Answer:
The pressure difference across hatch of the submarine is 3217.68 kpa.
Explanation:
Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.
Given:
Height of the hatch is 320 m
Surface gravity of the sea water is 1.025.
Density of water 1000 kg/m³.
Calculation:
Step1
Density of sea water is calculated as follows:
![S.G=\frac{\rho_{sw}}{\rho_{w}}](https://tex.z-dn.net/?f=S.G%3D%5Cfrac%7B%5Crho_%7Bsw%7D%7D%7B%5Crho_%7Bw%7D%7D)
Here, density of sea water is
, surface gravity is S.G and density of water is
.
Substitute all the values in the above equation as follows:
![S.G=\frac{\rho_{sw}}{\rho_{w}}](https://tex.z-dn.net/?f=S.G%3D%5Cfrac%7B%5Crho_%7Bsw%7D%7D%7B%5Crho_%7Bw%7D%7D)
![1.025=\frac{\rho_{sw}}{1000}](https://tex.z-dn.net/?f=1.025%3D%5Cfrac%7B%5Crho_%7Bsw%7D%7D%7B1000%7D)
kg/m³.
Step2
Difference in pressure is calculated as follows:
![\bigtriangleup p=rho_{sw}gh](https://tex.z-dn.net/?f=%5Cbigtriangleup%20p%3Drho_%7Bsw%7Dgh)
![\bigtriangleup p=1025\times9.81\times320](https://tex.z-dn.net/?f=%5Cbigtriangleup%20p%3D1025%5Ctimes9.81%5Ctimes320)
pa.
Or
![\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})](https://tex.z-dn.net/?f=%5Cbigtriangleup%20p%3D%283217680pa%29%28%5Cfrac%7B1kpa%7D%7B100pa%7D%29)
kpa.
Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.
Answer:
![Q_{cv}=-339.347kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%3D-339.347kJ)
Explanation:
First we calculate the mass of the aire inside the rigid tank in the initial and end moments.
(i could be 1 for initial and 2 for the end)
State1
![1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295](https://tex.z-dn.net/?f=1bar%2A%7C%5Cfrac%7B100kPa%7D%7B1%7D%7C%2A2%3Dm_1%2A0.287%2A295)
![m_1=232kg](https://tex.z-dn.net/?f=m_1%3D232kg)
State2
![8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350](https://tex.z-dn.net/?f=8bar%2A%7C%5Cfrac%7B100kPa%7D%7B1bar%7D%7C%2A2%3Dm_2%2A0.287%2A350)
![m_2=11.946](https://tex.z-dn.net/?f=m_2%3D11.946)
So, the total mass of the aire entered is
![m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg](https://tex.z-dn.net/?f=m_v%3Dm_2-m_1%5C%5Cm_v%3D11.946-2.362%5C%5Cm_v%3D9.584kg)
At this point we need to obtain the properties through the tables, so
For Specific Internal energy,
![u_1=210.49kJ/kg](https://tex.z-dn.net/?f=u_1%3D210.49kJ%2Fkg)
For Specific enthalpy
![h_1=295.17kJ/kg](https://tex.z-dn.net/?f=h_1%3D295.17kJ%2Fkg)
For the second state the Specific internal Energy (6bar, 350K)
![u_2=250.02kJ/kg](https://tex.z-dn.net/?f=u_2%3D250.02kJ%2Fkg)
At the end we make a Energy balance, so
![U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e](https://tex.z-dn.net/?f=U_%7Bcv%7D%28t%29-U_%7Bcv%7D%28t%29%3DQ_%7Bcv%7D-W%7Bcv%7D%2B%5Csum_i%20m_ih_i%20-%20%5Csum_e%20m_eh_e)
No work done there is here, so clearing the equation for Q
![Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20m_2u_2-m_1u_1-h_1%28m_v%29)
![Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20%2811.946%2A250.02%29-%282.362%2A210.49%29-%28295.17%2A9.584%29)
![Q_{cv}=-339.347kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%3D-339.347kJ)
The sign indicates that the tank transferred heat<em> to</em> the surroundings.
Answer:
4m/s
Explanation:
We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted
Mathematically
\
We also know that Power = force x velocity ..................(i)
The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load
=> power = mg x Velocity........(ii)
While hoisting the load at at constant speed only the potential energy of the mass increases
Thus Potential energy = Mass x g x H...................(iii)
where
g = accleration due to gravity (9.81m/s2)
H = Height to which the load is hoisted
Equating equations (ii) and (iii) we get
m x g x v = ![\frac{mgh}{t}](https://tex.z-dn.net/?f=%5Cfrac%7Bmgh%7D%7Bt%7D)
thus we get v = H/t
Applying values we get
v = 6/1.5 = 4m/s
Answer:
Indicators for ineffective system engineering are as follows
1.Requirement trends
2.System definition change backlog trends
3.interface trends
4.Requirement validation trends
5.Requirement verification trends
6.Work product approval trends
7.Review action closure trends
8.Risk exposure trends
9.Risk handling trends
10.Technology maturity trends
11.Technical measurement trends
12.System engineering skills trends
13.Process compliance trends