A Scalar can only be a positive quantity that has a magnitude but no direction ? a)-True b)-False
1 answer:
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Answer:
A) W' = 178.568 KW
B) ΔS = 2.6367 KW/k
C) η = 0.3
Explanation:
We are given;
Temperature at state 1;T1 = 360 °C
Temperature at state 2;T2 = 160 °C
Pressure at state 1;P1 = 10 bar
Pressure at State 2;P2 = 1 bar
Volumetric flow rate;V' = 0.8 m³/s
A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;
Specific volume;v1 = 0.287322 m³/kg
Mass flow rate of water vapour at turbine is defined by the formula;
m' = V'/v1
So; m' = 0.8/0.287322
m' = 2.784 kg/s
Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific enthalpy;h1 = 3179.46 KJ/kg
Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific enthalpy;h2 = 3115.32 KJ/kg
Now, since stray heat transfer is neglected at turbine, we have;
-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2 - h1)
Plugging in relevant values, the work of the turbine is;
W' = -2.784(3115.32 - 3179.46)
W' = 178.568 KW
B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific entropy: s1 = 7.3357 KJ/Kg.k
Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific entropy; s2 = 8.2828 KJ/kg.k
The amount of entropy produced is defined by;
ΔS = m'(s2 - s1)
ΔS = 2.784(8.2828 - 7.3357)
ΔS = 2.6367 KW/k
C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;
h2s = 2966.14 KJ/Kg
Energy equation for turbine at ideal process is defined as;
Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Again, Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2s - h1)
W' = -2.784(2966.14 - 3179.46)
W' = 593.88 KW
the isentropic turbine efficiency is defined as;
η = W_actual/W_ideal
η = 178.568/593.88 = 0.3
Answer:
Cost = $2527.2 per month.
Explanation:
Given that
Discharge ,Q = 130 L/min
So
Cost = $0.45 per cubic meter
1 month = 30 days
1 days = 24 hr = 24 x 60 min
1 month = 30 x 24 x 60 min
1 month = 43,200 min
Lets x
x = 0.13 x 43,200
So the total cost = 5616 x 045 $
Cost = $2527.2 per month.