CANKAYA UNIVERSITY

FACULTY OF ENGINEERING AND ARCHITECTURE

MECHANICAL ENGINEERING DEPARTMENT

ME 211 THERMODYNAMICS I

FALL 2015

HW # 4 SOLUTION

1) Air at 10 0C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s.

The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very

small compared with the inlet velocity. Determine (a) the mass flow rate of the air and (b) the

temperature of the air leaving the diffuser.

Solution

(b) SSSF conditions

Conservation of mass

e i

e i

m m

Single inlet single outlet

2 1m m m

First Law of Thermodynamics

2 2

CV CV e e e e i i i i

e i

1 1Q W m (h V gz ) m (h V gz )

2 2

CV CV 2Q W 0,V 0, PE 0

2

e e i i i

e i

10 m h m (h V )

2

Single inlet single outlet

2

1 1 2

2

2 1 1

1m(h V ) mh

2

1h (h V )

2

At the inlet

0

1

1

T 10 C

p 80kPa

2) Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of

the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes

in kinetic and potential energies are negligible, determine the necessary power input to the

compressor.

Solution

(b) SSSF conditions

Conservation of mass

e i

e i

m m

Single inlet single outlet

2 1m m m

First Law of Thermodynamics

2 2

CV CV e e e e i i i i

e i

1 1Q W m (h V gz ) m (h V gz )

2 2

KE 0, PE 0

CV CV e e i i

e i

Q W m h m h

Single inlet single outlet

CV CV 2 1

out C 2 1

out out

C out 2 1

Q W m(h h )

Q ( W ) m(h h )

Q mq

W mq m(h h )

1

1

1

1

2

1

p 100kPah 280.13kJ / kg

T 280 K

p 600kPah 400.98kJ / kg

T 400 K

3) The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions

of the steam are as indicated in Figure given below

(a) Compare the magnitudes of h, KE, PE

(b) Determine the work done per unit mass of the steam flowing through the

turbine.

(c) Calculate the mass flow rate of the steam.

Solution

SSSF

At the inlet:

0

1 sat

1 sat

p 2 MPa 20bar T 212.4 C

T T

At the outlet:

Let us compute h, KE, PE

b)

Conservation of mass

e i

e i

m m

Single inlet single outlet

2 1m m m

First Law of Thermodynamics

2 2

CV CV e e e e i i i i

e i

1 1Q W m (h V gz ) m (h V gz )

2 2

Single inlet single outlet

2 2

CV CV e e e e i i i i

2 2

CV CV e i e i e i

2 2

t e i e i e i

1 1Q W m (h V gz ) m (h V gz )

2 2

1Q W m (h h ) (V V ) g(z z )

2

1W m (h h ) (V V ) g(z z )

2

Neglecting heat loss,

4) CVQ 0

2 2

t 2 1 2 1 2 1

1W m (h h ) (V V ) g(z z )

2

or

2 2

t 2 1 2 1 2 1

W 1w (h h ) (V V ) g(z z ) kJ / kg

m 2

4) Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and

is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state

and the temperature drop during this process.

Solution

5) Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters the condenser

with a mass flow rate of 6 kg/min at 1 MPa and 70 0C and leaves at 35 0C. The cooling water

enters at 300 kPa and 15 0C and leaves at 25 0C. Neglecting any pressure drops, determine (a) the

mass flow rate of the cooling water required and (b) the heat transfer rate from the refrigerant to

water.

SSSF conditions

Conservation of mass

e i

e i

2 4 1 3

3 4 R

1 2 w

m m

m m m m

m m m

m m m

Rm mass flow rate of R-134a

wm =mass flow rate of water\

First Law of Thermodynamics

2 2

CV CV e e e e i i i i

e i

1 1Q W m (h V gz ) m (h V gz )

2 2

CV CVKE 0, PE 0,Q 0,W 0

e e i i

e i

4 4 2 2 1 1 3 3

R 4 3 W 1 2

m h m h

m h m h m h m h

m (h h ) m (h h )

Now we need to determine the enthalpies at all four states. Water exists as a compressed liquid at

both the inlet and the exit since the temperatures at both locations are below the saturation

temperature of water at 300 kPa.

At inlet 1:

0

1

1

T 15 C

p 300kPa

0

1 sat

1 sat

p 300kPa T 133.6 C

T T compressed liquid

At outlet 2:

0

1

1

T 25 C

p 300kPa

0

2 sat

2 sat

p 300kPa T 133.6 C

T T compressed liquid

State-3

0

3 sat

1 sat

p 1MPa 10bar T 39.39 C

T T

State-4

0

4 sat

4 sat

p 1MPa 10bar T 39.39 C

T T

Compressed liquid

So the heat lost by hot fluid is equal to heat gained by cold fluid

See CV on next page

W R

W w 2 1

Q Q

Q m (h h )

6) An insulated 8-m3 rigid tank contains air at 600 kPa and 400 K. A valve connected to the tank

is now opened, and air is allowed to escape until the pressure inside drops to 200 kPa. The air

temperature during the process is maintained constant by an electric resistance heater placed in

the tank. Determine the electrical energy supplied to air during this process.

Solution

USUF process

Conservation of mass:

2 1 i ecvi i

m m m m

Conservation of energy

𝑄𝐶𝑉 +2

ii i i

i

Vm h gz

2

=2

ee e e

e

Vm h gz

2

+[𝑚2 (𝑢2 +

𝑉22

2+ 𝑔𝑧2) − 𝑚1 (𝑢1 +

𝑉12

2+ 𝑔𝑧1)]

𝐶𝑉+ cvW

Assumptions

1) This is an unsteady process since the conditions within the device are changing during the

process, but it can be analyzed as a uniform flow process since the exit conditions remain

constant.

2) Kinetic and potential energies are negligible.

3 ) The tank is insulated and thus heat transfer is negligible.

4) Air is an ideal gas with variable specific heats. For this reason, we will use air tables.

We take the contents of the tank as the system, which is a control volume since mass crosses the

boundary. Noting that the microscopic energies of flowing and non- flowing fluids are

represented by enthalpy h and internal energy u, respectively, the mass and energy balances for

this uniform-flow system can be expressed as

Conservation of mass.

No mass is entering. So

2 1 i ecvi i

e 1 2 cv

m m m m

m m m

Conservation of energy

KE 0, PE 0, no mass is leaving since tank is insulated CVQ 0,

Electrical work is done on the CV

eW e em h +[𝑚2(𝑢2) − 𝑚1(𝑢1)]𝐶𝑉