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sukhopar [10]
3 years ago
7

Technician A says that a seal can be pried out of a bore using a sharp chisel. Technician B says that smaller metal-backed seals

can often be driven into place using a standard socket. Who is correct? Group of answer choicesA. A onlyB. B onlyC. BothD. Neither
Engineering
2 answers:
attashe74 [19]3 years ago
6 0

Answer:

The correct answer is letter "C": Both.

Explanation:

Industrial seals are used at interfaces between components to prevent leakage, to maintain heat, and to avoid contamination. The design, construction, and materials they use vary depending on industrial use but the most common are Polytetrafluoroethylene (PTFE), Nitrile Buna Rubber (NBR), and fluorocarbon.

Thus, using a sharp chisel could pry a seal out of a hole and a regular socket can often be used to force smaller metal-backed seals into place. Thus, technicians "A" and "B" are correct.

FrozenT [24]3 years ago
4 0

Answer:

the correct answer is c boyh

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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
sdas [7]

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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Explanation:

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