Homework Set 6
1. Air enters a compressor operating at steady state with
a pressure of 14.7 lbf/in^2 , a temperature of 80F, and a
volumetric flowrate of 1000 ft^3/min (cfm). At the exit,
the pressure is 100 lbf/in^2, and the temperature is 400F.
Find the diameter of the exit duct which gives rise to an
exit velocity of 700 ft/sec. Consider air to be an ideal gas.
Solution strategy:
There is a simple but important relationship between
the mass flowrate mdot (lbm/sec) and the volumetric flowrate
voldot (ft^3/sec). It is
mdot = (voldot)/v
where v is the specific volume. Since
mdot = (velocity x cross-sectional area)/v
it follows that
voldot = velocity x cross-sectional area
If there is steady flow and if there are no leaks in the wall
of a pipe or duct, then
mdot = constant
However, since v may vary with position along a pipe or duct,
it follows that
voldot is not a constant
To start the problem solution, we note that voldot = 1000
cfm at the inlet state (state 1). To find mdot, we calculate v1
from the ideal gas law:
v1 = R(T1)/(p1) = (53.33)(80 + 459.7)/(14.7)(144)
= 13.60 ft^3/lbm
and
mdot = [(voldot)1]/(v1) = 1000/13.60
= 1.225 lbm/sec
At the exit state (state 2), the same value of mdot prevails,
but there is a different value of voldot. First, we use the ideal
gas law to calculate v2:
v2 = R(T2)/(p2) = 3.18 ft^3/lbm
and
(voldot)2 = (mdot)(v2) = (1.225)(3.18)
= 3.89 ft^3/sec
= [velocity at 2] x [area at 2]
= (700 ft/sec) x [(pi)(D^2)/4]
Solve this equation for D in feet.
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2. Problem 4.24, page 157 in the 3rd edition and Problem
4.25, page 191 in the 4th edition.
For background on nozzles, read pages 131 - 133 in the
3rd edition and pages 157 - 160 in the 4th edition. These
readings contain a worked-out example.
A nozzle is a device whose main purpose is to change the
velocity in a pipe whose cross-sectional area varies in the
flow direction.
If 1 denotes the nozzle inlet and 2 denotes the nozzle exit,
the First Law for a flowing fluid reduces to (pages 132 and
158 in the respective editions):
0 = (Qdot)/(mdot) + (h1 - h2) + [(V1)^2 - (V2)^2]/2
where h is the specific enthalpy. The problem statement says
that the nozzle is well-insulated, so that Qdot = 0.
At state 1, p1 = 200 lbf/in^2 and T1 = 500F. The value of h1
can be read directly Table A-4E without interpolation. In
addition, the velocities at the inlet and the exit are given as
V1 = 200 ft/sec and 1700 ft/sec respectively. Also,
h2 = h1 + [(V1)^2 - (V2)^2]/2
The values of all the terms on the right-hand side of this
equation are known. However, the units must be carefully
dealt with. The units of h are Btu/lbm, while those of V^2
are (ft/sec)^2. Clearly, conversion factors are needed.
Let's see how to do this:
[(ft)^2]/[(sec)^2]} x 1 / 1
1 = (1/32.2){[(lb)f]/[(lb)m]}{[(sec)^2]/ft}
1 = 778.17 (ft-lbf)/Btu
So, to convert V^2 in (ft/sec)^2 to Btu/lbm divide by 32.2 and
778.17.
With this conversion factor, we can calculate h2. Then, with
this h2 and with p2 = 60 lbf/in^2, we have to go to Table A-4E
and look for T2.
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3. Problem 4.36, page 157 of the 3rd edition and problem 4.37,
page 192 of the 4th edition.
This problem is focused on turbines. A turbine is a powerproducing machine whose "product" is a rotating shaft.
Background material on turbines may be found on pages
133 - 135 of the 3rd edition and on pages 160 - 162 of the 4th
edition. The form of the energy equation for turbines is found
at the bottom of page 134 and at the bottom of page 161 in the
respective editions,
According to the problem statement, the turbine is wellinsulated, so that Qdot = 0. With this, the energy equation is
0 = (Wdot)/(mdot) + (h2 - h1) + [(V2)^2 - (V1)^2]/2
where Wdot is the power output of the rotating shaft. It is
given that Wdot = 10 MW = 10,000kW.
We begin by solving the energy equation for h1:
h1 = (Wdot)/(mdot) + h2 + [(V2)^2 - (V1)^2]/2
The value of h2 can be found from Table A-3 as hg at p =
0.06 bars. All other quantities on the right-hand side are
known, so that h1 can be found. With this value of h1 and
with T1 = 320C, Table A-4 is searched for p1.
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4. Problem 4.58, page 159 of the 3rd edition and problem 4.60,
page 194 of the 4th edition.
There are no work transfers in this problem and potential
energy changes are negligible. Also, since it was given that
cp = constant, it follows that
h2 - h1 = cp(T2 - T1)
The energy equation for this problem can then be written as:
(Qdot)/(mdot) = cp(T2 - T1) + [(V2)^2 - (V1)^2]/2
Before attempting to solve for Qdot, it is first necessary to
determine mdot. Since mdot is a constant (steady state, no
leaks), it may be evaluated anywhere along the pipe. It is
easiest to calculate mdot at the inlet (state 1). The essential
piece of information that is missing at state 1 is the specific
volume v1. From the ideal gas law,
v1 = R(T1)/p1
The molecular weight of carbon dioxide is 44.01 kg/kmole,
and R = (Rbar)/44.01. Once v1 has been found,
mdot = [(velocity)1][(pi)D^2/4]/v1
The last step in the solution is to evaluate the terms on the
right-hand side of the energy equation.
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