If water is boiling at 130 degrees Celsius inside a pressure cooker, the pressure is approximately 2.7 atmospheres
To determine the pressure inside a pressure cooker where water is boiling at 130 degrees Celsius, we use the properties of water and its boiling point at various pressures.
For water, the pressure at which the boiling point is elevated to 130 degrees Celsius can be found using steam tables.
At 130 degrees Celsius, the saturation pressure of water is approximately 2.7 atmospheres (atm). This is equivalent to :
= 2.7 x 101.325 kPa
= 273.5775 kPa, or about 2.7 bar (since 1 bar = 100 kPa)
A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the top, the pipe has a diameter of 30 mm and a pressure gauge indicates a pressure of 860 kPa. At the bottom the diameter is 85 mm and a pressure gauge reading is 1 MPa. Assume the losses are negligible and determine the flov rate. Does the flow direction matter?
Answer:
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
Explanation:
Given that
At top[tex]d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm[/tex]
[tex]\rho =900\dfrac{Kg}{m^3}[/tex]
We know that
[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2[/tex]
[tex]A_1V_1=A_2V_2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2[/tex]
[tex]V_2=8.02V_1[/tex]
[tex]Z_2=12 sin60^{\circ}[/tex]
[tex]\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}[/tex]
So [tex]V_1=1.30[/tex]m/s
We know that flow rate Q=AV
[tex]Q=A_1V_1[/tex]
By putting the values
[tex]A_1=\dfrac{\pi}{4}d^2[/tex]
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.
Heat in the amount of 100 KJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600K.Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.
Answer:
0.0833 k J/k
Explanation:
Given data in question
total amount of heat transfedded (Q) = 100 KJ
hot reservoir temperature R(h) = 1200 K
cold reservoir temperature R(c) = 600 k
Solution
we will apply here change of entropy (Δs) formula
Δs = [tex]\frac{Q}{R(h)}+\frac{Q}{R(c)}[/tex]
Δs = [tex]\frac{-100}{1200}+\frac{100}{600)}[/tex]
Δs = [tex]\frac{1}{12}[/tex]
Δs = 0.0833 K J/k
this change of entropy Δs is positive so we can say it is feasible and
increase of entropy principle is satisfied
Answer:
0.0837 kJ/K
Explanation:
Given:
Temperature of the cold reservoir T,cold = 600 K
Temperature of the hot reservoir T,hot= 1200 K
Heat transferred , Q=100 kJ
Now the entropy change for the cold reservoir
[tex]\bigtriangleup S,cold=-\frac{Q}{T,cold}[/tex]
[tex]\bigtriangleup S,cold=-\frac{-100}{600}[/tex]
[tex]\bigtriangleup S,cold=0.1667 kJ/K[/tex]
Now the entropy change for the cold reservoir
[tex]\bigtriangleup S,hot=-\frac{Q}{T,hot}[/tex]
[tex]\bigtriangleup S,hot=-\frac{100}{600}[/tex]
[tex]\bigtriangleup S,hot=-0.0833 kJ/K[/tex]
Therefore, the total entropy change for the two reservoir is
[tex]\bigtriangleup S=\bigtriangleup S,hot +\bigtriangleup S,cold[/tex]
thus,
ΔS=0.1667-0.0833
ΔS=0.0833 kJ/K
Since, the change of entropy is positive thus we can say it is possible and
increase of entropy principle is satisfied
What is pre-flush and post flush in petroleum engineering?
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Answer:
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Explanation:
What are the air-standard assumptions?
Answer:
The air-standard assumptions are:
The working fluid is air assumed to be perfect and it behave as an ideal gas .All process are internally reversible.The cycle is modeled as closed cycle with air cooled in the chiller heat exchanger and then re-circulated to the compressor. To avoid the complications, the combustion container are replaced by combustion heat exchanger .
________is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures
Answer:
Repairable component
Explanation:
Repairable component is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures.
________is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures
Repairable component
Three point bending is better than tensile for evaluating the strength of ceramics. a)-True b)- False
Answer:
a)-True
Explanation:
Three point bending is better than tensile for evaluating the strength of ceramics. It is got a positive benefit to tensile for evaluating the strength of ceramics.
In a nuclear reactor, more fuel is burned than is consumed. a)-True b)- False
Answer:
true
Explanation:
yes, it is true more fuel is burned than is consumed.
nuclear reactor generate electricity from nuclear fission and while nuclear fission nucleus splits into small parts to generate energy.
these energies can be used for my purposes mainly it is used for power generation.
we can also say that in nuclear reactor more fuel is consumed because of metal acting as a fissionable material.
Water flows in a pipe of diameter 0.5 m. The dianeter of the to 1,0 m. A U-tube manometer is of the enlargement with joining ercury levels 5 mm. Determine the flow rate as well as the pressure 3 Water pipe suddenly enlarges connected to either side pipes which contain water. The difference in m head loss as a result of the enlargement.
