Answer:
Explanation:
Given



From Clausius inequality
=Reversible cycle
=Irreversible cycle
=Impossible
(a)For 
Rejected heat 


thus it is Impossible cycle
(b)



Possible
(c)Carnot cycle




and maximum Work is obtained for reversible cycle when operate between same temperature limits

Thus it is possible
Answer:
A) See Attachment B) See Attachment C) 186.153
Explanation:
C) Per phase Voltage= 220/∛3
= 6.037
S=V²/Z
= (6.037)²/(0.01+i0.05)
= 62051.282-i310256.41
Active power losses per phase= 62.051kW
total Active power losses= 62.051×3
=186.153kW
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
Answer:
True.
Explanation:
According the engineering flow they don not possess flow energy when they are in rest.
When they are in motion they show a translation energy.
The features if fluids may be different according the variables of pressure and temperature.
Known :
D = 12 in = 1 ft
L = 850 ft
Q = 5.6 cfs
hA = 750 ft
hB = 765 ft
PA = 85 psi = 12240 lb/ft²
Solution :
A = πD² / 4 = π(1²) / 4
A = 0.785 ft²
<u>Velocity of water :</u>
U = Q / A = 5.6 / 0.785
U = 7.134 ft/s
<u>Friction loss due to pipe length :</u>
Re = UD / v = (7.134)(1) / (0.511 × 10^(-5))
Re = 1.4 × 10⁶
(From Moody Chart, We Get f = 0.015)
hf = f(L / d)(U² / 2g) = 0.015(850 / 1)((7.134²) / 2(32.2))
hf = 10 ft
PA + γhA = PB + γhB + γhf
PB = PA + γ(hA - hB - hf)
PB = 12240 + (62.4)(750 - 765 - 10)
PB = 10680 lb/ft²
PB = 74.167 psi