Bagnes, keyt leeann c.

CHE - 2202
23-05651

ChE 406 – CHEMICAL ENGINEERING THERMODYNAMICS

PROBLEM SET 1

1. Pressures up to 3000 bar are measured with a dead-weight gauge. The piston diameter is 4 mm. What
is the approximate mass in kg of the weights required?

( ) =1.2566 × 10 m
2
1m
2
π 4mm ×
100,000 Pa 8 πd 1000mm -5 2
3,000 bar × =3 ×10 Pa A= =
1 bar 4 4
8 -5 2
PA (3 ×10 Pa)(1.2566 ×10 m )
F=PA=mg ; m = = 2
= 384.29 kg
g 9.81 m/ s
2. The reading on a mercury manometer at 25°C (open to the atmosphere at one end) is 56.38 cm. The
local acceleration of gravity is 9.832 m/s2. Atmospheric pressure is 101.78 kPa. What is the absolute
pressure in kPa being measured? The density of mercury at 25°C is 13.534 g·cm 3.
3
1m g 1kg (100cm) kg
56.38 cm × =0.5638 m 13.534 3 × × 3
=13,534 3
100cm cm 1000g (1m) m

kg m 1000pa
P abs = pgh + p atm =13,534 × 9. 832 2 +0.5638m+101.78kpa ×
m
3
s 1kpa

1kpa
P abs = 176802.7732pa × =176.8028kp a
1000pa
3. An instrument to measure the acceleration of gravity on Mars is constructed of a spring from which is
suspended a mass of 0.40 kg. At a place on earth where the local acceleration of gravity is 9.81 m/s 2, the
spring extends 1.08 cm. When the instrument package is landed on Mars, it radios the information that
the spring is extended 0.40 cm. What is the Martian acceleration of gravity?

k=
F
x (
F=0.40kg 9.8 1
m
s
2)=3.924 N

3.924N n n (0.40kg)(X) m
k= =363.33 363.33 = x = 3.63 2
1m m m 1m s
1.08cm × 0,40 cm ×
100cm 100cm

4. A gas is confined in a 1.25(ft) diameter cylinder by a piston, on which rests a weight. The mass of the
piston and weight together is 250(lb m). The local acceleration of gravity is 32.169(ft)(s) 2. and atmospheric
pressure is 30.12(in Hg).

(A) What is the force in (lbf) exerted on the gas by the atmosphere, the piston, and the weight, assuming
no friction between the piston and cylinder?
3376.9 N 1psi F
for atm: 30.12 in hg × × =
1 inHg 6894.8 N π 12 in 2
(1.25 ft × )
4 1 ft
f atm = 2606.90 lbf

ft
( 250lbm ) (32.269 2
)
s
for weight: F = = 250 lbf
lbm ft
(32.269 )
lbf s2

F total = F atm + F weig h t = 2602.90 lbf + 250 lbf = 2856.90lbf

(B) What is the pressure of the gas in (psia)?

F 2856.90 lbf
P= = =16.67 psia
A π 2
(15 in )
4
(C) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston
and weight are raised 1.7(ft). what is the work done by the gas in (ft)(lb f)?
 w = Fd = ( 2856.90 lbf ) ( 1.7ft ) = 4856.73 lbf ft

(D) What is the change in potential energy of the piston and weight?
ft
(250lbm) (32.169 2
) (1.9ft)
s
pE= mgh = = 425 ft lbf
lbm ft
32.169
lbf s 2
5. A wind turbine with a rotor diameter of 77 m produces 1.5 MW of electrical power at a wind speed of
12 M/S. What fraction of the kinetic energy of the air passing through the turbine is converted to
electrical power? You may assume a density of 1.25 kg/m3 for air at the operating conditions.

( )
2
1 3 1 kg π (77m ) m 3
P wind = pA v P wind = 1.25 3 ( ) (12 )
2 2 m 4 s

6 1MW
Pwind = =5.03 × 10 W × = 5.03MW
1000000W

1.5Mw
= 0.30
5.03mw
6. The first accurate measurements of the properties of high-pressure gases were made by E. H. Amagat
in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mineshaft
and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height
of manometer required.
1 ×10 5 N m 2
400bar ×
P 1 bar
h= = =299.81m
pg kg m
(13600 3 )(9.81 2 )
m s
7. An automobile having a mass of 1250 kg is traveling at 40 m/S. What is its kinetic energy in kJ? How
much work must be done to bring it to a stop?
2
1 2 1 m 1KJ
KE = m v = ( 1250kg ) (40 ) =100000J × = 1000k j
2 2 s 1000J

W = KE = 1000KJ
8. At what absolute temperature do the Celsius and Fahrenheit temperature scales give the same
numerical value? What is the value? Show your solution.
9 5
° C × +32 = ° F - 32 ×
5 9
9 5
X × +32 = X-32 ×
5 9
X= -4 0
9. The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91%
efficiency for conversion of potential to electrical energy, and 8% loss of the resulting power in
transmission. what is the mass flow rate of water required to power a 200 W light bulb?
mgh
Pg =
t
200 W
Pg = =217.39W
1-0.08

217.39W =
(
m
m 9.81 2 ( 50m )
s )
×(0.91)
t
2
m M
217.39W = (446. 355 2 )
T s
m kg
= 0.49
T s
10. Verify that the SI unit of kinetic and potential energy is the joule.
2 2 2
1 2 1 m kg m m kg m
KE= m v = (kg) ( ) = 2 =N m=J pE = mgh = (kg) ( 2
) (m) = 2 =N m=J
2 2 s s s s