Answer:
The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3
Heat flux is 9.67×10^7 Btu/hrft^2
Explanation:
Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr
Area (A) = πD^2/4
Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft
A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2
Volume (V) = A × Length
L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft
V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3
Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3
Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2
Answer:
There is 0.466 KW required to operate this air-conditioning system
Explanation:
<u>Step 1:</u> Data given
Heat transfer rate of the house = Ql = 755 kJ/min
House temperature = Th = 24°C = 24 +273 = 297 Kelvin
Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin
<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.
COPr,c = 1 / ((To/Th) - 1)
COPr,c = 1 /(( 308/297) - 1)
COPr,c = 1/ 0.037
COPr,c = 27
<u>Step 3:</u> The power input cna be given as followed:
Wnet,in = Ql / COPr,max
Wnet, in = 755 / 27
Wnet,in = 27.963 kJ/min
Win = 27.963 * 1 KW/60kJ/min = 0.466 KW
There is 0.466 KW required to operate this air-conditioning system
Given :
Force, .
Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )
To Find :
The work done by force F .
Solution :
Displacement vector between point A and B is :
Now, we know work done is given by :
W = 12000 J
Therefore, work done by force is 12000 J .
Answer:
a) , b), , , c) , , ,
Explanation:
a) The total number of users that can be accomodated in the system is:
b) The length of the side of each cell is:
Minimum time for traversing a cell is:
The maximum time for traversing a cell is:
The approximate time is giving by the average of minimum and maximum times:
c) The total number of users that can be accomodated in the system is:
The length of each side of the cell is:
Minimum time for traversing a cell is:
The maximum time for traversing a cell is:
The approximate time is giving by the average of minimum and maximum times:
Answer:
D. a triangle and a T-Square
Explanation:
A T-Square is the best drawing tool to create squares. You would need a squares to create cubes.