What is the first component to install in steel stud framing system?

What should i do to my nissan 350z, twin charge it or have twin turbos, to twin turbos with a super charger? i want it to be ove

bulgar [2K]

Answer:

A

Explanation:

It just makes sense

4 0

3 years ago

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv

Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$ in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$\phi$ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

$= 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

$= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

= 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

$=\frac{1424500}{56.67 \pi}$

= 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

$=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

= 2.93 ksi

B). Angle of twist :

$\phi = \frac{TL}{JG}$

$=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

= 0.08002 rad

= 4.58°

6 0

3 years ago

The practice of honoring those who have been lost in battle dates back to which two ancient

Serjik [45]

It was between Iranians and filapinons nice work man it’s easy

3 0

3 years ago

Read 2 more answers

Why some types of aggregate are susceptible to damage from repeated freezing and thawing? Explain.

koban [17]

Answer:

Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.

When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.

8 0

4 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$