There are three basic methods of transmitting power as a. Electronical, mechanical, and fluid power b. Electical, mechanical, an
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For the given problem it is necessary to recap the concepts about Power, that is, is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time.
The equation for power can be written as
Where,
F= Net Force
V =Velocity
By the second newton law, force can be:
F = mg
Where m means the mass and g the gravity acceleration.
We can also write the equation as,
P = mgv
Replacing the values
P = 700*9.8*25
P = 17.15kW
Tenemos unidades en dos sistemas diferentes, por lo que convertimos los HP a Sistema internacional y tenemos que
1hp = 0.746kW
Then the fraction is,
Therefore the fraction of the engine power is being used to make the airplane climb is 22.984%
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.