Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ = 
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ = 
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction, 
, is given as follows;
= T₀ × 
= T₀ × 
× (s₄ + s₂ - s₁ - s₃)
= T₀ × 
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output, 
= 
+ ≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.
Answer:
بدلاً من ذلك يُشار إليه باسم مشاركة أو مشاركة شبكة ، الدليل المشترك هو دليل أو مجلد يمكن الوصول إليه من قبل العديد من المستخدمين على الشبكة. هذه هي الطريقة الأكثر شيوعًا للوصول إلى المعلومات ومشاركتها على شبكة محلية
Explanation:
Answer:
total amount of water after 2 min will be 84.4 kg/s
Explanation:
Given data:
one tank inflow = 0.1 kg/s
2nd tank inflow = 0.3 kg/s
3rd tank outflow = 0.03 kg/s
Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s
From third point, outflow is 0.03 kg/s
Therefore, resultant in- flow = 0.4 - 0.03
Resultant inflow is = 0.37 kg/s
Tank has initially 40 kg water
In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg
So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg
Answer:
a) the log mean temperature difference (Approx. 64.5 deg C)
b) the rate of heat addition into the oil.
The above have been solved for in the below workings
Explanation:
