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MissTica
3 years ago
8

A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype

will be operated in seawater. Determine the speed ofthe prototype to ensure Reynolds number similarity‘
Engineering
1 answer:
ella [17]3 years ago
8 0

Answer:

Explanation:

Given

scale i.e. L_r=1:15

Using Reynolds number similarity

(Re)_m=(Re)_p

(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p

Properties of air

\nu _{air}=1.57\times 10^{-4} ft/s

Properties of sea water

\nu _{sea}=1.26\times 10^{-5} ft/s

\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )

V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )

V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}

V_p=0.963\ ft/s

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3 years ago
Why is an integrated circuit (IC) referred to as a special device?"<br> points more than 10 plz help
sladkih [1.3K]

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3 years ago
Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b
coldgirl [10]

Answer:

A) q'_free = 3146.41 W/m²

B) q'_forced = 7521.41 W/m²

Explanation:

We are given;

Free convection coefficient; h_fr = 5 W/m²K

Force convection coefficient; h_forced = 30 W/m²K

Emissivity; ε = 0.95

Temperature of surrounding which is equal to temperature of air; T_s = T_air = 200°C = 473K

Initial temperature; T_i = 25°C = 298K

A) Now, since the convection feature is disabled, the mode of heat transfer associated with this condition is through free convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_free = q'_free convection + q'_radiation

q'_free convection = h_free(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Where, σ is stephan boltzmann constant and has a constant value of 5.67 × 10^(−8) W/m²K⁴

Thus, rewriting;

q'_free = q'_free convection + q'_radiation

We have;

q'_free = [h_free(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_free = [5(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_free = 875 + 2271.41

q'_free = 3146.41 W/m²

B) Now, in this case, since the convection feature is disabled, the mode of heat transfer associated with this condition is through forced convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_forced = q'_forced convection + q'_radiation

Where;

q'_forced convection = h_forced(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Thus, rewriting;

q'_forced = q'_free convection + q'_radiation

We have;

q'_forced = [h_forced(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_forced = [30(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_forced = 5250 + 2271.41

q'_forced = 7521.41 W/m²

7 0
4 years ago
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