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1 year ago

The source term will affect all algebraic equations.

Engineering

1 answer:

Drupady [299]1 year ago

8 0

The statement about the source term affecting all algebraic equations is; False

<h3>What is the source term in Algebraic Equations?</h3>

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.

Read more about Source Term at; brainly.com/question/28956378

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Complete Question is;

Select true or false.

The source term will affect all algebraic equations.

A.) True

B.) False

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xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

Which traits are common in all four career pathways of the Information Technology field? Check all that apply.

Komok [63]

Answer:

Accuracy and attention to detail, Problem solving and critical thinking skills, Knowledge of programming language .

Explanation:

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The IT industry's career paths can be categorized equally in the two primary field’s hardware and software areas

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In software, there is manufacturing, development, programming, software testing, and maintenance and support under software.

Computer operations, administration of databases, sales / marketing and data centre management are connected areas.

7 0

2 years ago

The combustion chamber has different shapes depending on the make and model of the engine. True or false

Alisiya [41]

Answer:

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3 years ago

A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen

arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

$z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2} }{4g}$ (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

$w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s$

if you consider the equation 1 for the free surface at the edge is equal to

$z_{s(R)} =h_{0} +\frac{w^{2}R^{2} }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft$

7 0

3 years ago

Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate

Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

$\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}$

Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

$t_{A_{1}$ = Inlet temperature of the gas = 1100 K

$t_{A_{2}$ = Outlet temperature of the gas = 733.33 K

$t_{B_{1}$ = Inlet temperature of the air = 300 K

$t_{B_{2}$ = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0

3 years ago