1. A concrete column used in bridge construction is cylindrical in shape with a diameter of 1 meter. The column is completely poured in a short interval of time and the hydration of concrete results in the equivalent of a uniform source strength of 0.7 W/kg. Determine the temperature at the center of the cylinder at a time when the outside surface temperature is 75 degree Celsius. The column is sufficiently long so that temperature variation along its length may be neglected. For concrete
Average thermal conductivity = 0.95 W/m K
Average density = 2300 kg/m3
a) 190.92 degree Celsius
b) 180.92 degree Celsius
c) 170.92 degree Celsius
d) 160.92 degree Celsius
Explanation: t max = t w + q g R2/4k. We get, q g = 0.7 W/kg.
2. The temperature distribution profile for a solid cylinder is
a) Parabolic
b) Linear
c) Ellipse
d) Hyperbolic
Explanation: It must be parabolic for maximum heat conduction.
3. For a solid cylinder, maximum temperature difference occurring at the center of the rod is given by
a) t W – q g R2/4K
b) q g R2/4K
c) t W + q g R2/4K
d) t W + q g R2/4KL
Explanation: As temperature distribution profile is parabolic so on integrating between boundary conditions we get the result.
4. Consider a convective heat flow to water at 75 degree Celsius from a cylindrical nuclear reactor fuel rod of 50 mm diameter. The rate of heat generatioN is 50000000 W/m3 and convective heat transfer coefficient is I kW/m2 K. The outer surface temperature of the fuel element would be
a) 625 degree Celsius
b) 700 degree Celsius
c) 550 degree Celsius
d) 400 degree Celsius
Explanation: t w = t a + q g R/ 2h.
5. For a cylindrical rod with uniformly distributed heat sources, the thermal gradient at half the radius location will be
a) Four times
b) Twice
c) One fourth
d) One half
Explanation: t = t w + q g (R2 – r2)/4k. (d t /d r) r = R/2 = 1/2(d t/d r) r = R.
6. The maximum temperature for cylindrical coordinate occurring at r = 0 is
a) t max = t a +q g R/h + q g R2/4k
b) t max = t a +q g R/4h + q g R2/4k
c) t max = t a +q g R/2h + q g R2/4k
d) t max = t a +q g R/6h + q g R2/4k
Explanation: As, t = t a +q g R/6h + q g R2/4k (R2 –r2).
7. In case of solid cylinder of radius R, the temperature distribution is given as
a) t – t w/t max – t w = 1 – (r/R)2
b) t – t w/t max – t w = 1 – (r/R)
c) t – t w/t max – t w = 1 – (r/R)3
d) t – t w/t max – t w = 1 – (r/R)4
Explanation: As we know for a long cylinder of radius R, t = t w + q g (R2 – r2)/4k. On integrating this we get the answer. Where, t w is outer surface temperature and t max is along cylinder axis.
8. Consider heat conduction through a solid sphere of radius R. There are certain assumptions
(i) Unsteady state conditions
(ii) One-dimensional radial conduction
(iii) Constant thermal conductivity
Identify the correct statements
a) i and iii
b) ii and iii
c) i, ii and iii
d) i and ii
Explanation: Statement 1 should be steady state condition.
9. An 8 cm diameter orange, approximately spherical in shape, undergoes ripening process and generates 18000 k J/m3 hr of energy. If external surface of the orange is at 6.5 degree Celsius, find out the temperature at the center of the orange. Take thermal conductivity = 0.8 k J/ m hr degree for the orange material
a) 13.5 degree Celsius
b) 12.5 degree Celsius
c) 11.5 degree Celsius
d) 10.5 degree Celsius
Explanation: q g = 5000 W/m3, k = 0.222 W/m K and t = t W + q g R 2/6K = 12.5 degree Celsius.
10. Consider the above problem, calculate the heat flow from the outer surface of the orange
a) 4.82 k J/hr
b) 5.82 k J/hr
c) 6.82 k J/hr
d) 7.82 k J/hr
Explanation: Q = 4/3 (π R 3 q g) = 1.34 J/s.