Answer:
fluid power
Explanation:
fluids commonly used in fluid power are Oil, Water, Air, CO², and Nitrogen gas, fluid power is commonly confused with hydraulic power, which only uses liquids, fluid power uses either liquids or gases
Answer:
V = 6.33 m/s
Explanation:
Given:
- The length of the wire L = 0.02 m
- The diameter of the wire D = 0.0005 m
- The calibration expression V = 0.0000625*h^2
- Environment temperature T_inf = 298 K
- Surface temperature T_s = 348 K
- The voltage drop dV = 5 V
- The electric current I = 0.1 A
Find:
- the velocity of Air
Solution:
- Calculate the surface area of the wire:
A = pi*D*L
A = pi*(0.0005)*(0.02) = 0.00003142 m^2
- The rate of energy in the wire P:
P = I*dV = 0.1*5 = 0.5 W
- Apply Newton's Law of Cooling:
P = h*A*(T_s - T_inf)
h = P /A*(T_s - T_inf)
Plug in the values:
h= 0.5/ 0.00003142*(348 - 298)
h = 318.27 W /m^2K
- Using the calibration relationship given, compute the velocity of air:
V = 6.25*10^-5 * h^2
V = 6.25*10^-5 * (318.27)^2
V = 6.33 m/s
Answer:
a) 
b) The flow would be going from section (b) to section (a)
Explanation:
1) Notation


For above conversions we use the conversion factor


head loss from section
2) Formulas and definitions
For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.
The formula is given by:

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

3)Part a
And on this case we have all the values in order to replace and solve for 


4)Part b
Analyzing the value obtained for
is a negative value, so on this case this means that the flow would be going from section (b) to section (a).
Answer:
D The answer would be D $270
Explanation:
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