Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N
Answer:
1) 
2) 
Explanation:
For isothermal process n =1

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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:
Temperature on the inside ofthe box
Explanation:
The power of the light bulb is the rate of heat conduction of the bulb, 
The thickness of the wall, L = 1.2 cm = 0.012m
Length of the cube's side, x = 20cm = 0.2 m
The area of the cubical box, A = 6x²
A = 6 * 0.2² = 6 * 0.04
A = 0.24 m²
Temperature of the surrounding, 
Temperature of the inside of the box, 
Coefficient of thermal conductivity, k = 0.8 W/m-K
The formula for the rate of heat conduction is given by:
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X
, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X
=
................ 2
Given that; V = 240V, f = 50Hz,
=
, I = 15A, so that;
From equation 2,
X
= 
= 16Ω
To determine the inductance of the inductor,
X
= 2
fL
L = 
= 
= 0.05091
The inductance of the inductor is 0.051H.