THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
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Textbook Question
Chapter 16.6, Problem 1P
Why is the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy?
Expert Solution & Answer
To determine
Reason behind the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy.
Answer to Problem 1P
The equilibrium criteria can be expressed in terms of properties alone when the Gibbs function is used.
Explanation of Solution
Conclusion:
The reacting system depends upon the heat transfer, change in entropy principle relation requires a detail knowledge about the heat transfer between the system and its surrounding, which is not practical.
Thus, the equilibrium criteria can be expressed in terms of properties alone when the Gibbs function is used.
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The crash cushion for a highway barrier consists of a
nest of barrels filled with an impact-absorbing material.
The barrier stopping force is measured versus the vehicle
penetration into the barrier. (Figure 1)
Part A
P Course Home
b My Questions | bartleby
Review
Determine the distance a car having a weight of 4000 lb will penetrate the barrier if it is originally traveling at 55 ft/s when it
strikes the first barrel.
Express your answer to three significant figures and include the appropriate units.
Figure
1 of 1
36
μΑ
S =
Value
Units
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Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the
condenser pressure is 8 kPa. The mass flow rate of steam entering the turbine is 50 kg/s.
Determine:
(a) the net power developed, in kW.
(b) the rate of heat transfer to the steam passing through the boiler, in kW.
(c) the percent thermal efficiency.
(d) the mass flow rate of condenser cooling water, in kg/s, if the cooling water undergoes a temperature
increase of 18°C with negligible pressure change in passing through the condenser.
4. The figure below shows a bent pipe with the external loading FA
228 lb, and M₁ = M₂ = 1 kip-ft. The force Fernal loading FA = 300 lb, FB:
parallel to the y-axis, and
and yc = 60°.
= 125 lb, Fc
=
acts parallel to the x-z plane, the force FB acts
Cartesian resultan Coordinate direction angles of Fc are ac = 120°, ẞc = 45°,
a. Compute the resultant force vector of the given external loading and express it in
EST
form.
b. Compute the resultant moment vector of the given external loading about the origin, O,
and express it in Cartesian vector form. Use the vector method while computing the
moments of forces.
c. Compute the resultant moment vector of the given external loading about the line OA
and express it in Cartesian vector form.
:00 PM EST
k
ghoufran@buffaternal du
2 ft
M₁
A
40°
FA
M2
C
18 in
1 ft
Fc
25
houfran@bald.edu - Feb 19,
3 ft
FB
Chapter 16 Solutions
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
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