150 mm wide. Assume the plates are negligibly thin, and that the natural convection air flow adjacent to the plates is not affected by the presence of the neighboring plates. The temperatures of the surface of the plates and the air are 90C and 14C, respectively. Consider the following: i) Determine the natural convection

Explanation:

Let's begin by listing out the given parameters:

L = 400 mm = (400 ÷ 1000) m = 0.40 m,

W = 150 mm = (150 ÷ 1000) m = 0.15 m,

T1 = 90 °C, T2 = 14 °C

I) To calculate the heat transfer coefficient for natural convection of air over a flat plate, we use the simplified formula

hc = C * [(ΔT ÷ L)^n]

For vertical plates, C = 1.42, n = 0.25

ΔT = Tu - Tl = (90 - 14) °C = 76 °C

hc = 1.42 * [76 ÷ 0.15]^0.25

hc = 1.42 * 4.744 = 6.73648

hc = 6.74 W/m²°C

II) q = hc A (ΔT)

A = L * W = 0.4 * 0.15 = 0.06

q = 6.74 * 0.06 * 76 = 30.7344

q = 31 W

this is true

explanation:

i looked it up

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: