1. Drew Peden
ME 321-402 Final Project
11.76
Known:
In a Rankine power system steam is passed over 100 thin walled tubes 0.01 m in diameter in the
condenser which is a shell-and-tube heat exchanger. The steam enters as a saturated vapor at 0.51 bar and
leaves as a saturated liquid.
πΜ π π‘ = 1.5
ππ
π
; ππ,π = 280 πΎ ; πΜ ππ€ = 15
ππ
π
; βΜ π = 5000
π
π2 β πΎ
Properties of Cooling Water
π π (
π½
ππ β πΎ
) π (
π
π β πΎ
) π (
ππ
π β π
)
Pr
4178 0.628 700 X 10β6 4.6
Assume:
- Heat exchanger is well insulated (no heat loss to surroundings)
- Constant Properties
- Negligible tube wall thermal resistance
- Fully developed flow throughout
- Modeled as a reversible process
Diagram:
Figure 1: Rankine Cycle
Figure 2: Temperature/Enthalpy Rankine Diagram
2. Find:
(a) What is the water outlet Temperature?
In an ideal Rankine cycle, the heat dissipated in the condenser can be modelled as:
(1) π = πΜ (β2 β β3)
Where πΜ is the mass flow rate of the hot working fluid; and β2 and β3 are the enthalpy values associated
with the hot working fluid at the given pressure at points 2 and 3, in this case 0.51 bar and conveniently
corresponding to the saturated liquid and saturated vapor states. These states are indicated by the
numbered locations in figures 1 and 2 and can be obtained from steam tables.
Plugging in values to equation 1:
πβππ‘ = 1.5
ππ
π
(2645.9 β 340.49)
ππ½
ππ
= 3458.115 ππ
πβππ‘ = π ππππ so to find the water outlet temperature we can use
(2) π ππππ = πΜ π€ π π,π€(ππ β ππ)
We know all values but ππ so just need to rearrange, plug in values and solve:
ππ =
π
πΜ π€ π π,π€
+ ππ =
3458115 π
15
ππ
π
(4178
π½
ππβπΎ
)
+ 280 πΎ
(b) What is the required tube length?
The general procedure will be to find the universal heat transfer coefficient (U) defined as:
(3) π =
1
1
β π
+
1
β π
Where β π and β π are the convection coefficients on the outside and inside of the thin walled tubes.
The outside transfer coefficient is known. In order to find the inside coefficient we must find the Nusselt
number, Nu, defined as:
(4) ππ’ =
βπ·
π π
The Nusselt number is a function of the Reynolds and Pruitt number. The Pruitt number of the cooling
water is known so the Reynolds number needs to be found. For a circular pipe with fully developed flow
the Reynolds number is
(4) π π π· =
4πΜ
ππ·π
ππ = 335.12 πΎ = 62Β° C
3. All the values are known so
π π π· =
4πΜ
ππ·π
=
4 (
15 ππ
100 π‘π’πππ π
)
π(0.01π)(700 X 10β6 ππ
π β π
)
= 27,283.7
For internal flow a Reynolds number π π β₯ 2300 is associated with turbulent flow. Looking at Table 8-41
it can be seen that the appropriate equation for the Nusselt number is
(5) ππ’ π· = 0.023π π4/5 ππ π
Where n = 0.4 for cold to hot convection. So the Nusselt number is
ππ’ π· = 0.023π π
4
5 ππ π = 0.023 (27283.7
4
5)4.60.4 = 150
The inside convection coefficient can be calculated from Equation 4 as
β π =
ππ’ β π π
π·
=
150 (0.628
π
π β πΎ
)
0.01 π
= 9,420
π
π2 β πΎ
Now that the convection coefficients are known the universal heat transfer coefficient can be calculated
from Equation 3:
π =
1
1
β π
+
1
β π
=
1
1
5000
+
1
9420
= 3,266
π
π2 β πΎ
To find the length of each tube we must use the effectiveness method from the equation
(6) πππ( ππ’ππππ ππ π‘ππππ πππ π’πππ‘π ) =
ππ΄
πΆ πππ
In order to do so we first need to find the heat capacitance C of each fluid which will give us πΆ πππ and a
heat capacitance ratio that will allow to find NTU.
