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3 years ago

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So I am a high school student and i was wondering if i was making a marble run with 4 changes in direction from the start how lo

ng of peices would I need and what degrees would they need to be, I understand this might not be able to be figued but I still would apreciate help, Thanks.

1 answer:

Akimi4 [234]3 years ago

4 0

Answer:

OMG GOD I HAVE WONDERED THAT

Explanation:

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A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s

monitta

Answer:

a. $\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$ which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

$v_1 = \dfrac{4}{3} \times \pi \times r^3$

$\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3$

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

$\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}$

We have;

$\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}$

$\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}$

$log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )$

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

$\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}$

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. $W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}$

$p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }$

p₂ = 100000/0.941 = 106.265 kPa

$W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J$

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0

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

3 0

3 years ago