Answer:
take gdrumnjjhtffnhfvhih
Explanation:
you have to do just opposite. is that clear?
Answer:
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Explanation:
What is this...
this is a multi-part question. once an answer is submitted, you will be unable to return to this part. an ideal cogeneration steam plant is to generate power and 8600 kj/s of process heat. steam enters the turbine from the boiler at 7 mpa and 500°c. one-fourth of the steam is extracted from the turbine at 600 kpa pressure for process heating. the remainder of the steam continues to expand and exhausts to the condenser at 10 kpa. the steam extracted for the process heater is condensed in the heater and mixed with the feedwater at 600 kpa. the mixture is pumped to the boiler pressure of 7 mpa.
This response is accompanied by the T-s or H-s graphic.
2. The steam extraction rate is 4.088 kg per second.
3. Steam leaving the boiler has a net power output of 1089.5 KJ/Kg.
4. The boiler's mass flow rate of steam is 16.352Kg/s.
5. The plant generates 11016.2KJ/s of net electricity.
6. 0.218 is the usage factor.
We must ascertain all of the system's thermodynamic coordinates in order to examine the issue. The optimal cogeneration steam plant technology can be seen in the second image that is attached to this response.
We may get the details for each point from a chart of the thermodynamic characteristics of water.
+ Steam from the boiler reaches the turbine at 7 MPa and 500 °C:
h₆=3410.56KJ/Kg
s₆=6.7993 KJ/Kg
As a result, this cogeneration steam system is suitable. s₆=s₇=s₈
At 600 kPa, one-fourth of the steam is removed from the turbine:
2773.74 KJ/Kg = h7(s7) (overheated steam)
+The remaining steam expands further and exhausts to the condenser at a pressure of 10 kPa.
(This is wet steam with title X=0.8198; h8(s8)=2153.58 KJ/Kg.)
s1=0.64925KJ/Kg and h1(P=10Kpa)=191.83 KJ/Kg (condensed water).
Pumping this flow to 600KPa results in:
s₂=s₁
h₂(s₂)=192.585KJ/Kg
+The heater condenses the steam that was taken for the process heater.
h₃(P=600KPa)=670.42KJ/Kg (condensed water) (condensed water).
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a tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm (0.4 in.). determine the magnitude of the load required to produce a change in diameter if the deformation is entirely elastic.
Answer:
let me know if this's helpful.
The load required to produce a change in diameter of 10 mm in a cylindrical brass rod is 1.57 N.
How to solveA tensile stress of 0.157 MPa is required to produce a change in diameter of 10 mm in a cylindrical brass rod. The deformation is entirely elastic, so the Poisson's ratio for brass is 0.34, and the modulus of elasticity is 97 GPa.
The load required to produce this tensile stress is 1.57 N.
To solve this problem, we can use the following equation:
Stress = (Change in diameter) * (π / (2 * Diameter))
Plugging in the values for the change in diameter, diameter, and Poisson's ratio, we get the following tensile stress:
Stress = (0.001) * (π / (2 * 0.010)) = 0.157 MPa
The load required to produce this tensile stress is then:
Load = Stress * Modulus of Elasticity
Plugging in the values for the tensile stress and modulus of elasticity, we get the following load:
Load = 0.157 MPa * 97 GPa = 1.57 N
Therefore, the load required to produce a change in diameter of 10 mm in a cylindrical brass rod is 1.57 N.
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What is the capacitance if an insulator with dielectric constant κ and thickness d/2 is slipped between the electrodes without changing the plate separation?.
let, A be the are of cross section of the electrodes.
The capacitance is C.
Let E represent the electric field between the electrodes.
Be the charge per square meter (sigma).
∉ be the permittivity of open space
and considering the fact that
Insulator's dielectric constant is K.
Between the electrodes is slid d/2.
between the electrodes is now an electric field.
E = sigma + sigma + sigma
E = sigma / ——1
we are aware,
E = V /(d/2)
V = E d/2
V=sigma /*d/2 ( from equation 1)
V = d/2 sigma ———2
once more, we are aware
C = Q/V
Sigma A/(sigma d/2) = C Using Q = sigma A and equation 2,
C = 2∉A/d
has a K-valued dielectric constant.
C = 2K∉A/d
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What is insurance an example of? Select one:
a. Risk avoidance
b. Risk assignment
c. Risk mitigation
d. Risk rejection
Answer:
Insurance is an example of C. Risk Mitigation.
Explanation:
Insurance is used to manage (or mitigate) risk. Companies and individuals purchase insurance to protect themselves from financial stress. For example, car insurance insures your car up to a certain percentage or value amount if something were to happen to your car while driving.
The source term will affect all algebraic equations.
The statement about the source term affecting all algebraic equations is; False
What is the source term in Algebraic Equations?
In the finite-element method of Analysis, we usually go from differential equations to a set of algebraic equations. Now, it is pertinent to note that each algebraic equation will tend to relate a nodal value with all other nodal values.
The assumed polynomial variation that will exist within each element is usually the basis for which we derive the algebraic equations. Thus, to derive the algebraic equations, we will have to assume a polynomial variation for the parameter values within each element.
The above can be done through interpolation of nodal parameter values in the post-processing stage. This is because the assumed polynomial variation within each element that is used for deriving the algebraic equations is also used for post-processing.
Finally, for each algebraic equation, the source term will tend to contribute to the coefficient that is taken to the right hand side of it and then entered into the corresponding row of the {f} vector.
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Complete Question is;
Select true or false.
The source term will affect all algebraic equations.
A.) True
B.) False