Consider a sphere made of stainless steel with diameter of 25 cm. It is heated to temperature of 300°C for some chemical tests. After finishing the tests, the sphere is cooled by exposing it to a flow of air at 1 atm pressure and 25°C with a velocity of 3 m/s. By the end of cooling process, the sphere's temperature drops to 200°C. The rate of heat transfer loss due to convection is closest to:.
a) 485 W
b) 513 W
c) 88 W
d) 611 w

263.69 W.

(None of the option).

Explanation:

So, from the question above we are given the following parameters or data or information which is going to allow us to solve this question and they are;

(1). diameter of 25 cm.

(2). Initial temperature of 300°C.

(3).temperature drops to 200°C = final temperature.

Step one: Calculate the Reynolds number.

Reynolds number = 3 × 0.25/1.562 × 10^-5 = 48015.365.

Step two: Calculate average heat transfer coefficient.

The average heat transfer coefficient = k/D { 2 + (0.4Re^1/2 + 0.06Re^2/3} px^0.4 × (u/uz)^1/4.

The average heat transfer coefficient = 0.10204 × [ 2 + (87.65 + 79.26) (0.8719) × 0.8909.

average heat transfer coefficient = 0.20204 ( 2 + 129.652).

average heat transfer coefficient = 13.43/m^2.k.

Step three: The rate of heat transfer loss due to convection = (average heat transfer coefficient ) × πD^2 × ( T1 - T2).

The rate of heat transfer loss due to convection= 13.43 × π(0.25)^2 × (300 - 200).

=>The rate of heat transfer loss due to convection = 263.69 W.

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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