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2 years ago

Visual aids are useful for all of the following reasons except

Engineering

1 answer:

Ksivusya [100]2 years ago

4 0

Answer:

Explanation:

their presence allows the speaker to take a rest from talking

I couldnt see the question

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Question 1 of 8.

Rasek [7]

The answer seems pretty obvious, all of the above

8 0

2 years ago

Read 2 more answers

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

3 0

3 years ago

In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W

GrogVix [38]

Answer:

The shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

Explanation:

Given:

Temperature $T = 1273.15$ K

Initial Pressure $P_{1} = 1.8$ MPa

Final pressure $P_{2} = 0.1$ MPa

From the table superheated,

$h_{i} = 4635$ $\frac{K J}{Kg}$ and $h_{f} = 2706.54$ $\frac{K J}{Kg}$

Work done by shaft is,

$W = h_{f} - h_{i}$

$W = 2706.54 - 4635$

$W = -1928.46 \frac{kJ}{kg}$

But here efficiency is 0.56,

So work generated per kg is,

Work = $0.56 \times(- 1928.46)$

Work = $-1.3$ $\frac{MJ}{kg}$

Therefore, the shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

6 0

2 years ago

Complex poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop transfer fu

Sergio [31]

The answer in image

Don't forget put heart ♥️

7 0

2 years ago

Because there is a one-to-many relationship between sales reps and customers in the TAL Distributors database, one sales rep can

stiv31 [10]

Answer:

True

Explanation:

The relationship between sales reps and customers is an example of a one-to-many relationship. This is because one sales rep can be associated with many customers but a customer must have one sales rep. It is impossible for a customer to have zero sales rep but this is quite possible for a sales rep to have either zero, one or even more customers. Therefore the statement is True.

7 0

2 years ago