Calculate Anode Heat Production: 65kV X-Ray Tube at 120mA & 0.8% Efficiency

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To calculate the rate of heat production in a 65kV X-ray tube operating at 120mA with an efficiency of 0.8%, first determine the power output by multiplying voltage and current, resulting in 7.8 kW. Since only 0.8% of this power is converted into X-rays, the remaining 99.2% contributes to heating the anode. Therefore, the heat production rate can be calculated as 7.8 kW multiplied by 0.992, yielding approximately 7.74 kW. This indicates that the majority of the energy is dissipated as heat rather than being converted into X-rays. Understanding this calculation is crucial for managing anode heat in X-ray tube operations.
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An X-ray tube operating at 65kV has a tube current of 120mA. The tube produces X-ray with an efficiency of 0.8%

calculate the rate of heat production in the anode

rate = ?


I really don't get this question, anyone?
 
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oH come on...someone knows the answer, or at least guide me! its bugging me not knowing.
 
ella_101 said:
An X-ray tube operating at 65kV has a tube current of 120mA. The tube produces X-ray with an efficiency of 0.8%

OK, one has voltage (65 kV) and current (120 mA), and voltage * current is power. Power is a rate of energy (production/transfer).

If X-rays are only 0.8% of the energy (0.008), then 0.992 of the energy/power goes into heating the anode.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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