Explanation:
Formula for steady flow energy equation for the flow of fluid is as follows.
![m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w](https://tex.z-dn.net/?f=m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%5D%20%2B%20z_%7B1%7Dg%5D%20%2B%20q%20%3D%20m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%20%2B%20z_%7B1%7Dg%5D%20%2B%20w)
Now, we will substitute 0 for both
and
, 0 for w, 334.9 kJ/kg for
, 2726.5 kJ/kg for
, 5 m/s for
and 220 m/s for
.
Putting the given values into the above formula as follows.
![1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0](https://tex.z-dn.net/?f=1%20%5Ctimes%20%5B334.9%20%5Ctimes%2010%5E%7B3%7D%20J%2Fkg%20%2B%20%5Cfrac%7B%285%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%20q%20%3D%201%20%5Ctimes%20%5B2726.5%20%5Ctimes%2010%5E%7B3%7D%20%2B%20%5Cfrac%7B%28220%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%200)
q = 6597.711 kJ
Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.
Answer: 1.6Hz
foe[vqefmvkeqmvkevkefmvqelkfveklveqv
It takes 7.0s for a piece of wood to be falling from a tall building in the velocity of 68.6 m/s
<u>Explanation:</u>
v = u + at
68.6 ms^-1 = 0 ms^-1 + (9.8 ms^-2 x t)
t = 68.6 ms^-1 / 9.8 ms^-2
t = 7.0s
It takes 7.0s for a piece of wood to be falling from a tall building in the velocity of 68.6 m/s
Since it is specified that the relationship is linear, you could think of a linear graph for this. The base year would be 1994. The first point is at (0,2523) while the second point is (2,2891). The slope would be
m = (2891 - 2523)/(2-0) = 184
The slope-intercept form is y = mx + b. Let's substitute (0,2523) to the equation such that
2523 = 184(0)+b
b = 2523
Hence, the equation of the linear graph is y = 184x + 2523. Now, at 2006, x = 2006 - 1994 = 12 years.
y = 184(12) + 2523 = 4,731 students
Answer:
e_12=1-Tc/Th
This is same as the original Carnot engine.
Explanation:
For original Carnot engine, its efficiency is given by
e = 1-Tc/Th
For the composite engine, its efficiency is given by
e_12=(W_1+W_2)/Q_H1
where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.
But the work done can be written as
W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.
e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1
But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel
each other.
e_12=1+Q_C2/Q_H1
but, Q_C2/Q_H2= -T_C/T'
⇒ Q_C2 = -Q_H2(T_C/T')
= Q_C1(T_C/T')
(T1 is the intermediate temperature)
But, Q_C1 = -Q_H1(T'/T_H)
so, Q_C2 = -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by
e_12=1-Tc/Th
This is same as the original Carnot engine.