Answer:
14.52 minutes
<u>OR</u>
14 minutes and 31 seconds
Explanation:
Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.
Specific heat at constant volume at 27°C = 0.718 kJ/kg*K
Initial temperature of room (in kelvin) = 283.15 K
Final temperature (required) of room = 293.15 K
Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg
Heat required at constant volume: 0.718 * (change in temp) * (mass of air)
Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ
Time taken for temperature rise: heat required / (rate of heat change)
Where rate of heat change = 10000 - 5000 = 5000 kJ/hr
Time taken = 1210.26 / 5000 = 0.24205 hours
Converted to minutes = 0.24205 * 60 = 14.52 minutes
Pretty sure it’s Cenozoic
Answer:
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Explanation:
Answer:
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0 it is a incompressible flow.
c) ap = (16/3) i + (32/3) j + (16/3) k
Explanation:
Given
V = xy² i − (1/3) y³ j + xy k
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0
then
∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z
⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.
c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z
⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)
⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k
⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k
At point (1, 2, 3)
⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k
⇒ ap = (16/3) i + (32/3) j + (16/3) k
Answer:
B) voltage at the sending end of the feeder = 2483.66 v
Explanation:
attached below is the the equivalent circuits and the remaining solution for option A
B) voltage = 2400 v
I = 
= 20.83 A
calculate voltage at sending end ( Vs )
Vs = 2400 + 20.83 ∠ -cos^-1 (0.8) ( 0.75*2 + 0.5 + j 2 + j2 )
hence Vs = 2483.66 ∠ 0.961
therefore voltage at the sending end = 2483.66 v
