Scientists use characteristics to compare stars. Match each characteristic on the left with the statement on the right that uses
1 answer:
You might be interested in
Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k
Explanation:
Temperature range → 0 to 80'c
respective voltage output → 0.2v to 0.5v
required temperature range 20'c to 40'c
Where T = 20'c respective voltage
Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)
so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v
Therefore, the formula for the circuit will be
The simplest circuit will be a op-amp
NOTE: Refer the figure attached
Vs is sensor output
Vr is the reference volt, Vr = 0.275v
choose R2, R1 such that it will maintain required ratio
The output Vo can be connected to voltage buffer if you required better isolation.
