Answer:
For water
Flow rate= 0.79128*10^-3 Ns
For Air
Flow rate =1.2717*10^-3 Ns
Explanation:
For the flow rate of water in pipe.
Area of the pipe= πd²/4
Diameter = 30/1000
Diameter= 0.03 m
Area= 3.14*(0.03)²/4
Area= 7.065*10^-4
Flow rate = 7.065*10^-4*1.12E-3
Flow rate= 0.79128*10^-3 Ns
For the flow rate of air in pipe.
Flow rate = 7.065*10^-4*1.8E-5
Flow rate =1.2717*10^-3 Ns
Answer:
Explanation:
Hello!
To solve this problem you must follow the following steps, which are fully registered in the attached image.
1. Draw the complete outline of the problem.
2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties.
3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.
3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.
4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values and finds the work done.
5. draw the pV diagram using the 300F isothermal line
Answer:
Gs = 2.647
e = 0.7986
Explanation:
We know that moist unit weight of soil is given as

where,
= moist unit weight of the soil
Gs = specific gravity of the soil
S = degree of saturation
e = void ratio
= unit weight of water = 9.81 kN/m3
From data given we know that:
At 50% saturation,
puttng all value to get Gs value;

Gs - 1.194*e = 1.694 .........(1)
for saturaion 75%, unit weight = 17.71 KN/m3

Gs - 1.055*e = 1.805 .........(2)
solving both equations (1) and (2), we obtained;
Gs = 2.647
e = 0.7986