Answer:
![a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa](https://tex.z-dn.net/?f=a.T_3%3D1723.8kPa%5C%5Cb.n%3D0.563%5C%5Cc.MEP%3D674.95kPa)
Explanation:
a. Internal energy and the relative specific volume at
are determined from A-17:
.
The relative specific volume at
is calculated from the compression ratio:
![\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825](https://tex.z-dn.net/?f=%5Calpha_r_2%3D%5Cfrac%7B%5Calpha_r_1%7D%7Br%7D%5C%5C%3D%5Cfrac%7B621.2%7D%7B16%7D%5C%5C%3D38.825)
#from this, the temperature and enthalpy at state 2,
can be determined using interpolations
and
. The specific volume at
can then be determined as:
![\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg](https://tex.z-dn.net/?f=%5Calpha_1%3D%5Cfrac%7BRT_1%7D%7BP_1%7D%5C%5C%5C%5C%3D%5Cfrac%7B0.287%5Ctimes%20300%7D%7B95%7D%20m%5E3%2Fkg%5C%5C0.906316m%5E3%2Fkg)
Specific volume,
:
![\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg](https://tex.z-dn.net/?f=%5Calpha_2%3D%5Cfrac%7B%5Calpha_1%7D%7Br%7D%5C%5C%3D%5Cfrac%7B0.906316%7D%7B16%7Dm%5E3%2Fkg%5C%5C%3D0.05664m%5E3%2Fkg)
The pressures at
is:
![P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa](https://tex.z-dn.net/?f=P_2%3DP_3%3D%5Cfrac%7BRT_2%7D%7B%5Calpha_2%7D%5C%5C%5C%5C%3D%5Cfrac%7B0.287%5Ctimes862%7D%7B0.05664%7D%5C%5C%3D4367.06kPa)
.The thermal efficiency=> maximum temperature at
can be obtained from the expansion work at constant pressure during ![s_2-s_3](https://tex.z-dn.net/?f=s_2-s_3)
![\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K](https://tex.z-dn.net/?f=%5Cbigtriangleup%20%5Comega_2_-_3%3DP%28%5Calpha_3-%5Calpha_2%29%5C%5CR%28T_3-T_2%29%3DP%5Calpha%28r_c-1%29%5C%5CT_3%3DT_2%2B%5Cfrac%7BP%5Calpha_2%7D%7BR%7D%28r_c-1%29%5C%5C%5C%5C%3D%28862%2B%5Cfrac%7B4367%5Ctimes%200.05664%7D%7B0.287%7D%282-1%29%29K%5C%5C%3D1723.84K)
b.Relative SV and enthalpy at
are obtained for the given temperature with interpolation with data from A-17 :![a_r_3=4.553 \ and\ h_3=1909.62kJ/kg](https://tex.z-dn.net/?f=a_r_3%3D4.553%20%5C%20and%5C%20%20h_3%3D1909.62kJ%2Fkg)
Relative SV at
is
![a_r_4=\frac{r}{r_c}\alpha _r_3](https://tex.z-dn.net/?f=a_r_4%3D%5Cfrac%7Br%7D%7Br_c%7D%5Calpha%20_r_3)
=![=\frac{16}{2}\times4.533\\=36.424](https://tex.z-dn.net/?f=%3D%5Cfrac%7B16%7D%7B2%7D%5Ctimes4.533%5C%5C%3D36.424)
Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;
![n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563](https://tex.z-dn.net/?f=n%3D1-%5Cfrac%7Bq_o%7D%7Bq_i%7D%5C%5C%3D1-%5Cfrac%7Bu_4-u_1%7D%7Bh_3-h_2%7D%5C%5C%3D1-%5Cfrac%7B65903-214.07%7D%7B1909.62-890.9%7D%5C%5C%3D0.563)
Hence, the thermal efficiency is 0.563
c. The mean relative pressure is calculated from its standard definition:
![MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa](https://tex.z-dn.net/?f=MEP%3D%5Cfrac%7B%5Comega%7D%7B%5Calpa_1-%5Calpa_2%7D%5C%5C%3D%5Cfrac%7Bq_i-q_o%7D%7B%5Calpha_1%281-1%2Fr%29%7D%5C%5C%3D%5Cfrac%7B1909.62-890.9-%2865903-214.7%29%7D%7B0.90632%281-1%2F16%29%7D%5C%5C%3D674.95kPa)
Hence, the mean effective relative pressure is 674.95kPa
Answer:
1.58 W
Explanation:
Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is
![A = \pi r^2 = \pi 12^2 = 452.389 m^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E2%20%3D%20%5Cpi%2012%5E2%20%3D%20452.389%20m%5E2)
From the intensity of the sound we can calculate the power at 12 m
![P = AI = 452.389 * 3.5\times10^{-3} = 1.58 W](https://tex.z-dn.net/?f=P%20%3D%20AI%20%3D%20452.389%20%2A%203.5%5Ctimes10%5E%7B-3%7D%20%3D%201.58%20W)
Janice is the one who learned something by watching what was going on in the world around her.
Answer
given,
wavelength (λ)= 500 n m
thickness of film= 10⁻⁴ cm
refractive index = μ = 1.375
distance traveled is double which is equal to 2 x 10⁻⁴ cm
a) Number of wave
![N = \dfrac{d}{\mu\lambda}](https://tex.z-dn.net/?f=N%20%3D%20%5Cdfrac%7Bd%7D%7B%5Cmu%5Clambda%7D)
![N = \dfrac{2 \times 10^{-6}}{1.375\times 500 \times 10^{-9}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cdfrac%7B2%20%5Ctimes%2010%5E%7B-6%7D%7D%7B1.375%5Ctimes%20500%20%5Ctimes%2010%5E%7B-9%7D%7D)
N = 2.91
N = 3
b) phase difference is equal to
Reflection from the first surface has a 180° (½λ) phase change.
There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.
net phase difference = ![180^0\times \dfrac{3}{2}](https://tex.z-dn.net/?f=180%5E0%5Ctimes%20%5Cdfrac%7B3%7D%7B2%7D)
= 270°