Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>
Answer:
The sentence excerpted from the e-mail uses passive voice.
Given the purpose of your message, this voice is appropriate.
Explanation:
Because the objective is to remedy the situation a passive voice is great because it emphasizes the action and the object instead of the subject.
We want to emphasize the document and the incorrect information, not our colleague.
Answer:
Tso = 28.15°C
Explanation:
given data
t2 = 21 mm
ki = 0.026 W/m K
t1 = 9 mm
kp = 180 W/m K
length of the roof is L = 13 m
net solar radiation into the roof = 107 W/m²
temperature of the inner surface Ts,i = -4°C
air temperature is T[infinity] = 29°C
convective heat transfer coefficient h = 47 W/m² K
solution
As when energy on the outer surface at roof of a refrigerated truck that is balance as
Q =
.....................1
Q =
.....................2
now we compare both equation 1 and 2 and put here value
solve it and we get
Tso = 28.153113
so Tso = 28.15°C
The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.
<h3>How to determine the amount of settlement?</h3>
For a layer of 3.8 m thickness, we were given the following parameters:
U = 50% = 0.5.
Sc = 7.3 cm.
For Sf, we have:
Sf = Sc/U
Sf = 7.3/0.5
Sf = 14.6
Therefore, Sf for a layer of 38 m thickness is given by:
Sf = 14.6 × 38/3.8
Sf = 146 cm.
At 50%, the time for a layer of 3.8 m thickness is:
= 1.5 year.
At 50%, the time for a layer of 38 m thickness is:
= 1.5 × (38/3.8)²
= 150 years.
For the thickness of 38 m, U₂ is given by:
![\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05](https://tex.z-dn.net/?f=%5Cfrac%7BU_1%5E2%7D%7BU_2%5E2%7D%20%3D%5Cfrac%7B%28T_v%29_1%7D%7B%28T_v%29_2%7D%20%3D%20%5Cfrac%7Bt_1%7D%7Bt_2%7D%20%5C%5C%5C%5CU_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B1.5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.01%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0025%7D%20%5C%5C%5C%5CU_2%3D0.05)
The new settlement after 1.5 year is:
Sc = U₂Sf
Sc = 0.05 × 146
Sc = 7.3 cm.
For time, t₂ = 5 year:
![U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09](https://tex.z-dn.net/?f=U_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.03%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0075%7D%20%5C%5C%5C%5CU_2%3D0.09)
The new settlement after 5 year is:
Sc = U₂Sf
Sc = 0.09 × 146
Sc = 13.14 cm.
Read more on clay layer here: brainly.com/question/22238205
Answer:
for a) F= 744.97 N
for a) F= 167.85 N
for a) F= 764.57 N
Explanation:
the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.
Then from steam tables
T= 120°C → P required=Pr= 198.67 kPa
then the pressure developed by the piston is
P = (m*g + F)/A
where m= mass of the piston ,g= gravity F= force required and A= area of the piston
then
Pr = P = (m*g + F)/A
F = Pr*A-m*g
since A= π/4*D²
F =π/4* Pr*D²-m*g
replacing values
F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²
F= 744.97 N
b) for T₂=80°C → Pr₂=47.41 kPa
F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²
F₂= 167.85 N
c) for m=0 (mass of the piston neglected) ,the force required is
F₃ =π/4*Pr*D² = π/4*198.67 *10³Pa*(0.07m)²= 764.57 N
F₃ =764.57 N