CHEMENG 2018: PROCESS FLUID MECHANICS
TUTORIAL 2: CONTROL VOLUME, CONTINUITY EQUATION, ENERGY EQUATIONS
(Due: 5 April 2023)
In your tutorial team, attempt all questions and submit your team’s best solutions to the following problems for assessment: 2.4, 2.5, 2.8, 2.11
Note that all numerical answers given are approximate only – and may even be incorrect!
2.2 Water (density = 1000 kg/m3) flows with a uniform velocity 2 m/s through the 6-m wide and 0.5-m high rectangular channel, as shown in Fig. 2.2
a. Determine the mass flow rate across section CD of the control volume ABCD by
integratingBout = A bvnˆdA= A bv.cosdA,whereparameterb=1. .
b. Repeat part (a) with b =1/, where is the fluid density.
c. Explain the physical meaning of the quantity calculated in part (b).
Ans: (a) 6000 kg/s; (b) 6 m3/s
Fig. 2.2 Fig. 2.3
2.3 A process gas (MW = 28) flows in a long, straight section of 300-mm diameter (D) pipe. The temperatures and pressures at sections (1) and (2) are as indicated in Fig. 2.3. If the average gas velocity at section (1) is 205 m/s, determine:
a. The average gas velocity at section (2);
b. The mass flow rates of gas (in kg/h) at section (1) and section (2).
Assume steady state and ideal gas behaviour.
a. Mass flow rate of the mixture;
b. Mixture density, and
c. Diameter of the exit pipe. Ans. (a) 173 kg/s; (b) 961 kg/m3; (c) 125 mm
C2.4 Two immiscible liquids are mixed in the Y-shaped pipe illustrated in Figure 2.4. The first liquid has a specific gravity of 1.05 and enters the device with a flowrate of 6000 L/min. The second liquid has a specific gravity of 0.85 and enters the device with a flow rate of 4800 L/min. If the mixture exits the device at a velocity of 15 m/s, calculate
Ans: 315 m/s; 50,500 kg/h
Page 1 of 3
iven:4)1-E) m
=8 π f ( –
) volumetricflow
=5 0 m m =0 .
r a t e :d e n s i n g
r a t e A:
=3 . 9 2 x
=7 5 m m =0 .
xA r e a x v e l o c i t y I
R 1 =05. 4 2=
0.0375)” x0.888
flow r a t e : r2
r =4(1-Er),°d
=fzuzA2, Ac =T R
2 =12ULTR2,
249, R i =f2UcTR2 2 R=u 2
Programming Help, Add QQ: 749389476
Q = 4800 L/min
r2 v=4.01−R2 m/s
where R is the radius of the inlet tube, and r the radial distance from the centreline of the tube. The diameter of the inlet tube is 50 mm, while the outlet tube is 75 mm in diameter. The velocity in the outlet tube can be assumed to be uniform. Calculate:
&2.5 Water flows steadily through a pump as shown in Fig. 2.5. The velocity in the outlet tube (section 2) is uniform, but the velocity distribution in the inlet tube is parabolic, with the velocity (in m/s) given by:
The velocity of water in the outlet tube.
The volumetric flow rate of water through the pump in (m3/h).
Ans: (a) 0.89 m/s; (b) 14 m3/h
2.6 In a rocket motor, 10 kg/s of liquid oxygen and 2 kg/s of liquid hydrocarbon fuel are fed into the combustion chamber, as shown in Fig. 2.6. The gaseous products of combustion flow out of the exhaust nozzle at a high velocity. The pressure and temperature of the gases at the nozzle exit plane are 100 kPa (abs) and 800oC, respectively. The nozzle exit has a diameter of 200 mm. Assuming one-dimensional steady flow, with the exhaust gases behaving as an ideal gas having MW of 17, calculate the exit gas velocity.
