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Chapter 1: Problem 23

Water is heated in an insulated, constant diameter tube by a \(5-\mathrm{kW}\) electric resistance heater. If the water enters the heater steadily at \(15^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\), determine the mass flow rate of water.

Short Answer

Expert verified
Answer: The mass flow rate of the water in the tube is approximately 0.7077 kg/s.

Step by step solution

01

Write down the given information

The given information can be written as: - Electric heater power: \(P = 5\,\mathrm{kW}\) - Water temperature at inlet: \(T_1 = 15^{\circ}\mathrm{C} \) - Water temperature at outlet: \(T_2 = 60^{\circ}\mathrm{C} \)
02

Convert temperatures to Kelvin

To work with temperatures in calculations, it is necessary to convert them to Kelvin because Kelvin is an absolute temperature scale: $$T_1(K) = T_1(^{\circ}\mathrm{C}) + 273.15$$ $$T_2(K) = T_2(^{\circ}\mathrm{C}) + 273.15$$
03

Write down the energy balance equation

The energy balance equation is given as follows: $$P = \dot{m}c\Delta T$$ Where: - \(P\) is the power (in Watts) - \(\dot{m}\) is the mass flow rate (in \(\mathrm{kg/s}\)) - \(c\) is the specific heat capacity of water at constant pressure (\(4.18\,\mathrm{kJ/kg\cdot K}\)) - \(\Delta T = T_2 - T_1\) is the temperature difference between the outlet and inlet
04

Convert the power to watts

The power of the electric heater is given in kilowatts, so we need to convert it to watts: $$P (\mathrm{W}) = P (\mathrm{kW}) \times 1000$$ Where \(P(\mathrm{W})\) is the power in watts, and \(P(\mathrm{kW})\) is the power in kilowatts.
05

Solve for the mass flow rate

We can rearrange the energy balance equation to solve for the mass flow rate: $$\dot{m} = \frac{P}{c\Delta T}$$ Plug in the values: $$\dot{m} = \frac{P(\mathrm{W})}{c\Delta T}$$ $$\dot{m} = \frac{P(\mathrm{W})}{c(T_2(K) - T_1(K))}$$
06

Calculate the mass flow rate

Use the given and calculated values to find the mass flow rate \(\dot{m}\): $$\dot{m} = \frac{5000}{4.18\times (333.15-288.15)}$$ $$\dot{m} = \frac{5000}{4.18\times45}$$ $$\dot{m} = 0.7077\,\mathrm{kg/s}$$ Thus, the mass flow rate of the water in the tube is approximately \(0.7077\,\mathrm{kg/s}\).

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Most popular questions from this chapter

A cylindrical fuel rod of \(2 \mathrm{~cm}\) in diameter is encased in a concentric tube and cooled by water. The fuel generates heat uniformly at a rate of \(150 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient on the fuel rod is \(5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the average temperature of the cooling water, sufficiently far from the fuel rod, is \(70^{\circ} \mathrm{C}\). Determine the surface temperature of the fuel rod and discuss whether the value of the given convection heat transfer coefficient on the fuel rod is reasonable.

A room is heated by a \(1.2 \mathrm{~kW}\) electric resistance heater whose wires have a diameter of \(4 \mathrm{~mm}\) and a total length of \(3.4 \mathrm{~m}\). The air in the room is at \(23^{\circ} \mathrm{C}\) and the interior surfaces of the room are at \(17^{\circ} \mathrm{C}\). The convection heat transfer coefficient on the surface of the wires is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rates of heat transfer from the wires to the room by convection and by radiation are equal, the surface temperature of the wire is (a) \(3534^{\circ} \mathrm{C}\) (b) \(1778^{\circ} \mathrm{C}\) (c) \(1772^{\circ} \mathrm{C}\) (d) \(98^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Why is the metabolic rate of women, in general, lower than that of men? What is the effect of clothing on the environmental temperature that feels comfortable?

Can a medium involve \((a)\) conduction and convection, (b) conduction and radiation, or \((c)\) convection and radiation simultaneously? Give examples for the "yes" answers.

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long \(25 \mathrm{~mm}\) diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((\dot{W} / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ \dot{W} / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) vs. \(V(\mathrm{~m} / \mathrm{s})\). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form of \(h=C V^{m}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.
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