Solucion al Problema #5 de la Asignacion #2

Solucion al Problema #5 de la Asignacion #2
Definimos los parametros conocidos
mass := 1
kg
v1 := 2
s
m
d1 := 6in
s
3
Vt := 1m
P2 := 20Pa
d2 := 2in
Dens := 1000
ang := 25deg
kg
3
m
1. Calculamos la velocidad de salida (v2)
Q1
Q2
v1⋅ A1
v2
v2⋅ A2
v1⋅
d1
π ⋅  
2
v1⋅
A1
A2
d2
π  
2
2
2
v1⋅
2
v2 := v1⋅ 
d1
d1 
2

 d2 
2
d2
v2 = 18
m
s
2. Calculamos la Presion de entrada (P1), utilizamos Bernoulli. La
altura de entrada y salida no varian, asumimos.
P1 := 30Pa
Given
P1 +
1
2
2
Dens ⋅ v1
P2 +
1
2
2
Dens ⋅ v2
5
P1 = 1.6 × 10 Pa
P1 := Find ( P1)
3. Buscamos las Fuerzas (Fx y Fy): Formato 1 usando el masico
Fx := 100N
Given
Fx + P1⋅ π ⋅ 
d1 

2
2
− P2⋅ π ⋅ 
Fy := 100N
d2 
2
 cos ( ang )
2
Fy − Vt⋅ Dens ⋅ g − P2⋅ π ⋅ 
d2 
2
 ⋅ sin ( ang )
2
 Fx 
  := Find ( Fx, Fy )
 Fy 
mass ⋅ ( −v1 + v2⋅ cos ( ang ) )
mass ⋅ v2⋅ sin ( ang )
Fx = −2.905⋅ kN
Fy = 9.814⋅ kN
3a. Buscamos las Fuerzas: Formato 2 usando las velocidades
Fy := 100N
Fx := 100N
Given
Fx + P1⋅ π ⋅ 
2
2
d2
− P2⋅ π ⋅   cos ( ang )

2
2
d1 
Fy − Vt⋅ Dens ⋅ g − P2⋅ π ⋅ 
d2 

2
 Fx 
  := Find ( Fx, Fy )
 Fy 
2
⋅ sin ( ang )
−v1⋅ v1⋅ π ⋅ 
2
2
d2
⋅ Dens + v2⋅ v2⋅ π ⋅   ⋅ Dens ⋅ cos ( ang )

2
2
v2⋅ v2⋅ π ⋅ 
Fx = −2.397⋅ kN
d1 
d2 
2
 ⋅ Dens ⋅ sin ( ang )
2
Fy = 10.084⋅ kN
Ambos resultados son similares, los ordenes de magnitud son en kN.
El programa hace las conversiones correctas y analiza
matematicamente las ecuaciones, por tanto los resultados dados aqui
son finales.