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Chapter 14: Problem 3

What is the driving force for \((a)\) heat transfer, (b) electric current flow, \((c)\) fluid flow, and \((d)\) mass transfer?

Short Answer

Expert verified
Answer: The driving forces for different types of transfers are: (a) Heat transfer - temperature difference (\(\Delta T = T_1 - T_2\)) (b) Electric current flow - voltage difference (\(\Delta V = V_1 - V_2\)) (c) Fluid flow - pressure difference (\(\Delta P = P_1 - P_2\)) (d) Mass transfer - concentration difference (\(\Delta C = C_1 - C_2\))

Step by step solution

01

(a) Driving force for heat transfer:

The driving force for heat transfer is the temperature difference between two objects or regions. In general, heat flows from the region at a higher temperature to the region at a lower temperature. This temperature difference can be represented as \(\Delta T = T_1 - T_2\), where \(T_1\) and \(T_2\) are the temperatures of the two regions.
02

(b) Driving force for electric current flow:

The driving force for the flow of electric current is the voltage difference (or potential difference) between two points in a circuit. It causes the flow of charged particles (electrons) from the region at a higher potential to the region at a lower potential. The potential difference can be represented as \(\Delta V = V_1 - V_2\), where \(V_1\) and \(V_2\) are the potentials of the two points in the circuit.
03

(c) Driving force for fluid flow:

The driving force for fluid flow is a pressure difference between two points within the fluid. The fluid flows from an area of high pressure to an area of low pressure. This pressure difference can be represented as \(\Delta P = P_1 - P_2\), where \(P_1\) and \(P_2\) are the pressures at the two points in the fluid.
04

(d) Driving force for mass transfer:

The driving force for mass transfer is the concentration difference between two regions in a system. Molecules or particles will move from an area of higher concentration to an area of lower concentration. The concentration difference can be represented as \(\Delta C = C_1 - C_2\), where \(C_1\) and \(C_2\) are the concentrations in the two regions.

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Most popular questions from this chapter

What is the relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\) known as? Under what conditions is it valid? What is the practical importance of it? \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) and what are the names of the variables in it? Under what conditions is it valid? What is the importance of it in engineering?

Show that for an ideal gas mixture maintained at a constant temperature and pressure, the molar concentration \(C\) of the mixture remains constant but this is not necessarily the case for the density \(\rho\) of the mixture.

The pressure in a pipeline that transports helium gas at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) is maintained at \(14.5\) psia by venting helium to the atmosphere through a \(0.25\)-in-internal-diameter tube that extends \(30 \mathrm{ft}\) into the air. Assuming both the helium and the atmospheric air to be at \(80^{\circ} \mathrm{F}\), determine \((a)\) the mass flow rate of helium lost to the atmosphere through the tube, (b) the mass flow rate of air that infiltrates into the pipeline, and \((c)\) the flow velocity at the bottom of the tube where it is attached to the pipeline that will be measured by an anemometer in steady operation.

A wall made of natural rubber separates \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) gases at \(25^{\circ} \mathrm{C}\) and \(750 \mathrm{kPa}\). Determine the molar concentrations of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) in the wall.

Carbon at \(1273 \mathrm{~K}\) is contained in a \(7-\mathrm{cm}\)-innerdiameter cylinder made of iron whose thickness is \(1.2 \mathrm{~mm}\). The concentration of carbon in the iron at the inner surface is \(0.5 \mathrm{~kg} / \mathrm{m}^{3}\) and the concentration of carbon in the iron at the outer surface is negligible. The diffusion coefficient of carbon through iron is \(3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of carbon by diffusion through the cylinder shell per unit length of the cylinder is (a) \(2.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (b) \(5.4 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (c) \(8.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(1.6 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\) (e) \(5.2 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\) 14-185 The surface of an iron component is to be hardened by carbon. The diffusion coefficient of carbon in iron at \(1000^{\circ} \mathrm{C}\) is given to be \(3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). If the penetration depth of carbon in iron is desired to be \(1.0 \mathrm{~mm}\), the hardening process must take at least (a) \(1.10 \mathrm{~h}\) (b) \(1.47 \mathrm{~h}\) (c) \(1.86 \mathrm{~h}\) (d) \(2.50 \mathrm{~h}\) (e) \(2.95 \mathrm{~h}\)
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