All categories

3 years ago

10.) A solid circular titanium control rod at 7,000 lb axial tension force where stress must not

Engineering

1 answer:

lina2011 [118]3 years ago

6 0

Answer: exceed 42,000 psi. Assume that the modulus of elasticity of the titanium is 16,500,000 psi

Explanation: sorry if im wrong and have a wonderful day!!!!

You might be interested in

A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The

attashe74 [19]

Answer:

$q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C$

Explanation:

$R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W$

So, heat rate can be calculated as follows:

$q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW$

Surface temperatures can be calculated as follows:

$T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C$

6 0

4 years ago

Which of the following approaches to lean engineering consists of teams of people from different departments sharing

AURORKA [14]

Answer:

PDSA, where you work as a group

Explanation:

8 0

3 years ago

A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h

sertanlavr [38]

Answer:

Explanation:

Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)

fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

8 0

3 years ago

A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-

777dan777 [17]

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

<u>Determine the design flow ins ft^3/s </u>

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

∴ T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

= 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276

hence ; Qt = 10^3.23276

= 1709.07 ft^3/s

4 0

3 years ago

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t

iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, $P_{1}$ = 1 bar

$v_{2}$ = 4 bar

pV¹°¹ = constant

From superheated propane table, at $P_{1}$ = 1 bar and $T_{1}$ = 27⁰C

$v_{1}$ = 0.557 m³/kg

$v_{2}$ = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

$V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1 \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557 \\\\V2 = 0.158 m^3/kg$

At = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

$W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\ \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\ \\W = -19.75KJ$

(b) Heat transfer = ?

$Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ$

8 0

4 years ago