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

. .

.