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Chapter 4: Problem 49

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

Expert verified
The maximum electric power is 2.787 kW.

Step by step solution

01

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: \(26^{\text{ o C}}\)
02

Convert mass flow rate to consistent units

First, convert the mass flow rate from \( \frac{\text{kg}}{\text{min}} \) to \( \frac{\text{kg}}{\text{s}} \): \[ \frac{10 \text{ kg}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{10}{60} \text{ kg/s} = \frac{1}{6} \text{ kg/s} \]
03

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}\.
04

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: \[ Q = \frac{1}{6} \text{ kg/s} \times 4.18 \frac{kJ}{kg \cdot\degree C} \times (26^{\text{ o}} C - 22^{\text{ o}} C)\]
05

Perform the calculation

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The electronic components of a computer consume \(0.1 \mathrm{~kW}\), of electrical power. To prevent overheating, cooling air is supplied by a 25-W fan mounted at the inlet of the electronics enclosure. At steady state, air enters the fan at \(20^{\circ} \mathrm{C}, 1\) bar and exits the electronics enclosure at \(35^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the enclosure to the surroundings and the effects of kinetic and potential energy can be ignored. Determine the volumetric flow rate of the entering air, in \(\mathrm{m}^{3} / \mathrm{s}\).

A feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, \(42^{\circ} \mathrm{C}\), and a mass flow rate of \(70 \mathrm{~kg} / \mathrm{s}\). A separate stream of water enters at inlet 2 as a two-phase liquid-vapor mixture at 7 bar with a quality of \(98 \%\). Saturated liquid at 7 bar exits the feedwater heater at 3 . Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), at inlet 2 .

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at \(3.2\) bar, \(10^{\circ} \mathrm{C}\), and is compressed at steady state to \(10 \mathrm{bar}, 70^{\circ} \mathrm{C}\). The volumetric flow rate of refrigerant entering is \(3.0 \mathrm{~m}^{3} / \mathrm{min}\). The power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

A tiny hole develops in the wall of a rigid tank whose volume is \(0.75 \mathrm{~m}^{3}\), and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) leaks in. Eventually, the pressure in the tank reaches 1 bar. The process occurs slowly enough that heat transfer between the tank and the surroundings keeps the temperature of the air inside the tank constant at \(25^{\circ} \mathrm{C}\). Determine the amount of heat transfer, in \(\mathrm{kJ}\), if initially the tank (a) is evacuated. (b) contains air at \(0.7\) bar, \(25^{\circ} \mathrm{C}\).

A well-insulated rigid tank of volume \(10 \mathrm{~m}^{3}\) is connected to a large steam line through which steam flows at 15 bar and \(280^{\circ} \mathrm{C}\). The tank is initially evacuated. Steam is allowed to flow into the tank until the pressure inside is \(p\). (a) Determine the amount of mass in the tank, in \(\mathrm{kg}\), and the temperature in the tank, in \({ }^{\circ} \mathrm{C}\), when \(p=15\) bar. (b) Plot the quantities of part (a) versus \(p\) ranging from \(0.1\) to 15 bar.
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