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3 years ago

12

How many minutes of daylight do we gain after winter solstice

1 answer:

serg [7]3 years ago

4 0

Answer:

Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.

Explanation:

Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.

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Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over it

irga5000 [103]

Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

$T_1$ = 27°C = (273 + 27) K = 300 K

$T_2$ = 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

$k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg$

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = $\dfrac{\rho \times v \times x }{\mu}$

reynolds number = $\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}$

reynolds number = $\dfrac{0.98}{18.6 \times 10^{-6}}$

reynolds number = 52688.11204

Prandtl number = $\dfrac{c_p \mu}{k}$

Prandtl number = $\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}$

Prandtl number = $\dfrac{0.018693}{0.02476}$

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate can be computed as follows:

Nusselt no = $\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} =161.4252008}$

$h =\dfrac{161.4252008 \times 0.02476}{ 0.4}$

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = $h\times A \times (T_2-T_1)$

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

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Identify the wavelength of this wave.

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A parsec is a measurement of: magnitude distance position speed

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