Problem 4.61
(Difficulty: 3)
4.61 Two types of gasoline are blended by passing them through a horizontal “wye” as shown. Calculate
the magnitude and direction of the force exerted on the “wye” by the gasoline. The pressure 𝑝3 =
145 𝑘𝑘𝑘.
Given: The pressure 𝑝3 = 145 𝑘𝑘𝑘.All the other parameters are shown in the figure.
Find: The force on the bend.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations: Continuity
Bernoulli equation;
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
The area for the inlet section and outlet section are:
𝐴1 =
𝜋 2 𝜋
𝐷 = × (0.2 𝑚)2 = 0.0314 𝑚2
4 1 4
𝐴3 =
𝜋 2 𝜋
𝐷 = × (0.2 𝑚)2 = 0.0314 𝑚2
4 3 4
𝐴2 =
𝜋 2 𝜋
𝐷 = × (0.1 𝑚)2 = 0.0079 𝑚2
4 2 4
The velocity at each section can be calculated by:
𝐿
𝑚3
30
30 × 10−3
𝑄1
𝑠
𝑠 = 0.955 𝑚
=
=
𝑉1 =
2
2
𝑠
𝐴1 0.0314 𝑚
0.0314 𝑚
𝐿
𝑚3
3.4
3.4 × 10−3
𝑄2
𝑠 =
𝑠 = 0.430 𝑚
𝑉2 =
=
2
2
𝑠
𝐴2 0.0079 𝑚
0.0079 𝑚
𝑉3 =
𝑄3 𝑄1 + 𝑄2
=
=
𝐴3
𝐴3
30 × 10−3
The pressure at the outlet is:
𝑚3
𝑚3
+ 3.4 × 10−3
𝑠
𝑠 = 1.064 𝑚
2
𝑠
0.0314 𝑚
The density of the gas:
𝑝3 = 145 𝑘𝑘𝑘
From the Bernoulli equation:
𝜌 = 680.3
𝑘𝑘
𝑚3
𝑝1 𝑉12 𝑝2 𝑉22 𝑝3 𝑉32
+
= +
= +
𝜌
2
𝜌
2
𝜌
2
𝑘𝑘
680.3 3
𝜌 2
𝑚 2
𝑚 2
𝑚
𝑝1 = 𝑝3 + (𝑉3 − 𝑉12 ) = 145 𝑘𝑘𝑘 +
× ��1.064 � − �0.955 � � = 145.08 𝑘𝑘𝑘
2
𝑠
𝑠
2
𝑘𝑘
680.3 3
2
2
𝜌 2
𝑚 × ��1.064 𝑚� − �0.43 𝑚� � = 145.33 𝑘𝑘𝑘
𝑝2 = 𝑝3 + (𝑉3 − 𝑉22 ) = 145 𝑘𝑘𝑘 +
2
𝑠
𝑠
2
The mass flow rates are
𝑚̇1 = 𝜌𝑉1 𝐴1 = 680.3
𝑚̇2 = 𝜌𝑉2 𝐴2 = 680.3
𝑘𝑘
𝑚
𝑘𝑘
× 0.955 × 0.0314 𝑚2 = 20.40
3
𝑚
𝑠
𝑠
𝑘𝑘
𝑚
𝑘𝑘
× 0.430 × 0.0079 𝑚2 = 2.311
3
𝑚
𝑠
𝑠
𝑚̇3 = 𝜌𝑉3 𝐴3 = 680.3
From the x momentum equation, we have:
𝑘𝑘
𝑚
𝑘𝑘
× 1.064 × 0.0314 𝑚2 = 22.72
3
𝑚
𝑠
𝑠
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 cos 30° 𝑚̇1 −𝑉2 cos 45° 𝑚̇2 + 𝑉3 𝑚̇3
𝑅𝑥 + 𝑝1 𝐴1 cos 30° + 𝑝2 𝐴2 cos 45° − 𝑝3 𝐴3 = −𝑉1 cos 30° 𝑚̇1 −𝑉2 cos 45° 𝑚̇2 + 𝑉3 𝑚̇3
𝑅𝑥 = −𝑉1 cos 30° 𝑚̇1 −𝑉2 cos 45° 𝑚̇2 + 𝑉3 𝑚̇3 + 𝑝3 𝐴3 − 𝑝1 𝐴1 cos 30° − 𝑝2 𝐴2 cos 45°
𝑅𝑥 = −197.4 𝑁
From the y momentum equation, we have:
𝐹𝑥 = 197.4 𝑁
𝑅𝑦 + 𝐹𝑠𝑠 − 𝐹𝐵𝐵 = 𝑉1 𝑚̇1 sin 30° − 𝑉2 𝑚̇2 sin 45°
𝑅𝑦 − 𝑝1 𝐴1 sin 30° + 𝑝2 𝐴2 sin 45° = 𝑉1 𝑚̇1 sin 30° − 𝑉2 𝑚̇2 sin 45°
𝑅𝑦 = 𝑉1 𝑚̇1 sin 30° − 𝑉2 𝑚̇2 sin 45° + 𝑝1 𝐴1 sin 30° − 𝑝2 𝐴2 sin 45°
𝑅𝑦 = 1475 𝑁
So the force can be computed by:
𝐹𝑦 = −1475 𝑁
𝐹 = �𝐹𝑥2 + 𝐹𝑦2 = �(197.4 𝑁)2 + (−1475 𝑁)2 = 1488 𝑁
The direction is calculated by (shown in the figure):
tan 𝛼 =
𝐹𝑦
= 7.4721
𝐹𝑥
𝛼 = 82.4°
𝐹𝑥
𝛼
𝐹𝑦
𝐹