91Ó°ÊÓ

A hair dryer may be idealized as a circular doct through which a mmall fan draws ambient air and withie which the air is heated as it flows over a coiled electric resistance wire. (a) If a dryer is designed to operate with an electric power consumption of \(P_{\text {tesc }}=500 \mathrm{~W}\) and to heat air from an ambient temperature of \(T_{i}=20^{\circ} \mathrm{C}\) to a discharge temperature of \(T_{n}=45^{\circ} \mathrm{C}\), at what volumetric flow rate \(\forall\) should the fan operate? Heat loss from the casing to the ambient air and the surroundings nuy be neglected. If the duct has a diameler of \(D=70 \mathrm{~mm}\), what is the discharge velocity \(V_{0}\) of the uir? The density and specific heat of the air maly be approximated as \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1007\) \(\mathrm{J} / \mathrm{Kg}+\mathrm{K}\), respectively. (b) Consider a dryer duct length of \(L=150 \mathrm{~mm}\) and a surface cmisuivity of \(e=0.8\). If the coeflicient associaled with heat transfer by natural convection from the casing to the ambient air is \(h=4 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\) and the temperature of the air and the surroundings is \(T_{\mathrm{w}}=\) \(T_{w e}=20^{\circ} \mathrm{C}\), confirm that the heat loss from the casing is, in fact, negligitle. The casing may be assumed to have an average surface temperature of \(T_{3}=40^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate Required Heat Transfer Rate

We use the formula for the heat transfer rate: \( Q = \dot{m} c_p \Delta T \), where \( Q \) is the heat transfer rate, \( \dot{m} \) is the mass flow rate, \( c_p \) is the specific heat capacity, and \( \Delta T \) is the temperature change. Here, the electric power \( P_{e} = 500 \) W is the rate of heat transfer. \( T_i = 20^{\circ}C \) and \( T_n = 45^{\circ}C \), so \( \Delta T = 25 \) K. Thus, \( \dot{m} = \frac{Q}{c_p \Delta T} = \frac{500}{1007 \times 25} = 0.0199 \) kg/s.

02

Calculate Volumetric Flow Rate

The volumetric flow rate \( \dot{V} \) is related to the mass flow rate \( \dot{m} \) and the air density \( \rho \) by the equation \( \dot{V} = \frac{\dot{m}}{\rho} \). With \( \rho = 1.10 \) kg/m³, \( \dot{V} = \frac{0.0199}{1.10} \approx 0.0181 \) m³/s.

03

Calculate Discharge Velocity

Use the equation \( \dot{V} = A \times V_0 \), where \( A \) is the cross-sectional area of the duct and \( V_0 \) is the discharge velocity. The area \( A = \frac{\pi D^2}{4} \) for \( D = 0.07 \) m. So, \( A \approx 0.00385 \) m². Then, \( V_0 = \frac{\dot{V}}{A} = \frac{0.0181}{0.00385} \approx 4.70 \) m/s.

04

Evaluate Heat Loss from the Casing

Use the formula for heat loss by convection \( Q_{c} = h A_s (T_s - T_\text{ambient}) \), where \( h = 4 \) W/m²-K, \( A_s = \pi D L \), \( T_s = 40^{\circ}C \), and \( T_\text{ambient} = 20^{\circ}C \). Substitute \( D = 0.07 \) m and \( L = 0.15 \) m to get \( A_s \approx 0.03299 \) m², and then \( Q_{c} = 4 \times 0.03299 \times 20 = 2.64 \) W. This is negligible compared to the power input of 500 W.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volumetric Flow Rate

Volumetric flow rate is a key concept in fluid dynamics, especially when calculating the flow of air through ducts such as in a hair dryer. It is defined as the volume of fluid passing through a cross-section per unit time and is denoted by \( \dot{V} \). Knowing the volumetric flow rate helps in understanding how efficiently a system delivers air or liquid.

To calculate volumetric flow rate, you use the relationship between mass flow rate and the fluid's density. The formula is:

• \( \dot{V} = \frac{\dot{m}}{\rho} \)

where \( \dot{m} \) is the mass flow rate and \( \rho \) is the density of the air. This equation is essential for designing systems where a specific amount of substance needs to be delivered in a given time.

In practice, this means that once you know how much mass of air (\( \dot{m} \)) is being moved through the hair dryer and the density of air (\( \rho = 1.10 \text{ kg/m}^3 \)), you can calculate how much volume is being handled. For the given problem, substituting the known values gives a volumetric flow rate of approximately 0.0181 m³/s, indicating the rate at which air passes through the dryer.

Specific Heat Capacity

Specific heat capacity is a thermal property that defines how much energy is required to change the temperature of a unit mass of a substance by one degree Celsius. It is a crucial factor in heating calculations, such as those involved in the hair dryer example. The specific heat capacity of air is denoted by \( c_p \), which for air is taken as 1007 J/(kg·K).

The specific heat capacity determines how much heat (energy) is needed to achieve a desired temperature change, \( \Delta T \). The formula used here is:

• \( Q = \dot{m} c_p \Delta T \)

where \( Q \) is the heat transfer rate, \( \dot{m} \) is the mass flow rate, and \( \Delta T \) is the temperature rise needed.

This formula reflects how efficient a system can heat a substance, like the air in a hair dryer. A higher specific heat capacity means more energy is needed to achieve the same temperature change. In this scenario, to raise the air temperature from 20°C to 45°C (\( \Delta T = 25 \) K), the formula helps in figuring out the power needed, given the mass flow rate, which directly relates to the volumetric flow rate calculation.

Natural Convection

Natural convection is a process whereby heat is transferred through a fluid without the aid of external forces, such as fans or pumps. It occurs due to the natural circulation that arises from the temperature difference between a surface and the fluid in contact with it.

In the case of the hair dryer, heat loss from the casing to the ambient environment is considered through natural convection. The formula used for this heat transfer is:

• \( Q_c = h A_s (T_s - T_{\text{ambient}}) \)

where \( h \) is the convective heat transfer coefficient, \( A_s \) is the surface area, \( T_s \) is the surface temperature, and \( T_{\text{ambient}} \) is the ambient temperature.

Natural convection is significant in scenarios where passive heat dissipation is needed. For this context, even though the casing's temperature is higher than the ambient air, the amount of heat loss calculated is relatively small (2.64 W), showing its negligible impact compared to the supplied electric power (500 W). Understanding this concept aids in identifying potential efficiency losses in thermal systems and ensuring designs are optimally balanced between passive and active heating methods.