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3 years ago

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A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a

stress state greater than 200 N/mm2. If the linkage is to be constructed from solid round stock, what is the minimum required diameter?

1 answer:

Fed [463]3 years ago

4 0

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = $2*10^{8} N/m^{2}$

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/( $2*10^{8}$ ) = $10^{-5}$ m^2

recall that area = $\pi d^{2} /4$

$10^{-5}$ = $\frac{3.142*d^{2} }{4}$ = $0.7855d^{2}$

$d^{2} = \frac{10^{-5} }{0.7855}$ = $1.273*10^{-5}$

$d = \sqrt{1.273*10^{-5} }$ = $3.57*10^{-3}$ m = 3.57 mm

<em>maximum diameter of the steel linkage d = 3.57 mm</em>

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Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload

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3 years ago

A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh

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Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

$T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C$

to kelvin = (5 + 273)K = 278 K

From the " thermophysical properties of gases at atmospheric pressure" table; At $T_f$ = 278 K ; by interpolation; we have the following

$\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}$ → v 13.93 (10⁻⁶) m²/s

$\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})}$ → k = 0.0245 W/m.K

$\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})}$ → ∝ = 19.6(10⁻⁶)m²/s

$\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720}$ → Pr = 0.713

$\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}$

The Rayleigh number for vertical cavity

$Ra_L = \frac{g \beta (T_1-T_2)L^3}{\alpha v}$

= $\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}$

= $8.38*10^5$

$\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24$

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

$Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}$

$NU_L= \frac{hL}{k}= 4.878$

$\frac{hL}{k}= 4.878$

$\frac{h*0.06}{0.0245}= 4.878$

$h = \frac{4.878*0.0245}{0.06}$

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

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4 years ago