For steel: cp = 135 J/kg-K a= 4.5x10(-6) /oK Density = 7.85 g/cm3 E = 207 GPa a.

Question

For steel: cp = 135 J/kg-K a= 4.5x10(-6) /oK Density = 7.85 g/cm3 E = 207 GPa

a. A 1 m long steel rod 10 cm in diameter is heated from room temperature (23C) to 300C. What happens to its length?

b. If instead of being free the rod ends are fixed at room temperature, and then the temperature is raised to 300C, what is the state of stress in the rod at 300C?

c. How much energy is stored in the material during the temperature rise for case (b)?

d. For case (c), briefly explain how the thermal energy is stored.

Explanation / Answer

a.) l = l0 ( 1+ aT) = 1 *(1 + 4.5*10-6 (300-23)) = 1.001 m

i.e lenth expands by 0.001m = 1mm

b.) At 300o C natural length lo = 1.001 m

l = 1-1.001 = -0.001

A = * (10/200)2 = 7.854 * 10-3 m2

= A E l/l0 = - 1.62 MPa

Stress is compressing the rod (compression)

c.) mass of rod = 1*7.854 * 10-3* 7.85*103 = 61.654kg

Cv = Cp - VT2a2/

= -1/AE

Cv = 135 + ALT2 *(4.5*10-6 )2/(A*207*109) = 135 - 0.02 T2 * 10-21

135 J/kgK

Energy = Cv*m*T = 135*61.654*(300-23) = 2.3 MJ

d.) Energy is mainly stored as internal vibration & compressive strain

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