PROBLEM 6.39
PROBLEM 6.47
PROBLEM 6.62
PROBLEM 6.123
PROBLEM 6.141
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and
exits at 120 kPa. The mass flow rate is 5.5 kg/s, and the power developed is 1120 kW. Stray
heat transfer and kinetic and potential energy effects are negligible. Assuming k = 1.4, determine
(a) the temperature of the air at the turbine exit, in K, and (b) the isentropic turbine efficiency.
KNOWN: Air expands adiabatically through a turbine operating at steady state. Operating data
are known.
FIND: Determine the exit temperature and the isentropic turbine efficiency.
ENGINEERING MODEL: (1) The control volume is at
state. (2) For the control volume, = 0 and kinetic and
potential energy effects can be neglected. (3) The air is modeled
as an ideal gas with constant specific heats: k = 1.4.
ANALYSIS: (a) Mass and energy rate balances reduce to give: 0 = + (h1 – h2). With
h1 – h2 = cp(T1 – T2)
T2 = T1
From Sec. 3.13.1; cp = kR/(k – 1) = (1.4)(8.314/28.97)/(1.4 – 1) = 1.004 kJ/kg∙K and
T2 = T1 = 1040 K – (1120 kW)/[(5.5 kg/s)(1.004 kJ/kg∙K)
= 837.2 K
(b) The isentropic efficiency is ηt = (h1 – h2)/(h1 – h2s) = cp (T1 – T2)/ cp (T1 – T2s). To get T2s we
note that for an isentropic process of an ideal gas with constant specific heats
=
→ T2s =
(1040 K) = 818.1 K
Thus, the isentropic efficiency is
ηt = (1040 – 837.2)/(1040 – 818.1) = 0.914 (91.4%)
(1)
p1 = 278 kPa
T1 = 1040 K
= 5.5 kg/s (2)
p2 = 120 kPa
= 1120 kW
T
s
(1)
278 kPa
120 kPa
1040 K .
(2s) (2) .
.
PROBLEM 6.152
PROBLEM 6.145
Air enters the compressor of a gas turbine power plant operating at steady state at 290 K, 100
kPa and exits at 330 kPa. Stray heat transfer and kinetic and potential energy effects are
negligible. The isentropic compressor efficiency is 90.3%. Using the ideal gas model for air,
determine the work input, in kJ per kg of air flowing.
KNOWN: Air is compressed adiabatically. The state is known at the inlet and the exit pressure
is specified. The isentropic compressor efficiency is known.
FIND: Determine the work per unit mass of air flowing.
ENGINEERING MODEL: (1) The control volume is at steady state. (2) and kinetic
and potential energy effects are negligible. (3) The air is modeled as an ideal gas.
ANAYSIS: The mass and energy rate balances reduce to give = (h1 – h2). With the
isentropic compressor efficiency: ηc = (h1 – h2s)/(h1 – h2)
= (h1 – h2s)/ηc
To find h2s, use Eq. 6.41 and data from Table A-22: pr2 = pr1(p2/p1) = 1.2311 (330/100) = 4.0626
Interpolating in Table A-22; h2s ≈ 408.5 kJ/kg. Also, at 290 K, h1 = 290.16 kJ/kg. Thus
= (290.16 – 408.5)/(0.903) = 131.05 kJ/kg (in)
Note: As indicated on the T-s diagram…
T2s ≈ 407.4 K and h2 = 290.16 + 131.05 = 421.2 kJ/kg → T2 ≈ 420 K
(1)
T1 = 290 K
p1 = 100 kPa
(2)
p2 = 330 kPa
T
100 kPa
(1)
(2) (2s)
290 K
s
330 kPa
Air
ηc=90.3%
420 K 407.4 K
. .
.