Answer:
attached below
Explanation:
Applying the state transition Formula
state - next = ( state *2 + in ) % 5
how this works : remainder of previous cycle is doubled to enable the calculation of the new remainder.
Input of current cycle is represented as either 1 or 0
since the dividing number = 5 . possible remainders = 1,2,3,4,0
<em>each remainder is represented as</em> :
S0 = zero remainder , S1 = 1 remainder , S2 = 2 remainder, S3 = 3 remainder,
S4 = 4 remainder
Answer:
A=False
B=False
C=False
D=False
E=False
F=False
Explanation:
A. In an isothermal process, only the reversibly heat transfer is 0, ![Q_{rev}=T (\Delta S)](https://tex.z-dn.net/?f=Q_%7Brev%7D%3DT%20%28%5CDelta%20S%29)
B. Consider the phase change of boiling water. Here, the temperature remains constant but the internal energy of the system increases.
C. This is not true even in reversible process, as can be inferred from the equation in part A.
D. This is only true in reversible processes, but not in all isothermal processes.
E. Consider the phase change of freezing water. Here, the surroundings are increasing their entropy, as they are taking in heat from the system.
F. This is not true if
, like in answer B. One case where this is true is in the reversible isothermal expansion (or compression) of an ideal gas.
Answer:
a)
, b) ![COP_{R} = 29](https://tex.z-dn.net/?f=COP_%7BR%7D%20%3D%2029)
Explanation:
a) The thermal efficiency is:
![\eta = \frac{\dot Q_{in} - \dot Q_{out}}{\dot Q_{in}}\times 100\,\%](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B%5Cdot%20Q_%7Bin%7D%20-%20%5Cdot%20Q_%7Bout%7D%7D%7B%5Cdot%20Q_%7Bin%7D%7D%5Ctimes%20100%5C%2C%5C%25)
![\eta = \frac{1\,MW-0.58\,MW}{1\,MW} \,\times 100\,\%](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B1%5C%2CMW-0.58%5C%2CMW%7D%7B1%5C%2CMW%7D%20%5C%2C%5Ctimes%20100%5C%2C%5C%25)
![\eta = 42\,\%](https://tex.z-dn.net/?f=%5Ceta%20%3D%2042%5C%2C%5C%25)
b) The coefficient of performance is:
![COP_{R} = \frac{\dot Q_{L}}{\dot W}](https://tex.z-dn.net/?f=COP_%7BR%7D%20%3D%20%5Cfrac%7B%5Cdot%20Q_%7BL%7D%7D%7B%5Cdot%20W%7D)
![COP_{R} = \frac{0.58\,MW}{0.02\,MW}](https://tex.z-dn.net/?f=COP_%7BR%7D%20%3D%20%5Cfrac%7B0.58%5C%2CMW%7D%7B0.02%5C%2CMW%7D)
![COP_{R} = 29](https://tex.z-dn.net/?f=COP_%7BR%7D%20%3D%2029)
Answer:
![0.015\ \text{W}](https://tex.z-dn.net/?f=0.015%5C%20%5Ctext%7BW%7D)
Explanation:
= Set temperature = ![70^{\circ}\text{C}](https://tex.z-dn.net/?f=70%5E%7B%5Ccirc%7D%5Ctext%7BC%7D)
= Air temperature = ![50^{\circ}\text{C}](https://tex.z-dn.net/?f=50%5E%7B%5Ccirc%7D%5Ctext%7BC%7D)
A = Surface area = ![30\ \text{mm}^2=30\times 10^{-6}\ \text{m}^2](https://tex.z-dn.net/?f=30%5C%20%5Ctext%7Bmm%7D%5E2%3D30%5Ctimes%2010%5E%7B-6%7D%5C%20%5Ctext%7Bm%7D%5E2)
h = Convection heat transfer coefficient = ![25\ \text{W/m}^2\text{K}](https://tex.z-dn.net/?f=25%5C%20%5Ctext%7BW%2Fm%7D%5E2%5Ctext%7BK%7D)
Heater power is given by
![P_e=hA(T_{set}-T_\infty)\\\Rightarrow P_e=25\times 30\times 10^{-6}(70-50)\\\Rightarrow P_e=0.015\ \text{W}](https://tex.z-dn.net/?f=P_e%3DhA%28T_%7Bset%7D-T_%5Cinfty%29%5C%5C%5CRightarrow%20P_e%3D25%5Ctimes%2030%5Ctimes%2010%5E%7B-6%7D%2870-50%29%5C%5C%5CRightarrow%20P_e%3D0.015%5C%20%5Ctext%7BW%7D)
The required heater power is ![0.015\ \text{W}](https://tex.z-dn.net/?f=0.015%5C%20%5Ctext%7BW%7D)