A nuclear fuel rod generates energy at a rate | vedclass

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(B) The energy generated per unit volume of the fuel rod is $P_v = 5 	imes 10^8 \,W/m^3$.The volume of the cylindrical rod is $V = \pi r^2 h = \pi

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$6$

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(B) The energy generated per unit volume of the fuel rod is $P_v = 5 	imes 10^8 \,W/m^3$.The volume of the cylindrical rod is $V = \pi r^2 h = \pi 	imes (4 	imes 10^{-3} \,m)^2 	imes 0.2 \,m = \pi 	imes 16 	imes 10^{-6} 	imes 0.2 \,m^3 = 3.2 \pi 	imes 10^{-6} \,m^3$.The total energy generated per second (power) by the fuel rod is $P = P_v 	imes V = (5 	imes 10^8) 	imes (3.2 \pi 	imes 10^{-6}) = 1600 \pi \,W$.The heat absorbed by the coolant per unit time is given by $Q = m_f c \Delta T$,where $m_f$ is the mass flow rate,$c$ is the specific heat,and $\Delta T$ is the temperature rise.Given $m_f = 0.2 \,kg/s$ and $c = 4 	imes 10^3 \,J \cdot kg^{-1} \cdot K^{-1}$,the rate of heat absorption is $Q = 0.2 	imes 4000 	imes \Delta T = 800 \Delta T \,W$.Equating the power generated to the heat absorbed: $800 \Delta T = 1600 \pi$.$\Delta T = \frac{1600 \pi}{800} = 2 \pi \approx 2 	imes 3.14 = 6.28 \,^{\circ}C$.Thus,the temperature rise is approximately $6 \,^{\circ}C$.

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