Question

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 420 K, 350 kPa, and velocity of 3 m/s. At the exit, the temperature is 300 K and the velocity is 460 m/s. Using the ideal gas model for air with constant ep=1.011 k/kg. K, determine: (a) the area at the inlet, in m2 (b) the heat transfer to the nozzle from its surroundings, in kW.

Answer 1

A) A1 ==0.2829 m^2

B)
`(dQ)/(dt) = -105.5 kW`

Step-by-step explanation:

A) we know from continuity equation

`(dm)/(dt) = (A_1 v_1)/(V_1)`

solving for A1

`A_1 = ((dm)/(dt) V_1)/(v_1)`

we know V = \frac{RT}{P} as per ideal gas equation, so we have

`A_1 = = ((dm)/(dt) (RT_1)/(P_1))/(v_1)`

`= (2.3 (0.287 * 450)/(350))/(3)`

=0.2829 m^2

b) the energy balanced equation is

`(dQ)/(dt) = (dm)/(dt) ( Cp(T_2 -T_1) + (V_2^2 - V_1^2)/(2))`

`= 2.3 ( 1.011(300 - 450) + [(460^2+3^2)/(2)])`

`(dQ)/(dt) = -105.5 kW`

Answer 2

(a)
`A_1=0.26 m^2`

(b)Q= -35.69 KW

Step-by-step explanation:

Given:

`P_1=350 KPa,T_1=420 K,V_1=3 m/s,T_2=300 K,V_2=460 m/s`

We know that foe air
`C_p=1.011(KJ)/(kg-k)`

Mass flow rate for air =2.3 kg/s

(a)

By mass balancing
`\dot{m}=\dot{m_1}\dot{m_2}`

`\dot{m}=\rho AV`

`\rho_1A_1V_1=\rho_2A_2V_2`

`\rho_1 =\frac {P_1}{RT_1},R=0.287(KJ)/(kg-K)`

`\rho_1 =\frac {350}{0.287* 420}`

`\rho_1=2.9(kg)/(m^3)`

`\dot{m}=\rho_1 A_1V_1`

`2.3=2.9* A_1* 3`

`A_1=0.26 m^2`

(b)

Now from first law for open(nozzle) system

`h_1+(V_1^2)/(2)+Q=h_2+(V_2^2)/(2)`

Δh=
`C_p(T_2-T_1)(KJ)/(kg)`

`1.011* 420+(3^2)/(2000)+Q=1.011* 300+(460^2)/(2000)`

Q=-15.52 KJ/s

⇒
`Q= -15.52* 2.3` KW

Q= -35.69 KW

If heat will loss from the system then we will take negative and if heat will incoming to the system we will take as positive.