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3 years ago

5

what would happen to an object if there were no inertia? Also would this mean that if there were no inertia that an object may t

ravel high speeds DEPENDING on the object's mass?

1 answer:

adelina 88 [10]3 years ago

4 0

There would be infinite acceleration, and the body would have no mass <span />

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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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