Answer:
efficiency =42.62%
AMOUNT OF POWER REJECTED IS 20.080 kW
Explanation:
given data:
power 20 hp
heat energy = 35kW
power production = 20 hp = 20* 746 W = 14920 Watt [1 hp =746 watt]
=42.62%
b)
= 20.080 kW
AMOUNT OF POWER REJECTED IS 20.080 kW
Answer:
the difference in pressure between the inside and outside of the droplets is 538 Pa
Explanation:
given data
temperature = 68 °F
average diameter = 200 µm
to find out
what is the difference in pressure between the inside and outside of the droplets
solution
we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is
σ = 2.69 × N/m
so average radius = = 100 µm = 100 ×
m
now here we know relation between pressure difference and surface tension
so we can derive difference pressure as
2π×σ×r = Δp×π×r² .....................1
here r is radius and Δp pressure difference and σ surface tension
Δp =
put here value
Δp =
Δp = 538
so the difference in pressure between the inside and outside of the droplets is 538 Pa
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
