Solutions to Mid-Term Exam for General Physics II,
Spring 2010.
1. 30%
(a) Water emerges from a faucet of diameter d0 in
steady, near vertical flow with speed v0.
Find the diameter of the falling water column as a
function of h, the distance below the faucet.
(b) Figure shows a Pitot tube used for
measuring aircraft speeds.
Assuming air flows by opening A unaffected,
while it is stopped at opening B, find the air
speed in terms of the pressure difference.
Solution
Equation of continuity:
 u A  const
Bernoulli’s principle:
p
(a)
1 2
 u   gh  const
2
Applying the equation of continuity to 2 cross sections, one at h = 0,
where the diameter of the water column is d0 & the other at h, where the
diameter is d :
2
 d0 
d 
   v  
 2
2
2
 v0  
 = density of water
Since the water is free falling, v 2  v02  2 gh ,
d
v0
d0 
v
v0
v02  2 gh
d0 
(b) Applying the Bernoulli’s principle:
4
v02
d0
v02  2 gh
pA 
1 2
 u  pB
2
So that
2.
u
2  pB  p A 

 = density of air
25%
Figure shows the cycle for a diesel engine.
Which point in the cycle has the highest
temperature and which has the lowest ?
Derive your results in detail.
Solution
From the ideal gas equation P V = n R T, we see that
a) For process 23,
Tmin = T2 , Tmax = T3.
b) For process 41,
Tmin = T1 , Tmax = T4.
Using the adiabatic relation T V  1  const , we have
c) For process 12,
d) For process 34,
Tmin = T1 , Tmax = T2.
Tmin = T4 , Tmax = T3.
From a), b), & d), we see that the overall maximum temperature is T3 .
From a), b), & c), we see that the overall minimum temperature is T1 .
3.
45%
You have 100 kg of steam at 100C, but no heat source to maintain it at that
condition. You also have a heat reservoir at 0C. Suppose you operate a reversible
heat engine with this system, so the steam gradually condenses and cools until it
reaches 0C.
(a) Calculate the total entropy change of the steam and subsequent water.
(b) Calculate the entropy change of the reservoir.
(c) Find the maximum amount of work that the engine can do.
Hint:
The latent heat of vaporization for the water is L v  2.3 MJ / kg .
The specific heat of water is c  4.2 kJ / kg  K .
Solution
(a)
Entropy change during the condensation at 100C is
100 kg   2.3 MJ / kg   0.617 MJ / K
m Lv
Sv  

Tv
273  100
Entropy change during the cooling from 100C to 0C is
Tw dQ
Tw m c dT
T
S w  

 m c ln w
Tv T
Tv
T
Tv
 100 kg  4.2 kJ / kg  K  ln
273
 131 kJ / K
373
Total entropy change of the fluid is
S f  Sv  S w  748 kJ / K   750 kJ / K .
(b) During the cooling, the amount of heat dumped to the reservoir is
Q  m  Lv  c T   100 kg   2.3 MJ / kg   4.2 kJ / kg  K 100 K  
 272,000 kJ
Entropy of the reservoir therefore increases by
Q 272000 kJ
Sr 

 996 kJ / K  1 MJ / K
Tr
273 K
(c)
Entropy change of entire system is
S  S f  Sr   748  996 kJ / K  248 kJ / K
Amount of unavailable energy is
Eunavailable  Tmin S   273 K  248 kJ / K   67,700 kJ
The maximum amount of work available is therefore
W  Q  Eunavailable   272,000  67,700 kJ  204,300 kJ
 200 MJ