(7) πΆ = πΜ π π
Because the hot fluid undergoes a phase change the temperature does not change. Specific heat is defined
as
πβππππ ππ ππππππ¦
πβππππ ππ π‘πππππππ‘π’ππ
so π π,β = β
πΆβ = πβΜ π π,β = 1.5
ππ
π
β = β
πΆ π = π πΜ π π,π = 15
ππ
π
(4178
π½
ππ β πΎ
) = 62,670
π
πΎ
4. These results show that πΆ π = πΆ πππ because it is smaller. The heat capacity ratio will give us our NTU:
(8) πΆπ =
πΆ πππ
πΆ πππ₯
=
62,670
β
= 0
We can now consult table 8-41
and observe that when πΆ π = 0, NTU is given by:
(9) πππ = βππ(1 β π)
The efficiency, Ξ΅, is needed still and is given by:
(10) π =
π
π πππ₯
Where q is the previously found heat transfer rate and π πππ₯ is:
(11) π πππ₯ = πΆ πππ( πβ,π β ππ,1) = 62,670
π
πΎ
(354.48 πΎ β 280 πΎ) = 4,667.66 ππ
So substituting in our efficiency into equation 9, our NTU becomes
πππ = βππ (1 β
3458.115 ππ
4.667.66 ππ
) = 1.35
We can finally solve for our tube length by substituting values into equation 6 and solving for L:
πππ =
ππ΄
πΆ ππ π
=
πππ·πΏ
πΆ πππ
πΏ =
πππ β πΆ πππ
πππ·
=
(1.35) (62,670
π
πΎ
)
3,266
π
π2 β πΎ
β π(0.01π)
= 824.57 π
(c)
With an accumulated fouling factor of π π = 0.0003
π2βK
π
, and using calculated length and inlet conditions,
what mass fraction of vapor is condensed?
Because the previous un-fouled system condensed all of the steam into water we can consider the former
condensed amount 100%. All we need to do is find the fouled efficiency and compare the ratios. The
overall heat transfer coefficient with fouling is
ππ = (
1
1
π
+ π π
) = (
1
1
3266
+ 0.0003
) = 1650
π
π2 β πΎ
824.57
100
π
π‘π’ππ
= 8.25
π
π‘π’ππ
5. Our new NTU can be found using equation 6:
πππ =
ππ΄
πΆ πππ
=
1650 β π β (0.01)(824.57)
62,670
= 0.682
We can find our effectiveness by rearranging equation 9:
π π =
β1
π πππ + 1 =
β1
π0.680 + 1 = 0.4944
The former efficiency was
π π’βπ =
3458.115
4.667.66
= 0.7409
So our mass fraction condensed is
(d)
With tube length from (b) and fouling from (c), explore the extent to which the water flow rate and inlet
temperature can be varied to improve the condenser performance. Show results graphically.
*Note: I am very confused with this problem, if my πΜ changes, it effects many other variables.
My argument will be that as πΜ π changes, so does Reynolds, Nusselt, convection coefficient, overall heat
transfer coefficient, NTU, and finally effectiveness.
Data in Appendix
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5 10 15 20 25 30 35
Effectiveness
Mass Flow Rate of Cooling Water, kg/s
Mass Flow Rate of Cooling Water vs. Effectiveness
π π’βπ
π π
=
0.7409
0.4944
= 0.667
6. 13.55
Known:
Two concentric spheres are separated by an air space.
π·1 = 0.8 π; π·2 = 1.2 π; π1 = 400 πΎ; π2 = 300 πΎ
Diagram:
Assume:
- Steady State
Find:
(a) If the surfaces are black,what is the net rate of radiation exchange between the surfaces?
This can be calculated with:
(11) π = π΄π(ππ
4 β ππ π’π
4 )
Where π β‘ π‘βπ ππ‘ππππ β π΅πππ‘π§ππππ πΆπππ π‘πππ‘ = 5.670 π
10β8 π
π2βπΎ4
. Solving equation 11:
π = π0.8π2 (5.670 π10β8
π
π2 β πΎ4 .) (400 πΎ4 β 300 πΎ4)
(b)
What is the net rate of radiation exchange if they are diffuse and gray and π1 = 0.5 and π2 = 0.05?
From Table 13.31
the heat flux for concentric spheres is
(12) π12 =
ππ΄( π1
4
β π2
4)
1
π1
+
1 β π2
π2
(
π1
π2
)
2
So
π = 1995 π
7. π12 =
(5.670 β 10β8) π0.82(100 πΎ)
1
0.5
+
1 β 0.05
0.05
(
0.8
1.2
)
2 = 191 π
(c)
What is the net rate of radiation exchange if π·2 is increased to 20 m, with π1 = 0.5 and π2 = 0.05, and
π·2 = 0.8 π?
