Question

A brass rod 175.00 mm long and 5.00 mm in diameter extends horizontally from a casting at 200°C. The rod is in an air environment with [infinity] 20°C and 30 W/m2·K. What is the temperature of the rod 43.75, 87.50, and 175.00 mm from the casting?

Answer 1

Answer:

(a) for  43.75 mm rod, the temperature is 121.97 ⁰C

(b) for  87.50 mm rod, the temperature is 80.17 ⁰C

(c) for  175.00 mm rod, the temperature is 53.46 ⁰C

Explanation:

Given;

L = 175 mm = 0.175 m

D = 5mm = 0.005

$T_b$ = 200°C

T∞ = 20°C

Heat transfer coefficient h = 30 W/m²·K

Thermal conductivity of brass K = 133 W/m.°C

$T_X = \frac{T-T_{infinity}}{T_b -T_{infinity}}$

where;

T is the temperature of the rod at different casting distance,

$T_X =\frac{Cosh[m(L-X)]+(h/mk)Sinh[m(L-X)]}{Cosh(mL) +(h/mk)Sinh(mL)}$

$m = \sqrt{\frac{4h}{kD} } = \sqrt{\frac{4*30}{133*0.005} } = 13.43 m^{-1}\\\\h/mk =\frac{30}{13.43*133} =0.0168$

Part (a) for  43.75 mm rod

$T_{43.75} =\frac{Cosh[13.43(0.175-0.04375)] +(0.0168)Sinh[13.43(0.175-0.04375)]}{Cosh(13.43*0.175)+(0.0168)Sinh(13.43*0.175)} \\\\T_{43.75} = \frac{2.9999+0.0475}{5.2918+0.0875} = 0.5665$

$0.5665 = \frac{T -20}{200-20} \\\\T =121.97^oC$

Part (b) for  87.50 mm rod

$T_{87.5} =\frac{Cosh[13.43(0.175-0.0875)] +(0.0168)Sinh[13.43(0.175-0.0875)]}{Cosh(13.43*0.175)+(0.0168)Sinh(13.43*0.175)} \\\\T_{87.5} = \frac{1.7735+0.0246}{5.2918+0.0875} = 0.3343$

$0.3343 = \frac{T -20}{200-20} \\\\T =80.17^oC$

Part (c) for  175.00 mm rod

$T_{175} =\frac{Cosh[13.43(0.175-0.175)] +(0.0168)Sinh[13.43(0.175-0.175)]}{Cosh(13.43*0.175)+(0.0168)Sinh(13.43*0.175)} \\\\T_{175} = \frac{1+0}{5.2918+0.0875} = 0.1859$

$0.1859 = \frac{T -20}{200-20} \\\\T =53.46^oC$