Explanation:
The given data is as follows.
Diameter = 0.1 m, = 1000 kPa
= 500 kPa
Change in pressure = 1000 kPa - 500 kPa = 500 kPa
Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to .
Q = 5 = = 0.0833
It is known that Q =
where, A = cross sectional area
V = speed of the fluid in that section
Hence, calculate V as follows.
V =
=
=
= 10.61 m/sec
Also it is known that Reynold's number is as follows.
Re =
=
= 1061032.954
As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.
L =
= 90.328 m
Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.
Answer:
When you add thermal energy to an object, these things can happen: Particles move faster (increased kinetic energy). Particles get farther apart (increased potential energy).
Explanation:
Answer:
The main difference b/w these two is that thermal energy deals with the heat while Temperature deals with the kinetic energy of the molecules .
Explanation:
In case of thermal energy , heat is the only concern for the object as in this scenario , bucket and swimming pool has larger surface area to absorb the heat directly from the sun while the temperature is relevant to the kinetic motion of the water molecules in the exhibited objects in the example .
Let me know if you need more explanation about this . thanks
Answer:
6.1 × 10⁻⁸ M
Explanation:
Let's consider the solution of fluorapatite.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
We can relate the solubility (S) with the solubility product constant (Ksp) using an ICE chart.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
I 0 0 0
C +5 S +3 S +S
E 5 S 3 S S
The Ksp is:
Ksp = [Ca²⁺]⁵ × [PO₄³⁻]³ × [F⁻] = (5 S)⁵ × (3 S)³ × S = 84,375 S⁹