Can just plug in additional value to equation 12 :
π12 =
(5.670 β 10β8) π0.82(100 πΎ)
1
0.5
+
1 β 0.05
0.05
(
0.8
20
)
2 = 982.6 π
What error introduced if π2 = 1?
Using equation 12 again we get:
π12 =
(5.670 β 10β8) π0.82(1.75 β 1010 )
1
0.5
+
1 β 1
1
(
0.8
20
)
2 = 997.5 π
(d)
For π·2 = 1.2 π and emissivities of π1 = 0.1,0.5, πππ 1.0, compute and plot the net rate of radiation
exchange as a function of π2 πππ 0.05 β€ π2 β€ 1.0.
π12 = 191 π
π12 = 982.6 π
100 β
982.6
997.5
β 100 = 1.5%
8. Data in Appendix
0
50
100
150
200
250
0 0.2 0.4 0.6 0.8 1
q, W
Ξ΅2
Rate of Radiation, Ξ΅1 = 0.1
0
200
400
600
800
1000
1200
0 0.2 0.4 0.6 0.8 1
q, W
Ξ΅2
Radiation Rate, Ξ΅1 = 0.5
0
500
1000
1500
2000
2500
0 0.2 0.4 0.6 0.8 1
q, W
Ξ΅2
Radiation Rate, Ξ΅1 = 1.0
9. 13.112
Known:
There is a double paned window separated by atmospheric air for which the critical Raleigh number for
the onset of convection is π π = 2000.
π1 = 22β ; π2 = β20β
Diagram:
Assume:
- Isothermal Surfaces
- Steady State
- Quiescent Air
- Air is dry and at 1 atm
Find:
(a)
What is the conduction heat flux across the air gap for the optimal spacing πΏ ππ between the panes?
For this first part we are just dealing with conduction heat flux through the air so need to first find the
properties of the air at the mean temperature:
π π =
π1 + π2
2
=
22 + (β20)
2
= 1β
From Table A-41
π β 106 (
π2
π
) πΌ β 106 (
π2
π
) π½ ( πΎβ1)(
1
π π
) π β 103 (
π
π β πΎ
)
13.67 19.2 0.003648 24.3
The critical Rayleigh Number, where free convection will turn turbulent, is given as 2000. Using the
definition of the Rayleigh number I think I just need to solve for x:
(13) π π =
ππ½(π2 β π1)π₯3
πΌ β π
π₯ = β
2000(19.2 β 10β6)(13.64 β 10β6)
9.8
π
π 2 (0.003648 πΎβ1)(295.15 β 253.15)
3
= 0.00704 π
10. So the conduction heat flux across it can be calculated with Fourierβs Law
(14) πΜ = π
βπ
βπ₯
Plugging in:
(b)
If the glass has an emissivity of π π = 0.90, what is the total heat flux across the gap?
We can look at Table 13.31
again to get the equation we need parallel planes:
(15) π12 =
ππ΄( π1
4
β π2
4)
1
π1
+
1
π2
β 1
With π1 = π2 = 0.9 and solving for heat flux instead of heat rate:
πΜ12 =
5.670 β 10β8 π
π2 β πΎ4 (295.15 πΎ4 β 253.15 πΎ4)
1
0.9
+
1
0.9
β 1
= 161.3
π
π2
So total heat flux is the conduction from part (a) plus the radiation:
(c)
What is the total heat flux if a special, low emissivity coating (π1 = 0.1) is applied to one of the panes at
its air glass interface? What is total heat flux if both panes are coated?
One Pane
Assuming the other pane still has an emissivity of 0.9 we can just plug into equation 15 with our new
value:
πΜ12 =
5.670 β 10β8 π
π2 β πΎ4 (295.15 πΎ4 β 253.15 πΎ4)
1
0.9
+
1
0.1
β 1
= 19.5
π
π2
πΜ = (0.0243
π
π β πΎ
)
42β
0.00704 π
= 145
π
π2
πΜ πππ‘ = 161.3 + 145 = 306.3
π
π2
11. Then add to the conduction:
Two Panes
Same procedure as with one pane but both emissivities equal to 0.1
πΜ12 =
5.670 β 10β8 π
π2 β πΎ4 (295.15 πΎ4 β 253.15 πΎ4)
1
0.1
+
1
0.1
β 1
= 10.4
π
π2
And add to conduction
πΜ πππ‘ = 19.5 + 145 = 164.5
π
π2
πΜ πππ‘ = 10.4 + 145 = 155.4
π
π2