Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
Answer:
Part A: (d/D=0.1)
DeltaV percent=42.6%
Part B:(d/D=0.01)
DeltaV percent=21.7%
Explanation:
We are going to use the following volume flow rate equation:

Above equation can be written as:


First Consider no wire i.e d/D=0
Above expression will become:


Part A: (d/D=0.1)




DeltaV percent=42.6%
Part B:(d/D=0.01)




DeltaV percent=21.7%
Answer:
gauge pressure is 133 kPa
Explanation:
given data
initial temperature T1 = 27°C = 300 K
gauge pressure = 300 kPa = 300 × 10³ Pa
atmospheric pressure = 1 atm
final temperature T2 = 77°C = 350 K
to find out
final pressure
solution
we know that gauge pressure is = absolute pressure - atmospheric pressure so
P (gauge ) = 300 × 10³ Pa - 1 ×
Pa
P (gauge ) = 2 ×
Pa
so from idea gas equation
................1
so
P2 = 2.33 ×
Pa
so gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 2.33 ×
- 1.0 ×
gauge pressure = 1.33 ×
Pa
so gauge pressure is 133 kPa
Answer:
Heat transfer = 2.617 Kw
Explanation:
Given:
T1 = 300 k
T2 = 440 k
h1 = 300.19 KJ/kg
h2 = 441.61 KJ/kg
Density = 1.225 kg/m²
Find:
Mass flow rate = 1.225 x [1.3/60]
Mass flow rate = 0.02654 kg/s
mh1 + mw = mh2 + Q
0.02654(300.19 + 240) = 0.02654(441.61) + Q
Q = 2.617 Kw
Heat transfer = 2.617 Kw