How much condensate would need to be removed per hour to accomplish a warm-up in 20 minutes?

To determine the amount of condensate that needs to be removed per hour to accomplish a warm-up in 20 minutes, we start with the understanding of the heat transfer involved in heating water or condensate.

The correct answer, which indicates approximately 308.67 kg/h, likely comes from a calculation that considers the specific heating requirements, the temperature change desired, and the properties of the condensate. The process relies on the specific heat capacity of water (as condensate is often primarily water) and the total heat energy needed to achieve the desired temperature increase within the set time frame.

In 20 minutes, a specific amount of heat must be transferred; therefore, calculating how much mass of condensate needs to be removed involves using the formula for heat transfer:

Q = m * c * ΔT

where Q is the heat energy required, m is the mass flow rate of the condensate (in kg/h), c is the specific heat capacity (approximately 4.18 kJ/kg·°C for water), and ΔT is the change in temperature in degrees Celsius.

Given that the duration is 20 minutes (which is one-third of an hour), the total amount of mass that needs to be removed over the hour can be