Question

Ten \(\mathrm{kg} / \mathrm{min}\) of cooling water circulates through a water jacket enclosing a housing filled with electronic components. At steady state, water enters the water jacket at \(22^{\circ} \mathrm{C}\) and exits with a negligible change in pressure at a temperature that cannot exceed \(26^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the water jacket to the surroundings, and kinetic and potential energy effects can be ignored. Determine the maximum electric power the electronic components can receive, in \(\mathrm{kW}\), for which the limit on the temperature of the exiting water is met.

Short answer

The maximum electric power is 2.787 kW.

Step-by-step solution

Identify the given information

The problem provides the following details: - Mass flow rate of cooling water: \(\frac{10 \text{ kg}}{\text{min}}\)- Temperature of water entering the water jacket: \(22^{\text{ o C}}\)- Maximum allowable temperature of water exiting the water jacket: \(26^{\text{ o C}}\)

Convert mass flow rate to consistent units

First, convert the mass flow rate from \( \frac{\text{kg}}{\text{min}} \) to \( \frac{\text{kg}}{\text{s}} \): \[ \frac{10 \text{ kg}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{10}{60} \text{ kg/s} = \frac{1}{6} \text{ kg/s} \]

Determine the specific heat capacity of water

The specific heat capacity of water \( c_p \) is approximately \ 4.18 \ \frac{kJ}{kg \cdot \degree C}\.

Apply the energy balance equation

Use the energy balance equation for steady-state conditions: \[ Q = \text{mass flow rate} \times c_p \times \text{ change in temperature}\] Substitute the known values: \[ Q = \frac{1}{6} \text{ kg/s} \times 4.18 \frac{kJ}{kg \cdot\degree C} \times (26^{\text{ o}} C - 22^{\text{ o}} C)\]

Perform the calculation

Solve the equation to find the maximum electric power: \[ Q = \frac{1}{6} \times 4.18 \times 4\ \frac{kJ}{s} = \frac{1}{6} \times 16.72 \frac{kJ}{s}= 2.787 kW\]

Key concepts

Mass Flow Rate

The mass flow rate is a fundamental concept in engineering thermodynamics. It represents the amount of mass passing through a given surface per unit time. In our problem, the mass flow rate of cooling water is given as \( 10 \frac{kg}{min} \). To simplify calculations, it's often useful to convert this into a consistent unit, such as seconds. We did this conversion: \[ \frac{10 \frac{kg}{min}}{60 \frac{min}{s}} = \frac{1}{6} \frac{kg}{s}\].

Understanding mass flow rate helps us determine how much mass enters or leaves a system, which is crucial for analyzing processes where mass is conserved. It's beneficial to remember:

• Use consistent units
• Convert as necessary to avoid mistakes in calculations

Specific Heat Capacity

Specific heat capacity, often denoted as \( c_p \), is another key concept. This defines the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. For water, \( c_p \) is relatively high, valued at approximately 4.18 \frac{kJ}{kg \cdot \degree C}\.

High specific heat capacity of water is what makes it effective in cooling systems. It allows water to absorb and carry away significant amounts of heat with only a slight increase in temperature. Here are some points to note:

• The specific heat capacity is substance-specific
• For calculations, always use the value suited for the substance in question at given conditions

Energy Balance Equation

The energy balance equation is vital for determining energy transfers in thermodynamic systems. In steady-state conditions, it can be succinctly written as:

\[ Q = \text{mass flow rate} \times c_p \times \text{ change in temperature} \]

This equation allows us to calculate the heat transfer rate (\( Q \)) by knowing the mass flow rate, specific heat capacity, and temperature change. Breaking this into parts for clarity:

• Mass flow rate: The amount of mass flowing per unit time
• Specific heat capacity: The amount of heat required for a temperature change of one degree
• Change in temperature: The difference between the final and initial temperatures

Applying this knowledge to our problem, we substitute known values to find:

\[ Q = \frac{1}{6} \frac{kg}{s} \times 4.18 \frac{kJ}{kg \cdot \degree C} \times (26^\text{ o}C - 22^\text{ o}C) = 2.787 \text{kw} \]

This calculation illustrates the importance of accurately applying the energy balance equation for solving engineering thermodynamics problems.