OA bloom is smaller than a bar
Answer: 1.98 × 10^4 N
Explanation:
Form similar triangle ADE and ABC
a/x= 2/3, a=2/3x
Width of the strip w= 2(4+a) = 8+2a
W= 8 +2 (2/3x)= 8+4/3x
Area of the strip = w Δx
(8 +4/3x) Δx
Pressure on the strip p= pgx= 10^3 ×9.81x= 9810x
But,
Force= Pressure × area= 9810x × (8+4/3x)Δx
Adding the forces and taking lim n to infinity
F total= lim n--> infinity E 9810x × (8+4/3x)Δx
Ftotal= Integral 2,0 9810x × (8+4/3x)Δx
F total= 9810 integral 2, 0 (8+4/3x)dx
= 9810(8+x^2/2 + 4/3x^3/3)2,0
=9810(4x^2 + 4/9x^3)
=9810(4x2^2 + 4/9×2^3-0)
=9810(16 + 32/9)
Hydrostatic force as an integral
Ft= 19.18 ×10^4N
Answer:
The heat transfer is 29.75 kJ
Explanation:
The process is a polytropic expansion process
General polytropic expansion process is given by PV^n = constant
Comparing PV^n = constant with PV^1.2 = constant
n = 1.2
(V2/V1)^n = P1/P2
(V2/0.02)^1.2 = 8/2
V2/0.02 = 4^(1/1.2)
V2 = 0.02 × 3.2 = 0.064 m^3
W = (P2V2 - P1V1)/1-n
P1 = 8 bar = 8×100 = 800 kPa
P2 = 2 bar = 2×100 = 200 kPa
V1 = 0.02 m^3
V2 = 0.064 m^3
1 - n = 1 - 1.2 = -0.2
W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ
∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ
Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ
Answer:
Cv = 0.026 cm²/min
t = 52.60 min
v% = 41.86 %
tv = 0.1375
t = 8.53 min
v = 53.61 %
Explanation:
given data
height = 2.54 cm
50 % consolidation = 12 min
solution
we get here first Cv value that is express as
Tv = .................1
here Tv for 50% is 0.196
put here value and we get
0.196 =
solve it we get
Cv = 0.026 cm²/min
and
for tv for 90 % consolidation is 0.848
put value in equation 1
0.848 =
solve it we get t
t = 52.60 min
and
v% will be here is
v% =
v% = 41.86 %
and
tv =
tv = 0.1375
so now put value in equation 1 we get
0.1375 =
solve it we get
t = 8.53 min
and
now put value of t 14 min in equation 1 will be
tv =
t = 0.225 min
and v will be after 14 min
0.0225 =
v = 53.61 %