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3 years ago

A civil engineer is likely to fit in which of the Holland occupational codes?

Engineering

2 answers:

SVETLANKA909090 [29]3 years ago

8 0

Answer:

OA. Realistic

Explanation:

RIC 4 17-2051.00 civil engineers

natulia [17]3 years ago

5 0

Answer:

Investigative

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Mention the types of water demend

amm1812

Answer:

domestic, public, commercial, and industrial uses.

8 0

2 years ago

Consider two different flows over geometrically similar airfoil shapes, one airfoil being twice the size of the other. The flow

yulyashka [42]

Answer

No;

The two flows are not dynamically similar

Explanation: Given

T∞,1 = 800k

V∞,1 = 200m/s

p∞,1 = 1.739kg/m³

T∞,2 = 200k

V∞,2 = 100m/s

p∞,2 = 1.23kg/m³

Size1 = 2 * Size2 (L1 = 2L2) Assumptions Made

α ∝√T

μ∝√T Two (2) conditions must be met if the two flows are to be considered similar.

Condition 1: Similar Parameters must be the same for both flows

Condition 2: The bodies and boundaries must be genetically true. Condition 2 is true

Checking for the first condition...

Well need to calculate Reynold's Number for both flows

And Check if they have the same Reynold's Number Using the following formula

Re = pVl/μ

Re1 = p1V1l1/μ1 Re2 = p2V2l2/μ2 Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2

Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2

Re1/Re2 = p1V1l1μ2/p2V2l2μ1

Re1/Re2 = p1V1l1√T2 / p1V1l1√T1

Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)

Re1/Re2 = 9837.2/3479

Re1/Re2 = 2,828/1

Re1:Re2 = 2.828:1

Re1 ≠ Re2,

So condition 1 is not satisfied Since one of tbe conditions is not true, the two flows are not dynamically similar

5 0

2 years ago

The TSS concentration in the primary treated effluent is 150 mg/L, and TSS removal is 63 percent. The sludge has an average soli

Umnica [9.8K]

Answer:

The capacity of the sludge pump is 0.217 m3/min

Explanation:

Solution is attached below

7 0

3 years ago

A vertical plate is partially submerged in water and has the indicated shape. The height is 3m, the width at the top is 4m, and

natka813 [3]

Answer: 1.98 × 10^4 N

Explanation:

Form similar triangle ADE and ABC

a/x= 2/3, a=2/3x

Width of the strip w= 2(4+a) = 8+2a

W= 8 +2 (2/3x)= 8+4/3x

Area of the strip = w Δx

(8 +4/3x) Δx

Pressure on the strip p= pgx= 10^3 ×9.81x= 9810x

But,

Force= Pressure × area= 9810x × (8+4/3x)Δx

Adding the forces and taking lim n to infinity

F total= lim n--> infinity E 9810x × (8+4/3x)Δx

Ftotal= Integral 2,0 9810x × (8+4/3x)Δx

F total= 9810 integral 2, 0 (8+4/3x)dx

= 9810(8+x^2/2 + 4/3x^3/3)2,0

=9810(4x^2 + 4/9x^3)

=9810(4x2^2 + 4/9×2^3-0)

=9810(16 + 32/9)

Hydrostatic force as an integral

Ft= 19.18 ×10^4N

7 0

3 years ago

Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −

sergij07 [2.7K]

Answer:

Given data:

Equation of the state $p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }$

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid $m^{3} / k g$

R = gas constant , $j/k g k$

a, b = Constants

Solution:

Specific heat difference, $\begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}$

According to cyclic reaction

$\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}$

Hence specific heat difference is

$c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}$

Equation of state, $p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}$

Differentiating the equation of state with respect to temperature at constant volume,

$\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}$

$\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}$

Differentiating the equation of the state with respect to volume at constant temperature.

$\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)$

Substituting both eq (3) and eq (4) in eq (2)

We get,

${cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

Specific heat difference equation,

$c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

7 0

2 years ago