Respuesta :
Answer:
(a)[tex]A_1=0.26 m^2[/tex]
(b)Q= -35.69 KW
Explanation:
Given:
[tex]P_1=350 KPa,T_1=420 K,V_1=3 m/s,T_2=300 K,V_2=460 m/s[/tex]
We know that foe air [tex]C_p=1.011\frac{KJ}{kg-k}[/tex]
Mass flow rate for air =2.3 kg/s
(a)
By mass balancing [tex]\dot{m}=\dot{m_1}\dot{m_2}[/tex]
[tex]\dot{m}=\rho AV[/tex]
[tex]\rho_1A_1V_1=\rho_2A_2V_2[/tex]
[tex]\rho_1 =\dfrac {P_1}{RT_1},R=0.287\frac{KJ}{kg-K}[/tex]
[tex]\rho_1 =\dfrac {350}{0.287\times 420}[/tex]
[tex]\rho_1=2.9\frac{kg}{m^3}[/tex]
[tex]\dot{m}=\rho_1 A_1V_1[/tex]
[tex]2.3=2.9\times A_1\times 3[/tex]
[tex]A_1=0.26 m^2[/tex]
(b)
Now from first law for open(nozzle) system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}[/tex]
Δh=[tex]C_p(T_2-T_1)\frac{KJ}{kg}[/tex]
[tex]1.011\times 420+\dfrac{3^2}{2000}+Q=1.011\times 300+\dfrac{460^2}{2000}[/tex]
Q=-15.52 KJ/s
⇒[tex]Q= -15.52\times 2.3[/tex] 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.
Answer:
A) A1 ==0.2829 m^2
B) [tex]\frac{dQ}{dt} = -105.5 kW[/tex]
Explanation:
A) we know from continuity equation
[tex]\frac{dm}{dt} = \frac{A_1 v_1}{V_1}[/tex]
solving for A1
[tex]A_1 = \frac{\frac{dm}{dt} V_1}{v_1}[/tex]
we know V = \frac{RT}{P} as per ideal gas equation, so we have
[tex]A_1 = = \frac{\frac{dm}{dt} \frac{RT_1}{P_1}}{v_1}[/tex]
[tex]= \frac{2.3 \frac{0.287 \times 450}{350}}{3}[/tex]
=0.2829 m^2
b) the energy balanced equation is
[tex]\frac{dQ}{dt} = \frac{dm}{dt} ( Cp(T_2 -T_1) + \frac{V_2^2 - V_1^2}{2})[/tex]
[tex]= 2.3 ( 1.011(300 - 450) + [\frac{460^2+3^2}{2}])[/tex]
[tex]\frac{dQ}{dt} = -105.5 kW[/tex]
