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Chapter 7: Problem 80

A 10 -cm-diameter, 30-cm-high cylindrical bottle contains cold water at \(3^{\circ} \mathrm{C}\). The bottle is placed in windy air at \(27^{\circ} \mathrm{C}\). The water temperature is measured to be \(11^{\circ} \mathrm{C}\) after \(45 \mathrm{~min}\) of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity.

Short Answer

Expert verified
The main factors affecting the convective heat transfer between a cylindrical bottle filled with water and air are the temperature difference between the solid surface and the fluid, the surface area of contact, and a heat transfer coefficient that characterizes the specific system. By following a step by step solution, we can estimate the average wind velocity based on the given data, such as initial and final water temperatures, air temperature, cooling duration, and dimensions of the bottle, and by using empirical correlations for forced convection in air.

Step by step solution

01

Define unknowns and given data

We are given: - Initial water temperature: \(T_{w1} = 3^{\circ}\mathrm{C}\) - Final water temperature: \(T_{w2} = 11^{\circ}\mathrm{C}\) - Air temperature: \(T_\infty = 27^{\circ}\mathrm{C}\) - Cooling duration: \(t = 45\) min - Diameter of the bottle: \(D = 10\) cm - Height of the bottle: \(H = 30\) cm - Neglected radiation effects We are asked to determine the average wind velocity \(V\).
02

Calculate the heat transferred

The mass of the water in the bottle can be calculated as follows: \(V_{water} = \pi(\frac{D}{2})^2 \times H\) To find the mass of the water, we use the density of water (\(\rho_{water}\)) and the volume of the water (\(V_{water}\)): \(m = \rho_{water} \times V_{water}\) Next, we can calculate the amount of heat transferred (\(Q\)) using the mass of the water, specific heat capacity of water (\(c_p\)), and temperature difference: \(Q = m \times c_p \times \Delta T\) Where \(\Delta T = T_{w2} - T_{w1}\).
03

Calculate the average heat transfer rate

The average heat transfer rate can be calculated as follows: \(\bar{q} = \dfrac{Q}{t}\)
04

Calculate the convective heat transfer coefficient

As we consider only convective heat transfer from the lateral surface area of the bottle, we can write the expression for the heat transfer rate as: \(\bar{q} = h A_s (T_\infty - T_s)\) Where \(h\) is the convective heat transfer coefficient, \(A_s\) is the lateral surface area of the bottle, \(T_\infty\) is the temperature of the air and \(T_s\) is the average surface temperature of the bottle. We can take the average surface temperature of the bottle \(T_s = \dfrac{T_{w1}+T_{w2}}{2}\). The lateral surface area of the bottle can be calculated as follows: \(A_s = \pi D \times H\) Solve the equation for the convective heat transfer coefficient \(h\): \(h = \dfrac{\bar{q}}{A_s (T_\infty - T_s)}\)
05

Estimate the average wind velocity

To solve for the wind velocity, we can use the relationship between the convective heat transfer coefficient (\(h\)) and the wind velocity, which can be given by an empirical correlation for forced convection in air: \(h = C \times V^n\) C and n are constants that depend on the flow regime (laminar or turbulent). For typical wind speeds and cylinder sizes, we can assume a turbulent flow regime and use \(C \approx 0.029\) and \(n = 0.8\). We can now solve for the wind velocity \(V\): \(V = (\dfrac{h}{C})^\frac{1}{n}\) By calculating the required values in each step, we will arrive at an estimate of the average wind velocity during the cooling process.

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Most popular questions from this chapter

Hot engine oil at \(150^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(2 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.5-\mathrm{m}-\) long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(0.2 \mathrm{~m}\) from the leading edge and the average convection heat transfer coefficient, and (b) repeat part ( \(a\) ) using the Churchill and Ozoe (1973) relation.

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m² \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m² \(\mathrm{K}\) (g) Rate of convective heat transfer, W

What is lift? What causes it? Does wall shear contribute to the lift?

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, 1.7-m-long smooth pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. The drag force exerted on the pipe by the air is (a) \(0.4 \mathrm{~N}\) (b) \(1.1 \mathrm{~N}\) (c) \(8.5 \mathrm{~N}\) (d) \(13 \mathrm{~N}\) (e) \(18 \mathrm{~N}\) (For air, use \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=1.269 \mathrm{~kg} / \mathrm{m}^{3}\) )

Conduct this experiment to determine the heat loss coefficient of your house or apartment in \(\mathrm{W} /{ }^{\circ} \mathrm{C}\) or \(\mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\). First make sure that the conditions in the house are steady and the house is at the set temperature of the thermostat. Use an outdoor thermometer to monitor outdoor temperature. One evening, using a watch or timer, determine how long the heater was on during a 3 -h period and the average outdoor temperature during that period. Then using the heat output rating of your heater, determine the amount of heat supplied. Also, estimate the amount of heat generation in the house during that period by noting the number of people, the total wattage of lights that were on, and the heat generated by the appliances and equipment. Using that information, calculate the average rate of heat loss from the house and the heat loss coefficient.
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