A 1 inch diameter tube at 922 K is used in nuclear reactor a

A 1 inch diameter tube at 922 K is used in nuclear reactor as part of a heat exchanger. It is exposed on the outside to liquid sodium with a temperature of 366 K at a velocity of 5 m/s. See the Appendix for properties of liquid metals. The tube is removed for inspection and you need to perform a microscopic analysis on the tube to look for signs of overheating the tube wall. At what angles around the tube relative to the oncoming flow would you concentrate your inspection? What is the worst-case local heat transfer coefficient you would expect for these conditions? What is the average heat transfer coefficient for the exterior of the tube? What if instead of liquid sodium, high-pressure water was used on the outside of the tube? What would the average heat transfer coefficient be? Which fluid is more effective as a coolant?

Solution

GIVEN:- (Tube temperature) Tt = 922k, (Liquid sodium temperature) Ts = 366k, (Velocity sodium) Vs = 5m/s, (Cross-section area of tube) At = 0.00051 m2, (Specific heat capacity of liquid sodium at constant pressure) Cp = 1.383 kJ/kg/K

SOLUTION:-

(a) Angle should be less than 90 degree. (ANSWER)

(b) The quantity of heat absorbed by per unit mass of liquid metal at constant pressure Qc = Cp (Tt-Ts)

i.e. Qc = 1383 (922-366) = 768948 J

For worst case local heat transfer coefficient(K), Qc = (k / rt) x At x (Tt-Ts) = (k/0.0127) x 0.00051 x (922-366)

So k = 34.439 kJ/m and K = k/rt = 2711.73 kJ/m2 (ANSWER)

(c) Now for coolant flowing at Vs --- Qc = h x (At) x (Tt-Ts) x Vs

i.e. 768948 = h x (0.00051) x (922-366) x 5,

So (Average heat transfer coefficent) h = 542.35 kJ/m3.s (ANSWER)

(d) For high pressure water instead of liquid sodium Qcw = 2329.64kJ

So Qcw = 2329640 = hw x 0.00051 x (922-366) x 5

So (Average heat transfer coefficent for water) hw = 1643.14 kJ/m3.s (ANSWER)

   Clearly, water is more effective as compared to liquid sodium. (ANSWER)