Answer:
Theoretically impossible
Explanation:
Given that:
The mass flow rate = 60 lbm/min
The differential pressure = 200 inches
The flow rate Q is can be expressed by the formula:
where;
K = Discharge coefficient
A = area
Δh = Head drop
g = gravity
From the given parameter, the area is unknown.
Therefore, the flow coefficient of the meter will be theoretically impossible to be determined.
Answer:
2800 [MPa]
Explanation:
In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip
Where is the magnitude of he maximum stress at the tip of the crack,
is the magnitude of the tensile stress,
is
the length of the internal crack, and
is the radius of curvature of the crack.
We have:
We replace:
We get:
Answer:
The phasor technique can't be applied directly in the following cases:
a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)
c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Explanation:
For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.
This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.
This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.
If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:
sin (ωt) = cos (ωt-90º)
This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.
This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).
Solution :
The isentropic efficiency of the turbine is given as :
The entropy relation for the isentropic process is given by :
Now obtaining the properties from the ideal gas properties of air table :
At
Calculating the relative pressure at state 2s :
Obtaining the properties from Ideal gas properties of air table :
At ,
Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:
So, at ,
Now calculating the work developed per kg of air is :
= 1757.57 - 975.66
= 781 kJ/kg
Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.