Question

hair dryer may be idealized as a circular duct through which a small fan draws ambient air and within which the air is heated as it flows over a coiled electric resistance wire.

(1) If a dryer is designed to operate with an electric power consumption of Pelec= 1500 W and to heat air from an ambient temperature of Ti= 20°C to a discharge temperature of To= 45°C, at what volumetric flow rate ∀˙ should the fan operate? Heat loss from the casing to the ambient air and the surroundings may be neglected. If the duct has a diameter of D= 50 mm, what is the discharge velocity Vo of the air? The density and specific heat of the air may be approximated as ρ= 1.10 kg/m³ and cp= 1007 J/kg·K, respectively.

(2) Consider a dryer duct length of L= 150 mm and a surface emissivity of ɛ= 0.8. If the coefficient associated with heat transfer by natural convection from the casing to the ambient air is h= 4 W/m²·K and the temperature of the air and the surroundings is ∞T∞=Tsur= 20°C, confirm that the heat loss from the casing is, in fact, negligible. The casing may be assumed to have an average surface temperature of Ts= 40°C.

Step 1

Find the mass flow rate of the air, in kg/s.

m˙= kg/s

Step 2

What is the volume flow rate of the air, in m³/s?

∀˙= m³/s

Step 3

What is the discharge velocity of the air, in m/s?

Vo= m/s

Step 4

Find the heat transfer from the casing by convection, in W.

qconv= W

Step 5

Find the heat transfer from the casing by radiation, in W.

qrad= W

Step 6

What is the total heat transfer from the casing, in W?

q= W

Answer 1

1.

Pelec= 1500 W

Ti= 20°C

To= 45°C

cp= 1007 J/kg·K

ρ= 1.10 kg/m3

V = ?

Pelec = m cp (T0 - Ti)

1500 = m * 1007 *(45 - 20)

m = 0.05958 kg/s ---- Step 1

V = m / ρ = 0.05958 / 1.1

V = 0.054163 m3/s   ---- Step 2

Flow rate = Area * Velocity

V = A * v

D = 50mm or 0.05m

A = pi/4 D2 = pi/4 * 0.052 = 0.00196349 m2

0.054163 = 0.00196349 * v

v = 27.5866652 m/s    ---- Step 3

h = 4 W/m2-K

Ts = 400C

T∞ = 20

Heat transfer by convection

Qconv = h As (Ts - T∞)

where,

As = pi * D * L = pi * 0.05 * 0.15 = 0.02356194 m2

Qconv = 4 * 0.02356194 (40 - 20)

Qconv = 1.8849555 W   ---- Step 4

Heat transfer by radiation

Qrad = ɛ * As * σ * (Ts4 - T∞4)

ɛ = 0.8

σ = 5.67 x 10 -8  

Qrad = 0.8 * 0.02356194 * 5.67 x 10 -8 * (3134 - 2934)

Qrad = 2.3810 W ---- Step 5

Qtotal = Qconv + Qrad

Qtotal = 1.8849555 + 2.3810

Qtotal = 4.2659 W   ---- Step 6

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