Answer:
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Explanation:
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Answer:
158
Explanation:
The input is 6.185/10.000 = 0.6185 of the full-scale of the A/D converter. If a 10V input is represented by a register value of 255, then the expected register value for this input is about ...
0.6185·255 = 157.7175 ≈ 158
_____
In hexadecimal, that is 9E.
Answer:
A) m' = 351.49 kg/s
B) m'= 1036.91 kg/s
Explanation:
We are given;
Pressure Ratio;r_p = 12
Inlet temperature of compressor;T1 = 300 K
Inlet temperature of turbine;T3 = 1000 K
cp = 1.005 kJ/kg·K
k = 1.4
Net power output; W' = 70 MW = 70000 KW
A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;
m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))
At, 100% efficiency, plugging in the relevant values, we have;
m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))
m' = 70000/199.1508
m' = 351.49 kg/s
B) At 85% efficiency, the formula will now be;
m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))
Where η is efficiency = 0.85
Thus;
m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))
m' = 70000/(1.005*(432.09129 - 364.9189)
m'= 1036.91 kg/s
Solution :
Given :
The power of the air‐conditioning (AC) unit is , W = 0.434 kW
The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by = 6.22
Therefore he heat removed is given by ,
Now if the electricity is valued at 0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24
= 6.48
Therefore the operating cost = $ 6.48 for 24 hours.