Exercícios de termodinâmica

6.11
Carbon dioxide at 200 kPa, 10oC flows at 1 kg/s in a 0.25 m2 cross sectional area pipe. Find the velocity and the volume flow rate. m .
= AV/v = V
.
/v v = RT/P = 0.1889 kJ/kg-K × 283.15 K/ 200 kPa = 0.267 m3/kg
V
.
= v m
.
= 0.267 × 1 = 0.267 m3/s
V = V
.
/ A = 0.267 (m3/s) / 0.25 m2 = 1.07 m/s

6.12
Air at 35°C, 105 kPa, flows in a 100 mm × 150 mm rectangular duct in a heating system. The volumetric flow rate is 0.015 m3/s. What is the velocity of the air flowing in the duct and what is the mass flow rate?
Solution:
Assume a constant velocity across the duct area with
A = 100 × 150 ×10-6 m2 = 0.015 m2 and the volumetric flow rate from Eq.6.3,
V .
= m
.
v = AV
V =
V .
A =
0.015 m3/s
0.015 m2
= 1.0 m/s
Ideal gas so note: v =
RT
P =
0.287 × 308.2
105 = 0.8424 m3/kg m .=
V . v =
0.015.0.8424 = 0.0178 kg/s

6.14 -Saturated vapor R-134a leaves the evaporator in a heat pump system at 10°C, with a steady mass flow rate of 0.1 kg/s. What is the smallest diameter tubing that can be used at this location if the velocity of the refrigerant is not to exceed 7 m/s? Solution:
Mass flow rate Eq.6.3: m
.
= V
.
/v = AV/v
Exit state Table B.5.1: (T = 10°C, x =1) => v = vg = 0.04945 m3/kg
The minimum area is associated with the maximum velocity for given m
.
AMIN = m . vg VMAX
=
0.1 kg/s × 0.04945 m3/kg
7 m/s = 0.000706 m2 = π4 D2
MIN
DMIN = 0.03 m = 30 mm

6.20
Liquid water at 15oC flows out of a nozzle straight up 15 m. What is nozzle Vexit?
Energy Eq.6.13: hexit + 12
V2
exit + gHexit = h2 + 12
V22
+ gH2
If the water can flow 15 m up it has specific potential energy of gH2 which must equal the specific kinetic energy out of the nozzle V 2 exit/2. The water does not change P or T so h is the same.
V 2 exit/2 = g(H2 – Hexit) = gH =>
Vexit = 2gH = 2 × 9.807 × 15 m2/s2 = 17.15 m/s

6.22
A nozzle receives 0.1 kg/s steam at 1 MPa, 400oC with negligible kinetic energy.
The