13

2. What is the Function of the Camshaft in an Internal Combustion Engine?

Engineering
1 answer:
mamaluj [8]3 years ago
8 0

Answer:

camshaft, in internal-combustion engines, rotating shaft with attached disks of irregular shape (the cams), which actuate the intake and exhaust valves of the cylinders.

Explanation:

I'm taking an engineering/tech class. I hope this helps! :)

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The component of a regenerative vapor power cycle that permits only liquid to pass through to a region of lower pressure is a?
Zepler [3.9K]

The component of a regenerative vapor power cycle that permits only liquid to pass through to a region of lower pressure is a Valve/trap.

<h3>What is vapor?</h3>
  • In physics, a substance in the gas phase at a temperature lower than its critical temperature is referred to as a vapor or vapor.
  • This means that the vapor can be condensed to a liquid by increasing the pressure on it without decreasing the temperature.
  • An aerosol is distinct from a vapor.
  • A suspension of minute liquid, solid, or both types of particles in a gas is known as an aerosol.
<h3>Why does vapor form?</h3>
  • Evaporation or sublimation are two processes that can be used to create it.
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Learn more about vapor here:

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3 0
2 years ago
A 3 m aluminum pole is kept at a residential site for construction
Aliun [14]

Answer:

I don't know sorry

Explanation:

5 0
3 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

4 0
3 years ago
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