First Law for Steady Open Systems
Consider two ways to compress 2 kg/s of air (an ideal gas with R = 0.287 kJ/kg·K and
constant heat capacities: cv = 0.718 kJ/kg·K, and cp = 1.005 kJ/kg·K) from an inlet Pin = 100
kPa and Tin = 20oC to an outlet Pout = 50 MPa. The first way uses a single compressor with
the input and output values shown above and a heat loss of 150 kW. In this case the outlet
temperature is Tout,1 =700oC. (1) Find the work in this case.
We start with the general equation for steady open systems.
inlet i
i
ii
outlet o
o
oou
cv gz
V
hmgz
V
hmWQ
dt
dE
22
22
We assume a steady problem so that dEcv/dt = 0 and we assume that changes in kinetic and
potential energies are zero since we have no data for velocities and elevations. Each device that
we analyze in this problem has one inlet and one outlet, so there is a single mass flow rate in the
steady system. This reduces the first law to the following equation, extended to apply to an ideal
gas with constant heat capacities for which h = cpT.
inoutpinoutuTTcmQhhmQW
For the single stage compression this equation gives the power as shown below. Here we note
that a heat loss is negative and a temperature difference in Celsius is the same as that in kelvins.
kJ
skW
CC
Kkg
kJ
s
kg
kWTTcmQW oo
inoutpu 1
20700
005.12
150
1,
kWWu517,1
In the second approach, two compressors are used. The first compressor has the same
inlet Pin = 100 kPa and Tin = 20oC given above. The first compressor has a heat loss of 75
kW and outlet conditions Pa = 700 kPa and Ta = 260oC. The outlet from this compressor is
then cooled in a heat exchanger to Pb = 680 kPa and Tb = 40oC before being sent to the
second compressor which has the same exit pressure, Pout = 50 MPa, but its outlet
temperature is Tout,2 =300oC. The second compressor also has a heat loss of 75 kW. (2)
What is the total work requirement for the two-compressor system?
For each compressor we can apply the equation used above for the single compressor. For the
first stage we have.
kJ
kJ
skW
CC
Kkg
kJ
s
kg
kWTTcmQW oo
inapu 4.557
1
20260
005.12
75
1.
The same equation can be applied to the second stage with different data.
kJ
kJ
skW
CC
Kkg
kJ
s
kg
kWTTcmQW oo
boutpu 6.579
1
40300
005.12
75
2,2.
Adding the work of each compressor gives the total work as -557.4 kW + (-596.6 kW) = –1,155 kW
In all cases the work is negative indicating a work input, which is expected for compression.
(3) What is the heat transfer in the heat exchanger?
We can apply the same simplified equation for the first law to the heat exchanger. However, the
heat exchanger has no useful work. We find the heat transfer by applying the given data to the
simplified equation
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