Answer:
390.242 MPa
Explanation:
Attached is the full solution.
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.
Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
The weight of the specimen in SSD condition is 373.3 cc
<u>Explanation</u>:
a) Apparent specific gravity = 
Where,
A = mass of oven dried test sample in air = 1034 g
B = saturated surface test sample in air = 1048.9 g
C = apparent mass of saturated test sample in water = 975.6 g
apparent specific gravity =
= 
Apparent specific gravity = 2.88
b) Bulk specific gravity 

= 2.76
c) Bulk specific gravity (SSD):


= 2.80
d) Absorption% :


Absorption = 1.44 %
e) Bulk Volume :


= 
Answer:
1) 
2) 
Explanation:
For isothermal process n =1

![V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B5%7D%7B%5B%5Cfrac%7B72%7D%7B80%7D%5D%5E%7B1%2F1%7D%20-%5B%5Cfrac%7B72%7D%7B180%7D%5D%5E%7B1%2F1%7D%7D)

calculate pressure ratio to determine correction factor

correction factor for calculate dpressure ration for isothermal process is
c1 = 1.03

b) for adiabatic process
n =1.4
volume of hydraulic accumulator is given as
![V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}](https://tex.z-dn.net/?f=V_o%20%3D%5Cfrac%7B%5CDelta%20V%7D%7B%5B%5Cfrac%7Bp_o%7D%7Bp_1%7D%5D%5E%7B1%2Fn%7D%20-%5B%5Cfrac%7Bp_o%7D%7Bp_2%7D%5D%5E%7B1%2Fn%7D%7D)
![V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B5%7D%7B%5B%5Cfrac%7B72%7D%7B80%7D%5D%5E%7B1%2F1.4%7D%20-%5B%5Cfrac%7B72%7D%7B180%7D%5D%5E%7B1%2F1.4%7D%7D)

calculate pressure ratio to determine correction factor

correction factor for calculate dpressure ration for isothermal process is
c1 = 1.15

Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation

under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that

hence , actual velocity will be equal to 8.39 m/s