A mass flow rate of 0.6 kg/s of cooling water is needed to remove heat from an X-ray tube operating at 3.0 MV, in which the temperature rise of the water is 30°K. What is the beam current in the X-ray tube?
A)12.8 mA
B)24.2 mA
C)25.1 mA
D)41.2 mA

Question

Quizplus · Accepted Answer

The heat removed by the cooling water can be calculated using the formula $Q = mc\Delta T$, where $Q$ is the heat removed, $m$ is the mass flow rate, $c$ is the specific heat capacity of water (approximately 4.18 J/g°C), and $\Delta T$ is the temperature rise. However, for this problem, we are more interested in the electrical power absorbed by the X-ray tube, which can be calculated from the heat removed since all the electrical power is assumed to be converted into heat.Given:- Mass flow rate, $m = 0.6$ kg/s- Temperature rise, $\Delta T = 30$ K- Voltage, $V = 3.0$ MV = $3.0 \times 10^6$ VFirst, convert the mass flow rate to grams per second (since the specific heat capacity is in J/g°C) and calculate the heat removed per second (which is power, $P$):$$P = m \times c \times \Delta T$$$$P = 0.6 \times 10^3 \times 4.18 \times 30$$$$P = 75240 \, \text{W} = 75.24 \, \text{kW}$$The electrical power absorbed by the X-ray tube is equal to the product of the beam current ($I$) and the voltage ($V$):$$P = IV$$Solving for $I$:$$I = \frac{P}{V} = \frac{75.24 \times 10^3}{3.0 \times 10^6} = 0.02508 \, \text{A} = 25.08 \, \text{mA}$$Thus, the beam current in the X-ray tube is approximately 25.1 mA.

A Mass Flow Rate of 0