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Tatiana [17]
3 years ago
12

Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w

ind profile you identified, calculate the 0-1 km and 0-6 km bulk wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at 1 km and 5 km. The atmospheric pressure at 1 km above sea level is typically very close to 850 mb. The pressure at 6 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-6 km wind shear values in knots (kts). For simplicity, assume that the surface winds are due south easterly, the 850 mb winds are due southerly, and the 500 mb winds are due westerly. Show your work.

Engineering
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

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Discuss why TVET Institutions need advice of the business community in order
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Answer:

Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community.  To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town).  This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace.  Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.

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4 0
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

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Answer:

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