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
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 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 .
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
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.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:
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.