Find the volume of water displaced and position of center of buoyancy for a wooden block of width 2.5m and of depth 1.5m. When i
1 answer:
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Answer:
Explanation:
C.1 Input-Output Model
It is a formal model that divides the economy into 2 sectors and traces the flow of inter-industry purchases and sales. This model was developed by Wassily Leontief in 1951. In simpler terms, the inter-industry model is a quantitative economic model that defines how the output of one industry becomes the input of another industrial sector. It is an interdependent economic model where the output of one becomes the input of another. For Eg: The Agriculture sector produces output using the inputs from the manufacturing sector.
The 3 main elements are:
Concentrates on an economy which is in equilibrium
Deals with technical aspects of production
Based on empirical investigations and assumptions
Assumptions
2 sectors - " Inter industry sector" and "final sector"
Output of one industry is the input for another
No 2 goods are produced jointly. i.e each industry produces homogenous goods
Prices, factor suppliers and consumer demands are given
No external economies or diseconomies of production
Constant returns to scale
The combinations of inputs are employed in rigidly fixed proportions.
Structure of IO model
See image 1
Quadrant 1: Flow of products which are both produced and consumed in the process of production
Quadrant 2: Final demand for products of each producing industry.
Quadrant 3: Primary inputs to industries (raw materials)
Quadrant 4: Primary inputs to direct consumption (Eg: electricity)
The model can be used in the analysis of the labor market, forecast economic development of a nation and analyze economic developments of various regions.
Leontief inverse matrix shows the output rises in each sector due to a unit increase in final demand. Inverting the matrix is significant since it is a linear system of equations with unique solutions. Thus, the final demand vector for the required output can be found.
C.2 Linear programming problems
Linear programming problems are optimization problems in which objective function and the constraints are all linear. It is most useful in making the best use of scarce resources during complex decision makings.
Primal LP, Dual LP, and Interpretations
Primal linear programming: They can be viewed as a resource allocation model that seeks to maximize revenue under limited resources. Every linear program has associated with it a related linear program called dual program. The original problem in relation to its dual is termed as a primal problem. The objective function is a linear combination of n variables. There are m constraints that place an upper bound on a linear combination of the n variables The goal is to maximize the value of objective functions that are subject to the constraints. If the primal linear programming has finite optimal value, then the dual has finite optimal value, and the primal and dual have the same optimal value. If the optimal solution to the primal problem makes a constraint into a strict inequality, it implies that the corresponding dual variable must be 0. The revenue-maximizing problem is an example of a primal problem.
Dual Linear Programming: They represent the worth per unit of resource. The objective function is a linear combination of m values that are the limits in the m constraints from the primal problem. There are n dual constraints that place a lower bound on a linear combination of m dual variables. The optimal dual solution implies fair prices for associated resources. Stri=ong duality implies the Company’s maximum revenue from selling furniture = Entrepreneur’s minimum cost of purchasing resources, i.e company makes no profit. Cost minimizing problem is an example of dual problems
See image 2
n - economic activities
m - resources
cj - revenue per unit of activity j
Answer:
where and
be the inner radius, outer radius of the annalus.
Explanation:
Let ,
and
be the inner radius, outer radius and length of the given annulus.
Temperatures at the inner surface, and at the outer surface,
.
Let q be the rate of heat transfer at the steady-state.
Given that, the heat flux at r=3cm=0.03m is
.
This heat transfer is same for any radial position in the annalus.
Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.
Now, according to Fourier's law:
where,
K=Thermal conductivity of the material.
T= temperature at any radial distance r.
A=Area through which heat transfer is taking place.
Here,
Variation of temperature w.r.t the radius of the annalus is
Putting the values from the equations (ii) and (iii) in the equation (i), we have
[as
, and
]
This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.
Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