Answer:
Q = 0.943[tex]m^{3}/s[/tex]
[tex]h_{L}[/tex] = 0.6605 m
Explanation:
Given :
Diameter, d₁ = 0.5 m
Area, A₁ = [tex]\frac{\pi }{4}\times 0.5^{2}[/tex]
= 0.19625 [tex]m^{2}[/tex]
Enlargement diameter, d₂ = 1 m
Enlargement Area, A₂ = [tex]\frac{\pi }{4}\times 1^{2}[/tex]
= 0.785 [tex]m^{2}[/tex]
Manometric difference, h = 35 mm
=35 X [tex]10^{-3}[/tex] m
From manometer , we get
[tex]P_{1}+\rho _{w}.g.z_{1}+\rho _{m}.g.h=P_{2}+\rho _{w}.g.(z_{1}+h)[/tex]
[tex]P_{1}-P_{2}=(\rho _{w}-\rho _{m}).g.h[/tex]
[tex]\frac{P_{1}-P_{2}}{\rho _{w}.g}=(1-\frac{\rho _{m}}{\rho _{w}})\times h[/tex]
= [tex](1-13.6)\times 35\times 10^{-3}[/tex]
= -0.441
Now from newtons first law,
[tex]\frac{P_{1}-P_{2}}{\rho _{w}.g}=\frac{V_{2}^{2}-V_{1}V_{2}}{g}[/tex]
-0.441 = [tex]\frac{Q^{2}}{9.81}\times (\frac{1}{A_{2}^{2}}-\frac{1}{A_{1}A_{2}})[/tex]
-0.441 = [tex]\frac{Q^{2}}{9.81}\times (\frac{1}{0.785^{2}}-\frac{1}{(0.19625\times 0.785)^{2}})[/tex]
Therefore. Q = 0.943 [tex]m^{3} /s[/tex]
Now V₁ = [tex]\frac{Q}{A_{1}}[/tex]
= [tex]\frac{0.943}{0.19625}[/tex]
= 4.80 m/s
V₂ = [tex]\frac{Q}{A_{2}}[/tex]
= [tex]\frac{0.943}{0.785}[/tex]
= 1.20 m/s
Therefore, heat loss due to sudden enlargement is given by
[tex]h_{L}=\frac{(V_{1}-V_{2})^{2}}{2g}[/tex]
[tex]h_{L}=\frac{(4.80-1.20)^{2}}{2\times 9.81}[/tex]
= 0.6605 m
An unknown gas (k=1.4, c v= 0.716 kJ/kg.K, c_p=1 kJ/kg.K, R = 0287 kJ/kg K) is trapped in a 1 m^3 piston-cylinder device at 1000 KPa and 1000 K. It then undergoes an isothermal (constant-temp) process in which 696 kJ of boundary work is delivered (positive work transfer). Determine the heat transfer. (Use the PG model).
Answer:
So heat transfer is 696 kJ
Explanation:
Given:
K = 1.4
[tex]C_{v}[/tex] = 0.716 kJ/kg
[tex]C_{p}[/tex] = 1 kJ/kg
R = 0.287 kJ/kg
V = 1 [tex]m^{3}[/tex]
P = 1000 kPa
T = 1000 K
Work delivered, δW = 696 kJ
It is isothermal process, so the initial and final temperature are same, that is T₁ = T₂ and the internal energy is zero (dU =0)
Therefore from 1st law of thermodynamcis,
δQ = dU + δW
= 0 + 696
= 696 kJ
So heat transfer is 696 kJ
Convert 25 mm into in.
Answer:
25 mm = 0.984252 inches
Explanation:
Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:
1 mm = 1/25.4 inches
From the question, we have to convert 25 mm into inches
Thus,
25 mm = (1/25.4)*25 inches
So,
[tex]25 mm=\frac{25}{25.4} inches[/tex]
Thus, solving we get:
25 mm = 0.984252 inches
What are the x and y coordinates of the centroid of the area?
Answer:
[tex]\\X_{C.G}=\frac{\int xdA}{A}\\Y_{C.G}=\frac{\int ydA}{A}[/tex]
Explanation:
The x co-ordinate of the centroid is given by:
[tex]x_{C.G}=\frac{\int xdA}{A}[/tex]
The y co-ordinate of the centroid is given by:
[tex]Y_{C.G}=\frac{\int ydA}{A}[/tex]
where
[tex]x^{}[/tex] is the x co-ordinate of a diffrential area [tex]dA[/tex]
and [tex]y^{}[/tex] is the x co-ordinate of a diffrential area [tex]dA[/tex]
See attached figure
A 5Kw solar system may produce enough energy to power your home. a)-True b)- False
Answer: True
Explanation:
Yes, it is true that 5 Kw solar system may produce enough energy to power your home as, on an average good quality of 5 KW solar system can produced 22 units per day enough to power all home appliances. As, a 5 KW solar system produced energy is basically depends on the three main factor that are:
Quality of the solar panel system.Location from where the solar system generated its energy.And also on the positioning of the solar system.If a host system is 80% efficient, the minimum horsepower rating of the motor should be ? if the hoist is to provide 20,000 ft-lb/s. a) 45.45hp b) 23,530hp c) 36.36hp d) 42.78hp
Answer:
answer is option A i.e.45.45 hp
Explanation:
Given data:
load =20000 ft lb/s
efficiency = 80%
we know that
1 hp = 550 ft lb/s
minimum horsepower rating can be obtained by using following formula
minimum horse power rating = [tex]\frac{load}{efficiency * 1 horse power} \\[/tex]
= [tex]\frac{20000}{0.8*550} = 45.45 hp[/tex]