ME200 HW week 5-1_Solution for 4.1, 4.5, 4.7, 4.13

4.1
Given: The mass flow rates into and out of a control volume are known, and the
initial mass is given.
Find: (a) Plot the amount of mass contained in the control volume as function of
time, for t ranging from 0 to 3 h. (b) Estimate the time, in h, when the tank is
nearly empty.
Schematic and given data:
Assumptions: as shown in the schematic, a control volume having one inlet and
one exit is under consideration.
Analysis: the inlet and exit mass flow rates are given,
(a) the mass rate balance,
dmcv
 mi  me  100(1  e 2t )  100
dt
 100e2t
(1)
The mass contained in the control volume at time t is obtained by interating
Eq.(1)
t
mcv(t)-mcv(0)=-100  e 2t dt  50e 2t |t0
0
→mcv(t)=50( e2t -1)+50=50e-2t
(2)
Using Eq.(2), the following plot is obtained for the amount of mass in the tank as a
function of time:
From the plot, we can see that initially, the amount of mass in the tank is
decreasing at a rapid rate. Eventually, the rate of change (decrease) of mass slows
to zero. The tank initially contains 50 kg and empties.
(b) From the plot, we see that the tank is nearly empty at t≈2.5 h.
4.5
Given: A pipe carrying an incompressible liquid contains an expansion chamber.
Find: (a) Develop an expression for the rate of change of liquid level in the
chamber in terms of certain quantities. (b) Compass the relative magnitudes of the
mass flow rates for specified valves for dh/dt, the rate of change of liquid level.
Schematic and given data:
Assumption: 1. The control volume is shown on the accompanying schematic.
2. The liquid is modeled as incompressible.
3. Flow is one-dimensional at 1,2.
Analysis: (a) The mass rate balance for the control volume is
dmcv
 m1  m2
dt
 D2
 D 12
 D 22
) L v , m1  g (
)V1 , m2  g (
)V2
With mcv  g (
4
4
4
The mass rate balance become
  D12 
  D2 2 
  D 2  dL
g
 g
 V1  g 
 V2

 4  dt
 4 
 4 
Or, solving for dL/dt and simplifying
dL D12 V1  D2 2 V2

dt
D2
(b) The mass flow rate expressions indicate m1~D12V1 , m2~D22V2
Thus,
dL
>0 →D12V1 > D22V2→ m1  m2
dt
dL
=0 →D12V1 = D22V2 → m1  m2
dt
dL
<0→D12V1 < D22V2 → m1  m2
dt
4.7
Given: Data are provided for a vegetable oil-filled spray can.
Find: Determine the mass flown rate per spray, and the mass remaining in the can
after a specified number of spray.
Schematic and given data:
Assumption: The control volume is shown in the accompanying schematic.
Analysis:
(a) Since each spray has a duration 0.25 s and consists of 0.25 g.
me =(0.25 g)/(0.25 s)=1 g/s.
(b) The mass rate balance, Eq.4.3, reduce to
mcv(t+  t)-mcv(0)=mi-me
Where mcv(0) is the initial amount mass within the can and me is the amount of
mass that exits. Thus, with mcv(0)=170 g and
me =(560 sprays)(0.25g/spray)=140 g.
The mass of vegitable oil remaining in the can after 560 sprays is
mcv= mcv(0)- me =170 g-140 g=30 g.
4.13
Given: Data are given for air entering and exiting from a fan.
Find: Determine at steady state (a) the mass flow rate, (b) the volumetric flow rate
at the inlet, and (c) the inlet and exit velocities.
Schematic and given data:
Assumption: 1. A control volume enclosing the fan is at steady state.
2. The air behaves as an ideal gas.
Analysis:
(a) The mass balance reduces to: m1=m2. Thus, using data at the exit of the fan
m
 AV 2  AV 2 p2
v2

RT2
 0.35m /s  105kPa 
3
=
 8.314 kJ 
 28.97 kg  K   291K 


103 N/m 2
1kJ
1kPa 103 N  m
=0.44 kg/s
(b) At the inlet
  8.314 

  28.97   289  
RT1


(AV)1= m v1= m (
)=0.44  
p1
101




=0.361 m3/s
(c) The cross-sectional area is
A
→ v1 
v2 
 D2
4

 AV 1
A
 AV 2
A
  0.6 
4

2
=0.2827 m2
0.361m3 /s
=1.28 m/s
0.2827m2
0.35m3 /s
=1.24 m/s

0.2827m2

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