Question

A brass rod 100 mm long and 5 mm in diameter extends horizontally from a casting at 200°e. the rod is in an air environment with t∞ = 20°c and h = 30 w/m2 ∙ k. what is the temperature of the rod 25, 50, and 100 mm from the casting?

Answer 1

When temperature is increased, the rate of dissolving increases. The kinetic energy of the molecules of the solute and solvent molecules is high thereby increasing their contact. An example is mixing powdered sugar to the water. When you add water to the sugar, the dissolving process is slow. However, when you increase the temperature of the water by boiling it, the sugar dissolves immediately. 

Answer 2

Answer:

156°C, 128°C, and 106.5°C

Explanation:

The temperature distribution can be calculated by the expression:

$\frac{T}{Tb} = \frac{cosh m(L-x) + (h/mk)sinh m(L-x)}{coshmL + (h/mk)sinhmL}$

Where, L is the length (100 mm = 0.1 m), x is the distance to the end of the rod, k is the conductivity of the brass which is 133 W/m.K, and m is:

m = $(\frac{hp}{kAc})^{1/2}$

The perimeter, p = π*D, where D is the diameter (5 mm = 0.005m), and the area, Ac = π*D²/4

m = $(\frac{4h}{kD})^{1/2}$

m = 13.43 m⁻¹

So:

coshmL = cosh(13.43*0.1) = 2.05

sinhmL = sinh(13.43*0.1) = 1.78

h/mk = 30/(13.43*133) = 0.0168

Tb is the difference of temperature of the casting and the surroundings: Tb = 200 - 20 = 180°C

Thus,

T = $180*(\frac{coshm(L-x) + 0.0168sinhm(L-x)}{2.05 + 0.0168*1.78} )$

So, substituing the values of x:

x= 25 mm = 0.025 m

T = $180*(\frac{cosh13.43(0.1 -0.025) + 0.0168sinh13.43(0.1-0.025)}{2.08} )$

T = 136°C

T = Trod - T∞

Trod = 136 + 20

Trod = 156°C

x = 50 mm = 0.05 m

T = $180*(\frac{cosh13.43(0.1 -0.05) + 0.0168sinh13.43(0.1-0.05)}{2.08} )$

T = 108°C

Trod = 128°C

x = 100 mm = 0.1 m

T = $180*(\frac{cosh13.43(0) + 0.0168sinh13.43(0)}{2.08} )$

T = 86.5°C

Trod = 106.5°C