Problemas Resueltos De Fox Mecanica De Fluidos

11.26.-Fox
h=3 km
M=1.35
T=30 C=303 K

Δx=htan∝
c=kRT½

c=1.4*287NmkgK*303K½
c=349ms
V=Mc=1.35*349ms=471ms∝=sin-1(1M)=sin-1(11.35)=47.79
Δx1=3000*tan47.79
Δx1=3.30 km

Ahora en

h=0 m
T=288.2 K
c=(1.4*287NmkgK*288.2K)½
c=340.29 ms
M=Vc=471ms*1340msM=1.38
∝=sin-1(1M)=sin-1(11.38)
∝=46.44
Δx=3000tan46.44
Δx2=3.154 km

Promedio Δx=Δx1+Δx22

Δx=3.21 Km

9.43 .- WHITE

To=300C
A1=A2=25 cm²=25x10-4 m
P1=120KPa
P2=60KPa
solucion;
P2P1=(T2T1)kk-1
12060=T2T13.5
T2T1=1.41
1.41=1+k-12*M22/ 1+k-12*M120.41=0.2M22 -0.141M12
M22=2.05+0.705M12

12060=1+0.2M12 3.51+0.2M223.5
0.21=0.2M12-0.242M22
0.7=0.2M12-0.496-0.121M12
M1=2.96M2=2.86
ahora T2 y T1?
300=T11+0.22.962
T1=169K
300=T21+0.22.862
T1=113K
M2=V/(kRT)½
V=2.86*1.4*287NmkgK*386K½
V2=1126.27ms
m=ῤVAm=PRT2*1126.27ms*25x10-4 m
m=1.2x105 *1287*1386*1126.7*2.5x10-4 m
m=1.52kgs
A=A1M1M2*(T1t2)(k+1)/2(k-1)
A=25x10-4 (2.962.86)0.713A=0.00093 m2

13.43.- Fox
He
Po=8 MPa
To=1000K
Me=3.5
Ae=100mm2

PoPe=(1+k-12*M2)k/(K-1)
8=Pe(1+1.66-12)*3.52)3.58=Pe*5.042.52
Pe=0.136MPa=136KPa

ToTe=1+k-12*M2
1000=Te(1+k-12*3.52)
1000=5.04Te
Te=198.4K

c=kRTe½
c=1.66*2077NmkgK*198.4K½
ce=827msVe=Me*ce=3.5*827=2894.8ms
ῤe=PeRTe=136x10³Nm2*kgm2077kgK*1198.4=0.33kgm3
m=ῤVA
m=0.33kgm3*2894.8ms*100mm2*m2106mm
m=0.096kg/s