During experiments on laser etching of microchips for IBM, a nuclear engineer places a discshaped sample 5 mm in diameter in a v
acuum chamber with surface temperature of 300 K. A 500 mW laser is then used to irradiate the sample uniformly. The sample surface can be considered ‘gray’ with an emissivity of 0.9 and can be considered ‘thin’ so that the disk temperature is uniform. In addition, a non-participating gas at a temperature of 27⁰C is used to cool the sample with a convection coefficient of 50 W/(m2K). Find the steady-state temperature of the disk
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Answer:
The lake can withdraw a maximum of cubic feet per year to provide water supply for the Triangle area.
Explanation:
The maximum amount of water that can be withdrawn from the lake is represented by the following formula:
(Eq. 1)
Where:
- Available amount of water for water supply in the Triangle area, measured in cubic feet per year.
- Inflow amount of water, measured in cubic feet per year.
- Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.
- Amount of water due to precipitation, measured in cubic feet per year.
- Amount of evaporated water, measured in cubic feet per year.
Then, we can expand this expression as follows:
(Eq. 2)
Where:
- Average watershed inflow, measured in cubic feet per second.
- Average flow to be released, measured in cubic feet per second.
- Yearly time, measured in seconds per year.
- Change in lake height due to precipitation, measured in feet per year.
- Change in lake height due to evaporation, measured in feet per year.
- Surface area of the lake, measured in square feet.
If we know that ,
,
,
,
and
, the available amount of water for supply purposes in the Triangle area is:
The lake can withdraw a maximum of cubic feet per year to provide water supply for the Triangle area.
Answer:
110 m or 11,000 cm
Explanation:
- let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
- let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
- let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively
M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s
C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c
<u>Using effectiveness-NUT method</u>
- <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>
C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c
C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c
From the result above cold fluid heat capacity rate is smaller
Dimensionless heat capacity rate, C = minimum capacity/maximum capacity
C= C<em>min</em>/C<em>max</em>
C = 5.016/8.62 = 0.582
.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>
Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)
Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW
.<em>3 Third, we determine the actual heat transfer rate, Q</em>
Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)
Q = (5.016 kW/°c)(80 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW
.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε
ε<em> </em>= Q/Qmax
ε <em>= </em>303.66 kW/702.24 kW
ε = 0.432
.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>
NTU =
NTU =
NTU = 0.661
<em>.6 sixth, we determine Heat Exchanger surface area, As</em>
From the question, the overall heat transfer coefficient U = 640 W/m²
As =
As =
As = 5.18 m²
<em>.7 Finally, we determine the length of the heat exchanger, L</em>
L =
L =
L= 109.91 m
L ≅ 110 m = 11,000 cm