EGR 334 Thermodynamics: Homework 12
Problem 4: 1
An 8 cu ft tank contains air at an initial temperature of 80 deg F and initial pressure of 100 psi. The tank develops a
small hole and air leaks from the tank at a constant rate of 0.03 lb/s for 90 s until the pressure of the air remaining in
the tank is 30 psi. Employing the ideal gas model, determine the final temperature of the air remaining in the tank.
----------------------------------------------------------------------------------------------- ----------------------------------------V = 8 ft3 State 1: T1 = 80 F = 540 R p1 = 100psi
R = 0.06855 btu/lb m-R
State 2: p2 = 30 psi
mdot = 0.03 lb/s for 90s.
mass balance:
dmsys
dt
 mi  me
dmsys  0.03dt
mf
 dm
sys
  0.03dt  0.03t  (0.03lbm ) / s (90s )  2.7lbm
mo
m2  m1  2.7lbm
energy balance:
dE
 Q  W  mi ei  me ee
dt
using the ideal gas law:
p1V1  m1RT1
m1 
(100lb f / in 2 )(8 ft 3 )
p1V1
Btu
144in 2

 4.00 lbm
RT1 (0.06855Btu / lbm  R )(540 R ) 778.17 ft  lb f
ft 2
Therefore:
m2  4.00  2.7  1.3 lbm
Using the ideal gas law again
p2V2  m2 RT2
(30lb f / in 2 )(8 ft 3 )
p2V2
Btu
144in 2
T2 

 498.4o R
Rm2 (0.06855Btu / lbm o R)(1.3 lbm ) 778.17 ft  lb f
ft 2
= 38.4 oF
EGR 334 Thermodynamics: Homework 12
Problem 4:6
The figure shows a mixing tank initially containing 3000 lb of liquid water. The tank is fitted with two inlet pipes,
one delivering hot water at a mass flow rate of 0.8 lb/s and the other delivering cold water at a mass flow rate of 1.3
lb/s. Water exits through a single exit pipe at a mass flow rate of 2.6 lb/s. Determine the amount of water in the
tank after1 hr.
-----------------------------------------------------------------------------mi = 3000 lbm of H2O
m1 = 0.8 lbm/s
m2 = 1.3 lbm/s
m3 = 2.6 lb/s
Mass balance:
dm
  mi   me
dt
dmsys
 m1  m2  m3
dt
dmsys  (m1  m2  m3 )dt
t 1hr
mf
 dm
sys
mi


(m1  m2  m3 )dt
t 0
m f  mi  (m1  m2  m3 )t
m f  (m1  m2  m3 )t  mi
m f  (0.8lb f / s  1.3lb f / s  2.6lb f / s)(1hr )
3600 s
 3000lb f  1200lb f
hr
EGR 334 Thermodynamics: Homework 12
Problem 4:11
As shown in the figure, air with volumetric flow rate of 15000 cu ft/min enters an air handling unit at 35 deg F, 1
atm. The air handling unit delivers air at 80 deg F, atm to a duct system with three branches consisting of two 26 in
diameter ducts and one 50 in duct. The velocity in each 26 in duct is 10 ft/s. Assuming ideal gas behavior for the
air, determine at steady state
a) the mass flow rate of air entering the air handling unit in lb/s.
b) the volumetric flow rate in each 26 in duct in cu. ft/min.
c) the velocity in the 50 in duct in ft/s.
---------------------------------------------------------------------Air: R = 0.06855 btu/lbm-R
State 1:
p1 = 1 atm
T1 = 35 deg F= 495 R
V1dot = (AV)1= 15000 ft3/min
State 2, 3, and 4: p2=p3=p4 = 1 atm
T2 = T3=T4 = 80 deg F.= 540 R
V2 = V3 = 10 ft/s
Assume Ideal Gas Law:
p1V1  m1RT1
as a rate equation
p1 V1  m1 RT1
m1 
m1 
p1 V1
RT1
2
(1atm)(15000 ft 3 / min) 14.7lb f / in
1btu
144in2
 1202lbm / min
(0.06855btu / lbm  R)(495R)
atm
778.17lb f  ft 1 ft 2
= 20 lbm/s
For state 2 and 3:
A2  A3 
d2
4

 (26in)2
4
 531in2
ft 2 60s
 2212.5 ft 3 / min
2
144in 1min
2
3
(1atm)(2212.5 ft / min) 14.7lb f / in
1btu
144in2
m2  m3 
 162.6 lbm / min
(0.06855btu / lbm  R)(540 R)
atm
778.17lb f  ft 1 ft 2
V3  V2  A2V2  (531in 2 )(10 ft / s)
using the mass balance at steady state:
0  m1  m2  m3  m4
m4  m1  m2  m3  1202  162.6  126.6  876.8lbm / min
V4 
m4 RT4 (876.8lbm / min)(0.06855btu / lbm  R)(540 R)
1atm
778.17lbf  ft 1 ft 2

 11932 ft 3 / min
2
p1
(1atm)
14.7lb f / in
btu
144in 2
AV4  V4

V4 
V4
V4

A4  d 42

4
4(11932 ft 3 / min) 144in 2
 875 ft / min  14.6 ft / s
 (50in) 2
1 ft 2
EGR 334 Thermodynamics: Homework 12
Problem 4:22
The figure shows a cylindrical tank being drained through a duct whose cross sectional area is 3 x 10 -4 m2. The
velocity of the water at the exit varies according to (2gz)1/2, where z is the water level in m, and g is the acceleration
of gravity, 9.81 m/s2. The tank initially contains 2500 kg of liquid water. Taking the density of the water as 10 3
kg/m3 determine the time in minutes when the tank contains 900 kg of water.
--------------------------------------------------------------Ae = 0.0003 m2
Ve = (2gz)1/2
mi = 2500 kg
ρ= 1000kg/m3
mf=900 kg
Mass flow balance:
dmsys
dt
 mi  me
where
mi  0
me   ( AeVe )  (1000kg / m3 )(0.0003m2 )(2  9.81m / s 2  z )1/ 2  (1.329kg / m1/ 2  s ) z1/ 2
Volume of Tank:
mtan k   ( Atan k z )  (1000kg / m3 )(1m 2 ) z  (1000kg / m) z
for an initial mass of 2500 kg:
zi 
mtank
2500kg

 2.5m
 Atank (1000kg / m3 )(1m 2 )
for a final mass of 900 kg:
zf 
mtank
900kg

 0.9m
 Atank (1000kg / m3 )(1m 2 )
therefore:
dmsys
 me
dt
d (1000kg / m) z
 (1.329kg / m1/ 2  s ) z1/ 2
dt
dz 1.329(kg / m1/ 2  s) 1/ 2

z  (0.001329 m1/ 2 / s ) z1/ 2
dt
1000kg / m
z  0.9
t t
1
1/ 2
 z1/ 2 dz  (0.001329 m / s)t 0 dt
z  2.5
2 1/ 2 z 0.9
z 
 (0.001329 m1/ 2 / s )(t  0)
z

2.5
1
2 (0.9m)1/ 2  (2.5m)1/ 2    (0.001329 m1/ 2 / s )t
t
1.265m1/ 2
 951.8s  15.9 min
0.001329 m1/ 2 / s