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Chapter 4: Problem 79

A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of \(0.42 \mathrm{gal} / \mathrm{min}\). The inlet pressure and temperature are \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\), and \(180^{\circ} \mathrm{F}\), respectively; at the exit the pressure is \(120 \mathrm{lbf} / \mathrm{in} .^{2}\) The pump requires \(1 / 35 \mathrm{hp}\) of power input. Water can be modeled as an incompressible substance with constant density of \(60.58 \mathrm{lb} / \mathrm{ft}^{3}\) and constant specific heat of \(1 \mathrm{Btu} / \mathrm{lb} \cdot{ }^{\circ} \mathrm{R}\). Neglecting kinetic and potential energy effects, determine the temperature change, in \({ }^{\circ} \mathrm{R}\), as the water flows through the pump. Comment on this change.

Short Answer

Expert verified
The temperature changes by 0.36 掳R, which is negligible.

Step by step solution

01

Determine Power Input in Consistent Units

First, convert the power input from horsepower (hp) to British Thermal Units per minute (Btu/min):\[ \text{Power} = \frac{1}{35} \text{hp} \times \frac{2545 \text{ Btu/hr}}{1 \text{ hp}} \times \frac{1 \text{ hr}}{60 \text{ min}} = 1.21 \text{ Btu/min} \]
02

Calculate the Mass Flow Rate

Convert the volumetric flow rate from gallons per minute (gal/min) to cubic feet per minute (ft鲁/min) and then to mass flow rate:\[ \text{Volumetric Flow Rate} = 0.42 \text{ gal/min} \times \frac{1 \text{ ft}^3}{7.48 \text{ gal}} = 0.0561 \text{ ft}^3/\text{min} \]\[ \text{Mass Flow Rate} = 0.0561 \text{ ft}^3/\text{min} \times 60.58 \text{ lb}/\text{ft}^3 = 3.40 \text{ lb/min} \]
03

Apply the Energy Balance

\[ \text{Temperature Change} = \frac{1.21 \text{ Btu/min}}{3.40 \text{ lb/min} \times 1 \text{ Btu/lb}^{\bullet{}}\text{R} } = 0.36^\text{R} \]
04

Comment on the Temperature Change

The temperature change of 0.36 掳R (Rankine) is very small.

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Key Concepts

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

Pump Efficiency
Understanding pump efficiency is critical for determining how well the pump converts the input mechanical energy into useful fluid flow. It is defined as the ratio of useful energy output to the input energy. Pump efficiency affects the energy consumption and cost. Higher efficiency means less energy is wasted. For this problem, knowing the power input and how it is used to circulate the water helps us understand why the temperature change is minimal.
Steady State
The term 'steady state' refers to a condition where all quantities within the system (like pressure, temperature, and flow rates) remain constant over time. This simplifies calculations as you don't need to consider changes over time. For the pump in this exercise, operating at steady state means that the inlet and outlet pressures, temperatures, and flow rates do not vary. Thus, the system is easier to analyze, focusing only on the flow through the pump.
Energy Balance
Energy balance involves accounting for all energy entering and leaving the system to understand how energy is conserved. For this pump problem, we apply the principle of energy conservation to calculate the temperature change. We know the pump requires 1.21 Btu/min of power input to maintain the steady state. Using the mass flow rate and specific heat, we find the temperature change by dividing the power input over the product of mass flow rate and specific heat. This leads to the small temperature change (0.36 掳R), indicating efficient energy use.
Temperature Change
The temperature change in the pump is calculated based on the energy added by the pump in terms of heat. We find it by dividing the power input by the product of mass flow rate and specific heat of the liquid. The result (0.36 掳R) indicates that the pump does not significantly heat the water, due to its efficient energy transfer. This small change is expected given the water's incompressible nature and the system's efficiency. Understanding this helps in designing systems where thermal effects are minimized, maintaining safe and functional performance in heating systems.

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Most popular questions from this chapter

Refrigerant \(134 \mathrm{a}\) enters a well-insulated nozzle at \(200 \mathrm{lbf} / \mathrm{in} .{ }^{2}, 220^{\circ} \mathrm{F}\), with a velocity of \(120 \mathrm{ft} / \mathrm{s}\) and exits at 20 lbf/in. \({ }^{2}\) with a velocity of \(1500 \mathrm{ft} / \mathrm{s}\). For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in \({ }^{\circ} \mathrm{F}\).

Refrigerant 134 a enters a compressor operating at steady state as saturated vapor at \(0.12 \mathrm{MPa}\) and exits at \(1.2 \mathrm{MPa}\) and \(70^{\circ} \mathrm{C}\) at a mass flow rate of \(0.108 \mathrm{~kg} / \mathrm{s}\). As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of \(0.32 \mathrm{~kJ} / \mathrm{s}\). Determine at steady state the power input to the compressor, in \(\mathrm{kW}\).

An 8-ft \({ }^{3}\) tank contains air at an initial temperature of \(80^{\circ} \mathrm{F}\) and initial pressure of \(100 \mathrm{lbf} / \mathrm{in}^{2}\). The tank develops a small hole, and air leaks from the tank at a constant rate of \(0.03 \mathrm{lb} / \mathrm{s}\) for \(90 \mathrm{~s}\) until the pressure of the air remaining in the tank is \(30 \mathrm{lbf} /\) in. \({ }^{2}\) Employing the ideal gas model, determine the final temperature, in \({ }^{\circ} \mathrm{F}\), of the air remaining in the tank.

A rigid copper tank, initially containing \(1 \mathrm{~m}^{3}\) of air at \(295 \mathrm{~K}, 5\) bar, is connected by a valve to a large supply line carrying air at \(295 \mathrm{~K}, 15\) bar. The valve is opened only as long as required to fill the tank with air to a pressure of 15 bar. Finally, the air in the tank is at \(310 \mathrm{~K}\). The copper tank, which has a mass of \(20 \mathrm{~kg}\), is at the same temperature as the air in the tank, initially and finally. The specific heat of the copper is \(c=0.385 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Assuming ideal gas behavior for the air, determine (a) the initial and final mass of air within the tank, each in \(\mathrm{kg}\), and (b) the heat transfer to the surroundings from the tank and its contents, in kJ, ignoring kinetic and potential energy effects.

Refrigerant \(134 a\) enters an insulated compressor operating at steady state as saturated vapor at \(-20^{\circ} \mathrm{C}\) with a mass flow rate of \(1.2 \mathrm{~kg} / \mathrm{s}\). Refrigerant exits at 7 bar, \(70^{\circ} \mathrm{C}\). Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in \(\mathrm{m}^{3} / \mathrm{s}\), and (b) the power input to the compressor, in \(\mathrm{kW}\).
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