Which of the following applies to a module?

What are the relevance of report writing

nasty-shy [4]

Answer:Report writing consists of the history and facts of a project or of any kind of event. It is useful to record past history and an overall summary of decisions. Report writing helps to solve problems as a path. Writing a report will guide you in a way that will modernize details of the improvements and upcoming plans.

6 0

2 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago

4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

hodyreva [135]

Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out

6 0

3 years ago

Find the current Lx in the figure

AleksandrR [38]

Explanation:

$\frac{1}{8} + \frac{1}{2} \\ 1.6 + 1.4 = 3 \\ \frac{1}{3} + \frac{1}{9} \\ 2.25 + 2 = 4.25 \: ohm$

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

$\frac{9}{9 + 3} \times 4 = 3\\ \frac{2}{2 + 8} \times 3 = 0.6a \\ = 0.6 \: milli \: amper$

6 0

3 years ago

Which of the following is the LEAST-Likely cause of tire wear? A) Underinflation B) Braking C) Acceleration D) Tire rotation

mihalych1998 [28]

Answer:

Tire rotation is the least likely cause of tire wear. So, the option D is correct.

Explanation:

Step1

Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.

Step2

On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.

Step3

Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.

Step4

Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.

Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.

4 0

3 years ago