Answer:
The air pressure in the tank is 53.9 
Solution:
As per the question:
Discharge rate, Q = 20 litres/ sec = 
(Since, 1 litre =
)
Diameter of the bore, d = 6 cm = 0.06 m
Head loss due to friction, 
Height, 
Now,
The velocity in the bore is given by:


Now, using Bernoulli's eqn:
(1)
The velocity head is given by:

Now, by using energy conservation on the surface of water on the roof and that in the tank :




The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%

W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000

<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
P_p = 27000 psf
Explanation:
given,
height of the retaining wall = h = 12 ft
internal angle of friction (∅)= 30°
unit weight = 125 pcf
Rankine passive earth pressure = ?
k_p is the coefficient of passive earth pressure


k_p = 3
Passive earth pressure


P_p = 27000 psf
Rankine passive earth pressure on the wall is equal to P_p = 27000 psf
Answer:
Technician A and B both are correct.
Explanation:
According to Technician A, we can stop noise by applying lubricant (Break friction).
According to Technician A, we can stop noise by set-up right position for break pad.
Answer:
Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s
Explanation:
du/u² = dt = ∫du/2.3183 = ∫dt
0.4313 u = t + c
(a) t = 0, u= 15m/s, c = 0.647
u = t+c/0.4313 = t + 0.647/0.4313
(a) when t= 1 u = 1+ 0.647/0.4313 = 3.8187m/s
(b) when t= 10 u = 10 + 0.647/0.4313 = 24.0858m/s
(c)when t= 1000 u = 1000 + 0.647/0.4313 = 3220.071m/s