Question

Air at 300 K and 100 kPa steadily flows into a hair dryer having electrical work input of 1500 W. Because of the size of the air intake, the inlet velocity of the air is negligible. The air temperature and velocity at the hair dryer exit are 80°C and 21 m/s, respectively. The flow process is both constant pressure and adiabatic. Assume air has constant specific heats evaluated at 300 K. Determine the air mass flow rate into the hair dryer, in kg/s.

Answer 1

Answer:

0.0280Kg/s

Explanation:

Given:

W = 1500

V2 = 300

V1 = 0

Q = 0 ( adiabatic)

T1 = 300

T2 = 20 m/s = converting to degrees we have 353°c

Let's use the energy equation

$Q - W = m_• *( Cp(dT) + 0.5*(V2^2 - V1^2) Q = 0$

[/tex] m_• = W / (Cp(dT) + 0.5*V2^2) [/tex]

$= 1500 / (1005(353 - 300) + 0.5*21^2)$

$= 1500 W / 53485.5 =0.0280 kg/s$

Answer 2

Answer:

The rate of air mass flow is given as $\dot{m}=0.02804 kg/s^2$

Explanation:

The steady state energy equation is given as

$\dot{Q}-\dot{W}=\dot{m}\left[(h_2-h_1)+\left(\dfrac{V_1^2-V_2^2}{2}\right)+g(z_1-z_2)\right]$

Here,

Q' is heat interaction per unit time with the hair dryer

W' is work interaction per unit time with the system

m' is mass flow rate into the hair dryer

Since the process is adiabatic so the value of Q' is 0 .

The elevations at inlet and outlet are same(z_1=z_2) so z_1-z_2=0

So the equation becomes

$0-\dot{W}=\dot{m}\left[(h_1-h_2)+\left(\dfrac{V_1^2-V_2^2}{2}\right)+g(0)\right]\\\dot{W}=-\dot{m}\left[(h_1-h_2)+\left(\dfrac{V_1^2-V_2^2}{2}\right)\right]\\\dot{W}=\dot{m}\left[(h_2-h_1)+\left(\dfrac{V_2^2-V_1^2}{2}\right)\right]\\\dot{W}=\dot{m}\left[C_p(T_2-T_1)+\left(\dfrac{V_2^2-V_1^2}{2}\right)\right]$

Now the value of power is given as 1500 W so the value of W' is 1500 W

m' is to be calculated

T_2 is the temperature at the exit given as 80 C or 80+273=363K

T_1 is given as 300 K

V_1 is given as negligible so it is 0

V_2 is given as 21 m/s

C_p is given as 1.005 kJ/kgK

So the equation becomes

$\dot{W}=\dot{m}\left[C_p(T_2-T_1)+\left(\dfrac{V_2^2-V_1^2}{2}\right)\right]\\1.5 kW=\dot{m}\left[1.005 kJ/kgK(363-300 )K+\left(\dfrac{21^2-0^2}{2}*(1/1000)\right)\right]\\1.5=\dot{m}(53.48)\\\dot{m}=\dfrac{1.5}{53.48}\\\dot{m}=0.02804 kg/s^2$

So the rate of air mass flow is given as $\dot{m}=0.02804 kg/s^2$