Answer:
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Explanation:
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An array of electronic chips is mounted within a sealedrectangular enclosure, and colling is implemented by attaching analuminum
heat sink (k=180 W/m*K). The base of the heat sinkhas dimensions of w1=w2=100mm, while the 6 fins are of thicknesst=10mm and pitch S=18mm. The fin length is Lf=50 mm, and thebase of the heat sink has a thickness of Lb=10mm.
If cooling is implemented by water flow through the heat sink,with u[infinity]=3 m/s and T[infinity]=17 C, what is the base temperatureTb of the heat sink when the power dissipation by the chips isPelec=1800W? The average convection coefficient for surfacesof the fins and the exposed base may be estimated by assumingparallel flow over a flat plate. Properties of the water maybe approximated as k=0.62 W/m*K, ν=7.73E-7 m2/s, andPr=5.2.
If cooling is implemented by water flow through the heat sink,with u[infinity]=3 m/s and T[infinity]=17 C, what is the base temperatureTb of the heat sink when the power dissipation by the chips isPelec=1800W? The average convection coefficient for surfacesof the fins and the exposed base may be estimated by assumingparallel flow over a flat plate. Properties of the water maybe approximated as k=0.62 W/m*K, ν=7.73E-7 m2/s, andPr=5.2.
1 answer:
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Answer:
a) , b)
, c)
, d)
, e)
Explanation:
a) The coefficient of performance of the reversible refrigeration cycle is:
The temperature of the hot reservoir is:
b) The coefficient of performance is:
c) The temperature of the hot reservoir can be determined with the help of the following relation:
d) The coefficient of performance is:
e) The temperature of the cold reservoir is:
Answer:
a. 0.11
b. 110 students
c. 50 students
d. 0.46
e. 460 students
f. 540 students
g. 0.96
Explanation:
(See attachment below)
a. Probability that a student got an A
To get an A, the student needs to get 9 or 10 questions right.
That means we want P(X≥9);
P(X>9) = P(9)+P(10)
= 0.06+0.05=0.11
b. How many students got an A on the quiz
Total students = 1000
Probability of getting A = 0.11 ---- Calculated from (a)
Number of students = 0.11 * 1000
Number of students = 110 students
So,the number of students that got A is 110
c. How many students did not miss a single question
For a student not to miss a single question, then that student scores a total of 10 out of possible 10
P(10) = 0.05
Total Students = 1000
Number of Students = 0.05 * 1000
Number of Students = 50 students
We see that 5
d. Probability that a student pass the quiz
To pass, a student needed to get at least 6 questions right.
So we want P(X>=6);
P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)
=0.08+0.12+0.15+0.06+0.05=0.46
So, the probability of a student passing the quiz is 0.46
e. Number of students that pass the quiz
Total students = 1000
Probability of passing the quiz = 0.46 ----- Calculated from (d)
Number of students = 0.46 * 1000
Number of students = 460 students
So,the number of students that passed the test is 460
f. Number of students that failed the quiz
Total students = 1000
Total students that passed = 460 ----- Calculated from (e)
Number of students that failed = 1000 - 460
Number of students that failed = 540
So,the number of students that failed is 540
g. Probability that a student got at least one question right
This means that we want to solve for P(X>=1)
Using the complement rule,
P(X>=1) = 1 - P(X<1)
P(X>=1) = 1 - P(X=0)
P(X>=1) = 1 - 0.04
P(X>=1) = 0.96
