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AE 2010 Thermodynamics Problem Set #5: 1st Law WITH SOLUTION NEW VERSION Georgia Institute Of Technology

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AE 2010 Thermodynamics Problem Set #5: 1st Law WITH SOLUTION NEW VERSION Georgia Institute Of Technology

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AE 2010 Thermodynamics Problem Set #5: 1st Law WITH SOLUTION NEW
VERSION Georgia Institute Of Technology

• Always indicate any assumptions you make. If you use any results or equations from
the class notes or text in you solutions, please note and reference them (but you
better be sure they are applicable to the problem at hand).
• Show all your work, no credit for just answers. When applicable, try to solve the
problem algebraically first. Only use numbers/values in the final steps of your
solution – and be sure to include units when you insert numbers.
• If the problem statement is given in ENGLISH units, the answer must also be in
English units; if the problem statement is in SI units, the answer must be in SI units.

1. Adiabatic Compression

A closed vessel is filled with xenon, a noble gas, initially at a volume of 7550 cm3, a
temperature of 20.0 C and pressure of 0.55 bar. The xenon is compressed adiabatically until
it reaches 355 C and 3.7 bar. You may assume that xenon is calorically perfect with a Cp
=2.5R (where R is the gas constant). Determine the volume of the xenon at the final
condition and how much work was required to compress the xenon.

2. Pump

An oil pump in a jet engine takes oil from a sump (reservoir) at 320 K and 0.90 bar and raises
the pressure to 10.0 bar with an outlet temperature of 321 K. At these conditions, the oil has a
density of 880 kg/m3 and a specific heat of 1.95 kJ/kgK. You may assume the pump is
adiabatic and that oil is an incompressible fluid. How much power (in units of Watts) is
required to operate the pump if the oil flow rate is 120. liters/hour?




sump Oil (T1=320K, p1=90 kPa)




Oil discharge
(T2=321K, p2=1 MPa)

, Solution for Problem Set #5: 1st Law

W12
Problem 1. Adiabatic Compression
Xe
Given: Closed and insulated container with Xe being
compressed with initial and final conditions
shown in figure, (cp/R)Xe = 2.5 Q12=0

Find: a) V2 T1=20.0C T2=355C
b) W12 p1=0.55 bar p2=3.7 bar
V1=7550 cm3
Assume: 1) Xe thermally and cal. perf. gas (given)
2) no work but compression work
Analysis:
a) V2
Since we know both initial and final state, can use an EOS to find V. Ideal Gas EOS:

m1 = m2
p1V1 p2V2
=
RT1 RT2
V2 = p1 p2 V1
T1 T2
This EOS requires using absolute temperatures:
0.55bar 3.7bar
V = 7550cm3 = 2410cm3
2
293K 628K

b) W12

We can’t use W=pdV since don’t know if the process is a quasi-equilibrium one (or even
how p is a function of V during the process). But we can use conserv. of energy (1st Law)
for our control mass:

E2 − E1 = W12 + Q12
U 2 −U 1 = W12 + 0

since the only energy change in the system is internal energy and the system is adiabatic.

w/ Xe a tpg and cpg W12 = U 2 −U1 = mXecv, Xe (T2 − T1 )

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