Answer:
The correct answer is "4.8137 m³". The further explanation is given below.
Explanation:
Firstly we have to calculate the concentration of Se:
Concentration the fish can take:
According to the general dilution principle will be:
⇒
The volume that can take the farmer will be:
On converting this into m³, we get
=
Finally the volume the farmer can remove would be:
Answer:
(1) 5.74
(2) 5.09
(3) 3.05×10⁻⁵ kg/s
(4) 0.00573 kW
Explanation:
The parameters given are;
Working temperature, = -15°C = 258.15 K
Temperature of the cooling water, = 30°C = 303.15 K
(1) The Carnot coefficient of performance is given as follows;
(2) For ammonia refrigerant, we have;
s₂ = s₁ = 4.9738 kJ/(kg·K)
0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738
∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89
h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg
(3) The circulation rate is given by the mass flow rate, as follows
The refrigeration capacity = 105 kJ/h
The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg
Therefore;
= 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s
(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg
The rating power = Work done per second = W×
∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.
Answer
Assuming
At 10000 m height temperature T = -55 C = 218 K
At 1000 m height temperature T = 0 C = 273 K
R = 287 J/kg K
V₂ = V₁ ×1.1222
V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s
V₂ = 1.1222 × 147.5 = 165.49 m/s
so, the jetliner need to increase speed by ( V₂ -V₁ )
= 165.49 - 147.5
= 17.5 m/s