Download Solutions to second midterm

College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 483
Alternative Energy Engineering II
Spring 2010 Number: 17724 Instructor: Larry Caretto
Solutions to Second Midterm Exam
a. If the incoming solar
radiation reaching the
surface is 800 W/m2,
how much radiation
would be would be
adsorbed?
1
0.9
0.88
0.8
Emissivity
1. The spectral emissivity of
a surface is shown in the
figure at the right. Answer
the following questions
for a surface that has this
spectral emissivity.
0.7
0.6
0.5
0.4
0.3
0.24
0.2
3.5 m
0.1
0
0
2
4
6
8
10
12
14
16
To do this calculation we
have to determine the mean
Wavelength (m)
absorptivity of the surface
for incoming solar
temperature with an effective temperature of 5800 K. At this temperature, the boundary point of
this distribution occurs at T = (3.5 m)(5800 K) = 20,300 mK. At this point the fraction of black
body radiation below the T boundary is 0.9859 (from interpolation in Table 6.2 on page 118 of
Hodge.) This gives the average solar absorptivity as follows.
 solar  solar  0.88[ f (T  20,3200 m  K )  0]  0.241  f (T  20,300 m  K )
 0.880.9859  0  0.241  0.9859  0.8710
Multiplying the incoming solar radiation by this solar absorptivity gives the absorbed solar
radiation as (0.8710)(800 W/m2) = 696.8 W/m2 .
b. If the surface is opaque, what happens to the radiation that is not absorbed?
The sum of the absorbed plus transmitted plus reflected radiation must equal zero. An opaque
surface does not transmit radiation, so the radiation that is not absorbed is reflected.
c. What is the emissive power of the surface if its temperature is 340 K?
To do this calculation we have to determine the mean emissivity of the surface at its temperature
of 340 K. At this temperature, the boundary point spectral emissivity distribution occurs at T =
(3.5 m)(340 K) = 1,190 mK. At this point the fraction of black body radiation below the T
boundary is 0.0.00201 (from interpolation in Table 6.2 on page 118 of Hodge.) This gives the
average emissivity at T = 340 K as follows.
 340 K   340 K  0.88[ f (T  1,190 m  K )  0]  0.241  f (T  1,190 m  K )
 0.880.00201  0  0.241  0.00201  0.2413
The emissive power of the surface is given by the following calculation:
Jacaranda (Engineering) 3333
E-mail: [email protected]
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Phone: 818.677.6448
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Second midterm exam
ME 483, L. S. Caretto, Spring 2010
E   T 4  0.2413
Page 2
5.6704 x10 8W
340K 4 = 182.8 W/m2 .
2
4
m K
d. What is the total amount of radiant energy leaving the surface?
This is the sum of the emissive power plus the reflected energy. The reflected energy is the
incoming solar energy (800 W/m 2) minus the energy absorbed (696.8 W/m2), which is 103.2
W/m2. Adding this to the emissive power of the surface gives the total radiant energy leaving the
surface as 103.2 W/m2 + 182.8 W/m2 = 286.1 W/m2 .
2. The surface of problem one is to be used as an absorber plate in a solar collector.
Measurements on this collector show that the temperature of the lower glass plate (the
one closest to the absorber plate) is 320 K when the absorber plate temperature is 340 K.
The convection coefficient for the air gap between the absorber plate and the first glass
plate is 3.5 W/m2K and the emissivity of the glass plate is 0.85.
a. What is the heat transfer, per unit area of the collector, between the absorber plate and
the lower glass plate?
The equation below for the heat transfer between the absorber plate and the lower glass plate is
taken from slide 34 of the lecture presentation for March 22-24.
Qtop  h p  g 2 Ac TP  Tg 2  

Ac TP4  Tg42
1
1

1
P
 g2


Qtop
Ac

8

3.5W
340 K  320 K
m2  K


5.6704 x10 W
340 K 4  320 K 4 107.8W
2
4
m K


1
1
m2

1
0.2413 0.85
Q/A = 107.8 W/m2
b. Is the heat transfer in part a greater than, less than, or equal to Q top? Briefly explain
your answer.
As the equation implies, this heat transfer is equal to Qtop. There is a series resistance between
the absorber plate and the ambient temperatures. The heat transfer through each resistor in the
thermal circuit is the same.
c. What is the combined heat transfer coefficient, UP-g2, for the heat transfer between the
absorber plate and the lower glass plate?
The heat transfer coefficient is simply Up-g2 = Q/[Ac(Tp – Tg2)]; it is computed from the heat transfer
found in part a as follows.
Qtop
107.8W
Ac
m2
U


Ac TP  Tg 2  TP  Tg 2  340 K  320 K
Qtop

U
5.39 W
m2  K
d. Is your answer to part c greater than, less than, or equal to the heat transfer
coefficient, Utop? Briefly explain your answer.
This Up-g2 value will be greater than the overall heat transfer coefficient, Utop. Here are two
possible arguments. In the first argument we compare the equations for the final value of Utop =
Qtop/[Ac(Tp – Tambient)] with the equation for Up-g2 of the first gap, Up-g2 = Qtop/[Ac(Tp – Tg2)]. We see
that Up-g2 = Utop(Tp – Tambient)/ (Tp – Tg2). Since (Tp – Tambient) is greater than (Tp – Tg2), Up-g2 > Utop.
Second midterm exam
ME 483, L. S. Caretto, Spring 2010
Page 3
The second argument notes that Up-g2 = 1/Rp-g2 and Utop = 1/Rtop = 1/(Rp-g2 + Rg2-g1 + Rg1-a).
Comparing these equations gives 1/Utop = 1/U p-g2 + Rg2-g1 + Rg1-a. Since all the resistances are
positive, this is only possible if 1/Utop > 1/U p-g2 which means that U p-g2 > Utop.
3. Answer the following questions for the solar collector described on the attached rating
sheet.
a. What is the efficiency of the collector for heating hot water (inlet temperature = 40oC)
when the incident solar radiation is 800 W/m 2 and the ambient temperature is 20oC?
From the efficiency equation data with an incoming solar radiation of I = 800 w/m 2 and a
temperature difference, P = Tin – Ta = 40oC – 20oC = 20oC, we get the following result.
  0.709  3.26090
P
P2
20
20 2
 0.01203
 0.709  3.26090
 0.01203
 = 62.1%
I
I
800
800
b. If the collector fluid is water ( = 1000 kg/m3, cp = 4184 J/kgK), what is the outlet
temperature of the fluid under the conditions of part a using the test flow rate from the
collector rating sheet?

Tout

62.1%  8002W 2.411 m 2 1 J
Qu
IAc
W s
m
 Tin 
 Tin 
 40o C 
3
m c p
m c p
48 mL 1000 kg m
4184 J
3
6
s
m
10 mL kgo C
Tout = 46.07oC
c. What fraction of a monthly heating load of 90 therms can this collector supply if the
average solar radiation during a 30-day month is 4 kWh/m2/day. For this month,
/()n = 0.94 and the average ambient temperature is 17.8oC? Assume that F’R/FR =
0.97 for this month.
Using the f-chart approach we can find the value of FRUc = 3.975 W/m2oC and FR()n = 0.706 as
the negative slope and intercept from the data sheet for the collector. The total load of 90
therms/month = (90x105Btu/month)(1.055 kJ/Btu)(GJ/106 kJ) = 9.495 GJ/month. The total
radiation for the month will be (30 days/month)(4 kWh/m2/day)(0.0036 GJ/kWh) = 0.432
GJ/m2/month. In compatible units the product of FRUc = (3.975 W/m2oC)(10-9 GJ/Ws)(3600
s/hr)(24 hr/day)(30 days/mo) = 0.01030 GJ/m2·mo.

F ' t
X  Ac  FRU c R
Tref  Ta
FR D


GJ
mo
o
o 
  2.411m2  0.01030



0
.
97
100
C

17
.
8
C   0.2086
2
9.495 GJ
 m  mo


0.432 GJ 

'



F  H i ,total
2
m 2  mo 
Y  Ac  FR  n R
  2.411 m 0.7150.97 0.94  9.495 GJ   0.0715
FR  n D 



mo 



For these values of X and Y: f = 1.029Y – 0.065X – 0.245Y2 + 0.0018X2 + 0.0215Y3 =
1.029(0.0715) – 0.065(0.2086) – 0.245(0. 0715)2 + 0.0018(0. 2086)2 + 0.0215(0.0715)3 = 0.0589.
Thus, 5.89% of the monthly heating load is supplied by solar .