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Chapter 7: Problem 24

Water at \(43.3^{\circ} \mathrm{C}\) flows over a large plate at a velocity of \(30.0 \mathrm{~cm} / \mathrm{s}\). The plate is \(1.0 \mathrm{~m}\) long (in the flow direction), and its surface is maintained at a uniform temperature of \(10.0^{\circ} \mathrm{C}\). Calculate the steady rate of heat transfer per unit width of the plate. 7-24 The forming section of a plastics plant puts out a continuous sheet of plastic that is \(1.2 \mathrm{~m}\) wide and \(2 \mathrm{~mm}\) thick at a rate of \(15 \mathrm{~m} / \mathrm{min}\). The temperature of the plastic sheet is \(90^{\circ} \mathrm{C}\) when it is exposed to the surrounding air, and the sheet is subjected to air flow at \(30^{\circ} \mathrm{C}\) at a velocity of \(3 \mathrm{~m} / \mathrm{s}\) on both sides along its surfaces normal to the direction of motion of the sheet. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in \(2 \mathrm{~s}\). Determine the rate of heat transfer from the plastic sheet to the air.

Short Answer

Expert verified
Solution: Using the calculated values from the previous steps, we can find the rate of heat transfer for both cases: For the water over the plate: \(h_{water} \approx \frac{23.26 \times 30.0 \times 33.3^{\circ} \mathrm{C}}{1.0 \mathrm{~m}} = 23,228.38 \mathrm{~W/m}\) Rate of heat transfer per unit width \(= 23,228.38 \mathrm{~W/m} \times 33.3^{\circ} \mathrm{C} \times 1.0 \mathrm{~m} = 772,593.54 \mathrm{~W/m^2}\) For the air over the plastic sheet: \(h_{air} \approx \frac{23.26 \times 3.0 \times 60^{\circ} \mathrm{C}}{2 \mathrm{~s}} = 2081.4 \mathrm{~W/m}\) Total rate of heat transfer \(= (2081.4 \mathrm{~W/m} \times 60^{\circ} \mathrm{C}) \times 1.2 \mathrm{~m} = 149,860.8 \mathrm{~W/m^2}\) The rate of heat transfer per unit width of the plate for the water is 772,593.54 W/m², and for the air over the plastic sheet is 149,860.8 W/m².

Step by step solution

01

Calculate the temperature difference for water and air

First, we need to determine the temperature difference between the water flowing above the plate and the plate's surface temperature, and between the plastic sheet and the air. For the water over the plate: Temperature difference = \(43.3^{\circ} \mathrm{C} - 10.0^{\circ} \mathrm{C} = 33.3^{\circ} \mathrm{C}\) For the plastics plant: Plastic sheet temperature \(=90^{\circ} \mathrm{C}\) Air temperature \(=30^{\circ} \mathrm{C}\) Temperature difference = \(90^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C} = 60^{\circ} \mathrm{C}\)
02

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient \((h)\) can be estimated for both cases (water over the plate and air over the plastic sheet) using empirical correlations related to the flow conditions, geometry, and fluid properties. For simplicity, we assume the average convective heat transfer coefficient over the length of the plate and the plastic sheet. For the water over the plate, we will use the Dittus Boelter Equation: \(h_{water} \approx \frac{23.26 \times 30.0 \times 33.3^{\circ} \mathrm{C}}{1.0 \mathrm{~m}}\) For the air over the plastic sheet, we will use the same equation: \(h_{air} \approx \frac{23.26 \times 3.0 \times 60^{\circ} \mathrm{C}}{2 \mathrm{~s}}\)
03

Calculate the rate of heat transfer for both cases

Now we can calculate the rate of heat transfer per unit width by multiplying the convective heat transfer coefficient with the temperature difference and the length of the plate. For the water over the plate: Rate of heat transfer per unit width \(= h_{water} \times (33.3^{\circ} \mathrm{C}) \times 1.0 \mathrm{~m}\) For the air over the plastic sheet: Total rate of heat transfer \(= (h_{air} \times 60^{\circ} \mathrm{C}) \times 1.2 \mathrm{~m}\) Use the calculated values of \(h_{water}\) and \(h_{air}\) to find the heat transfer rate for both cases.

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Most popular questions from this chapter

Engine oil at \(105^{\circ} \mathrm{C}\) flows over the surface of a flat plate whose temperature is \(15^{\circ} \mathrm{C}\) with a velocity of \(1.5 \mathrm{~m} / \mathrm{s}\). The local drag force per unit surface area \(0.8 \mathrm{~m}\) from the leading edge of the plate is (a) \(21.8 \mathrm{~N} / \mathrm{m}^{2}\) (b) \(14.3 \mathrm{~N} / \mathrm{m}^{2}\) (c) \(10.9 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.5 \mathrm{~N} / \mathrm{m}^{2}\) (e) \(5.5 \mathrm{~N} / \mathrm{m}^{2}\) (For oil, use \(\nu=8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}\) )

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.1-\mathrm{m}\)-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(5 \mathrm{~cm}\) from the leading edge and \((b)\) the average convection heat transfer coefficient over the entire plate.

What is the effect of streamlining on \((a)\) friction drag and \((b)\) pressure drag? Does the total drag acting on a body necessarily decrease as a result of streamlining? Explain.

Reconsider Prob. 7-67E. Using EES (or other) software, investigate the effects of air temperature and wind velocity on the rate of heat loss from the arm. Let the air temperature vary from \(20^{\circ} \mathrm{F}\) to \(80^{\circ} \mathrm{F}\) and the wind velocity from \(10 \mathrm{mph}\) to \(40 \mathrm{mph}\). Plot the rate of heat loss as a function of air temperature and of wind velocity, and discuss the results.

Warm air is blown over the inner surface of an automobile windshield to defrost ice accumulated on the outer surface of the windshield. Consider an automobile windshield \(\left(k_{w}=\right.\) \(0.8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R})\) with an overall height of 20 in and thickness of \(0.2\) in. The outside air ( \(1 \mathrm{~atm})\) ambient temperature is \(8^{\circ} \mathrm{F}\) and the average airflow velocity over the outer windshield surface is \(50 \mathrm{mph}\), while the ambient temperature inside the automobile is \(77^{\circ} \mathrm{F}\). Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as a flat plate surface.
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