Answer:
14.506°C
Explanation:
Given data :
flow rate of water been cooled = 0.011 m^3/s
inlet temp = 30°C + 273 = 303 k
cooling medium temperature = 6°C + 273 = 279 k
flow rate of cooling medium = 0.02 m^3/s
Determine the outlet temperature
we can determine the outlet temperature by applying the relation below
Heat gained by cooling medium = Heat lost by water
= ( Mcp ( To - 6 ) = Mcp ( 30 - To )
since the properties of water and the cooling medium ( water ) is the same
= 0.02 ( To - 6 ) = 0.011 ( 30 - To )
= 1.82 ( To - 6 ) = 30 - To
hence To ( outlet temperature ) = 14.506°C
Answer:
a) 180 m³/s
b) 213.4 kg/s
Explanation:
= 1 m²
= 100 kPa
= 180 m/s
Flow rate

Volumetric flow rate = 180 m³/s
Mass flow rate

Mass flow rate = 213.4 kg/s
Answer:
This is an asynchrnous 3-bit counter. Just note that this design is different and works differently than its synchronous counterpart. It's an easier design than its synchronous counterpart, and is not as reliable because it has delays.
Answer:
0.75 in
Explanation:
The 1 inch measure has 16 divisions.
Find the measure of 1 division as: 1/16 = 0.0625 in
The side length of the square has 12 divisions
1 division = 0.0625 in
12 divisions = 0.0625*12 = 0.75 in
The correct answer choice is A. 0.75.
Answer choices B is incorrect because from the diagram, it is clear that the side length of this square is less that 1 inch.
Answer choice C is incorrect because 0.34 in will mean 0.34/0.0625 divisions. This is 5 divisions, yet the square side length covers 12 divisions.