A water tank filled with solar-heated water at 40°C is to be used for showers in a field using gravity-driven flow.
1 answer:
You might be interested in
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q = <em>d</em>T
=
(T₂ - T₁)
From the above equations, the underlying assumption is that remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let be heat constant of calorimeter
Q₂ = ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m ' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
Given Information:
Frequency = f = 60 Hz
Complex rated power = G = 100 MVA
Intertia constant = H = 8 MJ/MVA
Mechanical power = Pmech = 80 MW
Electrical power = Pelec = 50 MW
Number of poles = P = 4
No. of cycles = 10
Required Information:
(a) stored energy = ?
(b) rotor acceleration = ?
(c) change in torque angle = ?
(c) rotor speed = ?
Answer:
(a) stored energy = 800 Mj
(b) rotor acceleration = 337.46 elec deg/s²
(c) change in torque angle (in elec deg) = 6.75 elec deg
(c) change in torque angle (in rmp/s) = 28.12 rpm/s
(c) rotor speed = 1505.62 rpm
Explanation:
(a) Find the stored energy in the rotor at synchronous speed.
The stored energy is given by
Where G represents complex rated power and H is the inertia constant of turbo-generator.
(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.
The rotor acceleration is given by
Where M is given by
So, the rotor acceleration is
(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.
The change in torque angle is given by
Where t is given by
So,
The change in torque in rpm/s is given by
The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by
Where P is the number of poles of the turbo-generator.
Answer:
Do you mean 4m^3 and 3.0 tones?
Explanation:
solution:
Mass = m = 3.0 tones
- 1 ton = 1,000 kg
= 3.0 × 1,000
= 3,000 kg
volume = v = 4m^3
Required:
Mass density of oil = p = ?
We know that;
The answer is:
750kg / m^3