Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
Answer:
El principio de arquimedes es un principio de la hidroestática que explica lo que experimenta un cuerpo ... la teoría y después como es costumbre pasaremos a ver problemas resueltos sobre éste principio. ... Si la magnitud del peso del cuerpo es menor a la magnitud de empuje. ... Alguien me ayuda con este Ejercicio :.
Explanation:
Answer:
The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C
Explanation:
given information:
mass, m = 3 kg
initial temperature, T₁ = 40°C
current, I = 10 A
voltage, V = 50 V
time, t = 30 min = 1800 s
Heat for the system because of the resistance is
Q = V I t
where
V = voltage (V)
I = current (A)
t = time (s)
Q = heat transfer to the system (J)
so,
Q = V x I x t
= 50 x 10 x 1800
= 900000
= 9 x 10⁵ J
the heat transfer in the closed system is
Q = ΔU + W
where
U = internal energy
W = work done by the system
thus,
Q = ΔU + W
9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.
ΔU = 9 x 10⁵ J = 900 kJ
then, the energy change in the system is
ΔU = m c ΔT
ΔT = ΔU / m c, c = 4.186 J/g°C
= 900 / (3 x 4.186)
= 71.67°C
so,the final temperature (T₂)
ΔT = T₂ - T₁
T₂ = ΔT + T₁
= 71.67°C + 40°C
= 111.67°C
Given Information:
Balanced Y-Y three phase circuit
Phase Voltage = Van = 120 < 10° V
Load Impedance = Zy = 20 +j15 Ω = 25 < 36.86° Ω
Required Information:
Load Voltages = Vab, Vbc, Vca = ?
Line currents = Ia, Ib, Ic = ?
Answer:
Vab = 208 < 40°
Vbc = 208 < -80°
Vca = 208 < 160°
Ia = 4.8 < -26.86°
Ib = 4.8 < -146.86°
Ic = 4.8 < 93.14°
Explanation:
Since it is balanced 3 phase system, all phases have equal magnitude and 120° phase shift
Van = 120 < 10° V
Vbn = 120 < -110° V
Vcn = 120 < 130° V
In a Y connected system, the phase voltage and line voltage are related as
Vab = 
(Van) <+30°
Vab = *120<10°+30°
Vab = 208 < 40°
Vbc = 208 < -80°
Vca = 208 < 160°
In a Y connected circuit, Iphase = Iline
Ia = Van/Zy
Ia = 120 < 10°/25 < 36.86°
Ia = 4.8 < -26.86°
Ib = 4.8 < -146.86°
Ic = 4.8 < 93.14°
