Air at 300 K and 100 kPa flows into a hair drier having electrical work input (h

Question

Air at 300 K and 100 kPa flows into a hair drier having electrical work input (heating coil and fan motor) of 1500 W. Because of the size of the air intake, the inlet velocity of air is negligible. The air temperature and velocity at the hair dryer exit are 80oC and 21 m/s, respectively. The flow process both constant pressure and adiabatic. Assume air has constant specific heats at 300 K. Cp=1.005 kj/kg.K and R= 0.287kj/kg.K . a) Determine the air mass flow rate into the hair drier, in kg/s b) Determine the air volume flow rate at the hair drier exit, in m3/s c) Determine the diameter of the drier exit in mm

Explanation / Answer

P1 = 100 Kpa and T1 = 300 K

Work input = 1500W = 1.5 KW

T2 = 80 C = 353 K

now, m.h1 + Work = m.h2 + m.V^2/2

where h = enthalpy = Cp.T

so, m x 1.005 x 300 + 1.500 = m x { 1.005 x 353 + 21^2/2000 }

a) so masss flow rate = m = 0.028 kg/s

b) now P = d.Rair.T

where d = density of air

so, 100 = d x 0.287 x 353

so, d = 0.987 m^3/Kg

so volume flow at exit = m/d = 0.028/0.987 = 0.0283 m^3/s

so, dia = 1.7167 mm

c)volume flow at exit = area x velocity = pi/4 x d^2 x 21 = 0.0283

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.