Abrass 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?

156°C, 128°C, and 106.5°C

Explanation:

The temperature distribution can be calculated by the expression:

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 =

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

m =

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 =

So, substituing the values of x:

x= 25 mm = 0.025 m

T =

T = 136°C

T = Trod - T∞

Trod = 136 + 20

Trod = 156°C

x = 50 mm = 0.05 m

T =

T = 108°C

Trod = 128°C

x = 100 mm = 0.1 m

T =

T = 86.5°C

Trod = 106.5°C

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

2b2t

explanation:

2b2t