ENME315: HOMEWORK
ENME 315 CHAPTER-10
HOMEWORK PROBLEMS
10-22
Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling)
temperature of Tsat = 100C in a teflon-pitted stainless steel pan placed on an electric
burner. The water level drops by 10 cm in 30 min during boiling. The inner surface
temperature of the pan is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the pan are
negligible. 3 The boiling regime is nucleate boiling (this assumption will be checked
later).
Properties The properties of water at the saturation temperature of 100C are (Tables
P = 1 atm
10-1 and A-9)
 l  957.9 kg/m 3
100C
 v  0.60 kg/m 3
  0.0589 N/m
Water
Ts
Prl  1.75
h fg  2257  10 3 J/kg
 l  0.282  10 3 kg  m/s
Heating
c pl  4217 J/kg  C
Also, C sf  0.0058 and n = 1.0 for the boiling of water on a teflon-pitted stainless
steel surface (Table 10-3). Note that we expressed the properties in units specified
under Eq. 10-2 connection with their definitions in order to avoid unit manipulations.
Analysis The rate of heat transfer to the water and the heat flux are
(957.9 kg/m 3 )(  (0.2 m) 2 /4  0.10 m)
 0.003344 kg/s
t
t
15  60 s
Q  m evap h fg  (0.03344 kg/s)(2257 kJ/kg)  7.547 kW
m evap 
m evap

V

As  D 2 / 4   (0.20 m) 2 / 4  0.03142 m 2
q  Q / A  (7547 W)/(0.031 42 m 2 ) = 240,200 W/m 2
s
The Rohsenow relation which gives the nucleate boiling heat flux for a
specified surface temperature can also be used to determine the surface temperature
when the heat flux is given. Assuming nucleate boiling, the temperature of the inner
surface of the pan is determined from Rohsenow relation to be
 g ( l   v ) 
q nucleate   l h fg 




1/ 2 

 c p ,l (T s  Tsat ) 
n
 C h Pr 
sf fg
l


3
1/2
ж 4217(T - 100) ч
ц3
й9.81(957.9- 0.60)щ
s
з
ч
ъ з
240,200 = (0.282ґ 10 )(2257ґ 10 )к
ч
ч
кл
ъ ззи0.0058(2257ґ 103 )1.75ш
0.0589
ы
- 3
3
It gives
Ts = 105.3C
which is in the nucleate boiling range (5 to 30C above surface temperature).
Therefore, the nucleate boiling assumption is valid.
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ENME315: HOMEWORK
10-32
Water is boiled at a temperature of Tsat = 100°C by hot gases flowing through a tube
submerged in water. The maximum rate of vaporization is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are
negligible.
Properties The properties of water at Tsat = 100°C are σ = 0.0589 N/m (Tables 10-1)
and, from Table A-9, ρl = 957.9 kg/m3, ρv = 0.5978 kg/m3, hfg = 2257 × 103 J/kg.
Analysis The maximum rate of vaporization occurs at the maximum heat flux.
For a horizontal cylindrical heating element, the coefficient Ccr is determined from
Table 10-4 to be
𝑫 𝒈(𝝆 −𝝆 ) 𝟏/𝟐
𝟗.𝟖𝟏(𝟗𝟓𝟕.𝟗−𝟎.𝟓𝟗𝟕𝟖) 𝟏/𝟐
𝑳 ∗= 𝟐 ( 𝒍𝝈 𝒗 )
= (𝟎. 𝟎𝟓𝟎/𝟐) [
]
= 𝟗. 𝟗𝟖> 1.2
𝟎.𝟎𝟓𝟖𝟗
𝑪𝒄𝒓 = 𝟎. 𝟏𝟐 (since 𝑳 ∗ > 1.2 and thus large cylinder)
Then the maximum heat flux is determined from
q max  C cr h fg [g v2 (  l   v )]1 / 4
 0.12( 2257  10 3 )[0.0589  9.81  (0.5978) 2 (957.9  0.5978)]1 / 4  1.0155  10 6 W/m 2
The heat transfer surface area is
𝐴𝑠 = 𝜋𝐷𝐿 = 𝜋(0.05 m)(10 m) = 1.5708 m2
Then, the rate of heat transfer during nucleate boiling becomes
𝑄̇boiling = 𝐴𝑠 𝑞̇ max = (1.5708 m2 )(1.0155 × 106 W/m2 ) = 1.5952 ×
106 W
The maximum rate of vaporization of water is determined from
𝑚̇vapor =
𝑄̇boiling
ℎ𝑓𝑔
1.5952×106 J/s
= 2257×103 J/kg = 0.707 𝑘𝑔/𝑠
Discussion The rate of vaporization can be increased by increasing the tube diameter,
thereby increasing the heat transfer surface area.
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