Ans: 2 km/s
2.7 Water flows through a hydraulic turbine at the rate of 10 m3/s. The inlet pressure and velocity are 250 kPa and 10 m/s, respectively. The discharge pressure and velocity are – 90 kPa and 1.0 m/s, respectively. Find the maximum power developed by the turbine.
Ans: 3.9 MW
Page 2 of 3
V = 15 m/s
Q = 6000 L/min
&2.8 Water at 20°C is pumped at a rate of 3000 L/min from a lower reservoir to an upper open
tank, as shown in Fig. 2.8. The specific energy loss (ghL) by friction in pipe flow is given by:
gh = 0.25(L D)(v2 2) , where L is the pipe length (in m), D is the pipe diameter (in m), and v L
is the average velocity of water through the pipe (in m/s). The piping system consists of 30 m of 250-mm ID pipe on the discharge side of the pump and 8 m of the same pipe on the suction side. Other relevant data are indicated in the diagram.
Calculate the power (in kW) that must be delivered to the water.
Calculate the pressures at the suction (inlet) and discharge (outlet) ends of the pump.
Ans: (a) 14.7 kW (b) 83.5 kPa & 378 kPa
water siphon consists of a 75-mm inside diameter pipe connected to a large tank as shown Fig. 2.10 below. Assume velocity of water is uniform through the tube.
Determine the maximum rate of discharge, and state the conditions under which this applies.
If the actual rate of discharge is 900 L/min, calculate the total head loss, hL,(in m of water) and the rate of energy loss (in W).
For the same operating condition as in (b) determine the pressure (in kPa) at location (2) of the siphon. Hint: a new control volume enclosing sections (2) and (3) would be useful for this part.
Ans: (a) 1440 L/min (b) 0.91 m; 134 W (c) – 30 kPa A2 = 0.007 m2
p2 = 35 kPa
A3 = 0.02 m2 v3 =6m/s
A1 = 0.1 m2 p1 = 70 kPa Q1
C2.11 Water flows through the horizontal branching pipe in the pipe network shown in Fig. 2.11 at a rate (Q1) of 24.0 m3/min as measured at section (1). Assuming viscous effects are negligible, determine the water velocity at section (2), the pressure at section (3), and the flow rate at section (4).
Ans: v2 = 9.3 m/s; p3 = 60 kPa; Q4 = 12.9 m3/min Page 3 of 3
I =1000kg/m
m fiirpump= 100 x 9.01×36.04627 =35361.4W
3.53614 x 10″kn
Z A =0 , Z B
00x,3 ix Y
=3 0 0 0 2 / m i n
+2x+ Wpump=+
=0 , V A =V B =0
5 x ( ) x*
(ing)(36,04627)
=0.025/Yx)Yig
:36.04627m
+ 8 1 =3 0 m
o f P i p e =2 5 0 m m
0 x 1 0 =1 0 0
p i p e =3 0
程序代写 CS代考 加QQ: 749389476
g+ *, Wpump:
P y – 4 3 =1 0 0 0
4 3 =4 A + P
44 =353613.90
43 =23.8823kPa
rpump=*g++
.81×36.04627
+ 2 x =2 g + +
z3= -(12-3)=
h1 =0.25(5(2,1
+ Wpump=fgt
Px-P3 =f9Wpump
2.30 =3774
浙大学霸代写 加微信 cstutorcs
A 1 =0 . 1
+0 , z 1 hy=
p 1 =7 0 k P a
& 1 =0 . 4 m
A s =0 . 0 2 m V i =6 m / s
#h0 2 1 =z z , t =
=2″000 kg/m3 Vi =4m/s
4 2 =3 5 k P a
Q :velocityxcross
10 a KymsX
doxc0Pax =6 0 k 4 a
=9 . 2 7 m / s
=V 3 x A 3
Q 4 =0 . 2 1 5 1
&4 =0.215/my,x . . Q G =1 2 . 9 m m i n
– 10.0639 + 0 . 1 2 =0.4 0.1849