Water is to be heated from \(10^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) as it flows through a 2 -cm-internal-diameter, 13 -m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of \(5 \mathrm{~L} / \mathrm{min}\), determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.

Short Answer

Expert verified
Answer: To find the power rating of the electric resistance heater, follow these steps: 1. Calculate the mass flow rate of water. 2. Calculate the energy transfer to the water. 3. Determine the power rating of the resistance heater. To estimate the inner surface temperature of the pipe at the exit, divide the energy transfer rate by the heat transfer coefficient and the pipe's exit area. Add the resulting difference to the exit water temperature.

Step by step solution

01

Calculate mass flow rate of water

First, we need to calculate the mass flow rate of water. Knowing the flow volume rate (\(5 \mathrm{~L} / \mathrm{min}\)) and the density of water (ρ = approximately \(1000 \mathrm{~kg} / \mathrm{m^3}\)), we can convert the volume flow rate to mass flow rate (\(\dot{m}\)) using the following equation: \(\dot{m} = \rho \times \mathrm{Volume \, Flow \, Rate}\). Convert the volume flow rate from L/min to m³/s.
02

Calculate energy transfer to water

We need to determine the energy transfer between the heater and the water (\(\dot{Q}\)). To do this, we use the energy equation: \(\dot{Q} = \dot{m} \times c \times (T_{out} - T_{in})\), where \(c\) is the specific heat capacity of water (\(c = 4.18 \mathrm{~kJ} / (\mathrm{kg} \cdot \mathrm{K})\), \(T_{out} = 80^\circ \mathrm{C}\), and \(T_{in} = 10^\circ \mathrm{C}\).
03

Calculate the power rating of the resistance heater

Now, we need to find the power rating of the resistance heater (\(P\)). The electric power and energy transfer have the same value because of the perfect insulation and steady operation. So, we have \(P = \dot{Q}\).
04

Estimate the inner surface temperature of the pipe at the exit

To estimate the inner surface temperature of the pipe at the exit, we note that the water heat transfer rate is constant along the pipe. Divide the energy transfer rate by the heat transfer coefficient (h) and the pipe's exit area (A). This will give us the difference between the inner surface temperature (\(T_s\)) and the water temperature (\(T_{out}\)). We can then add the difference to the exit water temperature (\(80^\circ \mathrm{C}\)) to get the inner surface temperature. Note that you may need to assume a value for the heat transfer coefficient based on given conditions or typical values for similar systems.

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Most popular questions from this chapter

Air at \(20^{\circ} \mathrm{C}(1 \mathrm{~atm})\) enters into a 5-mm-diameter and 10-cmlong circular tube at an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tube wall is maintained at a constant surface temperature of \(160^{\circ} \mathrm{C}\). Determine the convection heat transfer coefficient and the outlet mean temperature. Evaluate the air properties at \(50^{\circ} \mathrm{C}\).

Hot air at atmospheric pressure and \(85^{\circ} \mathrm{C}\) enters a \(10-\mathrm{m}\)-long uninsulated square duct of cross section \(0.15 \mathrm{~m} \times\) \(0.15 \mathrm{~m}\) that passes through the attic of a house at a rate of \(0.1 \mathrm{~m}^{3} / \mathrm{s}\). The duct is observed to be nearly isothermal at \(70^{\circ} \mathrm{C}\). Determine the exit temperature of the air and the rate of heat loss from the duct to the air space in the attic. Evaluate air properties at a bulk mean temperature of \(75^{\circ} \mathrm{C}\). Is this a good assumption?

Laid water is flowing between two very thin parallel 1 -m-wide and 10 -m-long plates with a spacing of \(12.5 \mathrm{~mm}\). The water enters the parallel plates at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.58 \mathrm{~kg} / \mathrm{s}\). The outer surface of the parallel plates is subjected to hydrogen gas (an ideal gas at \(1 \mathrm{~atm}\) ) flow width-wise in parallel over the upper and lower surfaces of the two plates. The free-stream hydrogen gas has a velocity of \(5 \mathrm{~m} / \mathrm{s}\) at a temperature of \(155^{\circ} \mathrm{C}\). Determine the outlet mean temperature of the water, the surface temperature of the parallel plates, and the total rate of heat transfer. Evaluate the properties for water at \(30^{\circ} \mathrm{C}\) and the properties of \(\mathrm{H}_{2}\) gas at \(100^{\circ} \mathrm{C}\). Is this a good assumption?

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a 12 -cm-diameter and \(8.5-\mathrm{m}\)-long tube at \(75^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\), and is cooled by a refrigerant evaporating outside at \(-10^{\circ} \mathrm{C}\). If the average heat transfer coefficient on the inner surface is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the exit temperature of water is (a) \(18.4^{\circ} \mathrm{C}\) (b) \(25.0^{\circ} \mathrm{C}\) (c) \(33.8^{\circ} \mathrm{C}\) (d) \(46.5^{\circ} \mathrm{C}\) (e) \(60.2^{\circ} \mathrm{C}\)

How does the friction factor \(f\) vary along the flow direction in the fully developed region in (a) laminar flow and (b) turbulent flow?

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