CHEMENG 1007/1007UAC: INTRODUCTION TO PROCESS ENGINEERING
TUTORIAL 10
MOMENTUM EQUATION
(Due: 12 June 2025)
Attempt all questions and submit your best solutions for assessment.
3.1 An oil (SG = 0.86) flows through the horizontal tee connection as shown in Fig. 3.1. The
discharge flow rate Q3 at section (3) is 2.0 m3/s and each pipe section has an inside diameter
of 0.5 m. Assume the flow is steady, uniform, incompressible and frictionless.
a. Find the flow rate of oil entering the line at section (2).
b. Calculate the pressures at sections (3) and (2).
c. Determine the components, magnitude and direction of the force exerted by the oil on the
tee.
Ans: (a) 0.82 m3/s (b) 171 kPa; 208 kPa (c) Rx = –45.3 kN; Ry = 7.3 kN;–81o from y-axis
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3.2 Calculate the magnitude and direction of the force required to hold the flow device in Fig. 3.2
stationary. The fluid is water at 20oC, and the device and its content weighs approximately
600 kg. Assume that flow is steady, uniform and frictionless, and that pressures at all outlets
are atmospheric. Gravity is perpendicular to the x-y plane (of the paper);
Ans: (a) 7.1 kN; - 14o from x-axis
Page 1 of 1
Area 0.812 0.196
Question 1
z 59 0.86
Q3 2m31s
i I IT
x
V3 10.2m s
10
v 6m s
Q P 2001pA
Q 1.178 a
Assumptions
steady state flow
uniform
Incompressible fluid
No friction loss frictionless
a Find Qz
A a 0.5
4
0.196
Q v A
65 7m
1 178
Qin Qout
Q Q2 Qs
1.178 Q2 2 m S
i Q2 0 8 2 m s
b Find ps and Pa
Find v2
V2 Q2 A
0.82
4 18 mis
Find v3
V3 03 A
2 1 76
10.2m s
Find p
0.86 1000 density water 861
Formula
P Ipv P3 P V22
Pz P p v12 V32
2
P3 200 000 0.5 860 x 6 10.2
171 KPA
P2 P p v v22
0.5 860 62 4.18
200 000
208kpA
C Roc PV Pn A2 PV P A
78 6,2 200000 0.196
45300 N
45.3 KN
Fy paV2A2 vaxson P3V3 As v3 sin
20800064.18 0.196 sino 171000 10.2 0196 10.20518
7300 N
7 3 kW
Angle aratan
2 2.3
2 Diagram
y
x
T 20 C O
p
1 8919 controlvolume
p 0 RPAgang
2
Assumptions
steady state flow
uniform
frictionless
Component
Momentum Momentumin
Force Pin Pout out
Foc P A PaA cos30 P Afios30 Pa Alsin 60
E PV A Vi PV2 Az V2 cos 30 PV3 A3 V3 sin60
V4 cos 30
PVy Ay
Fx PV A V PV Az V2 x cos 30 PV3Az v3 sin6
PV 4 Ay Vux cos 30 P A P2Azcos 30
1012 2 OC DC
1000 2 5 20 1 2 5 sin 60
1000 1 5 0.667 1.5 cos 30
6000 0.25 5000 0.25 cos 30
1000 1948.6 541.3 1299.7 1500 1732
6939 N
Y component
Fy PaAz Sin 30 P3 Az cos 60 RAY sin 30
Pv A o upV2 Az Va sin 30 PV3 As Uxcos 60
PV4A4 V4 sin 30
PV2 A
Fy PV
Afo 2 Vax sin 30 PusAz
P2A2 sin 30
V3 cos 60 PVuAy Vex son 30
1000 3 0.25 3 son 30 1000 2.5 0.1
2.5 cos 60 000 1 5 0.667 1.5 son 30
8000 0 25 sin 30
1125 312.5 750 375 1000
1687 125 N
F total
T.FI
8
nas
1138 N
7 1
Angle are tan
1 7
13.7
