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3 years ago

What is ONE DIFFERENCE between civil structural engineering

Engineering

1 answer:

Rudiy273 years ago

3 0

Answer:

One of the differences is that civil engineering focuses on design elements while structural engineering is more concern on inspecting the materials used for construction. The structural engineers are the one who are supposed to ensure that the materials used for construction can support the design of the structure.

Explanation:

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a piping system has an internal air pressure of 1,500 kpa. In addition to being subject to the air pressure, the piping supports

Alik [6]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

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4 years ago

9.19 Generate Bode magnitude and phase plots (straight-line approximations) for the following voltage transfer functions. (a) H(

Norma-Jean [14]

Answer:

attached below

Explanation:

8 0

3 years ago

1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem

BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁ = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

$\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

$\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}$

$P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa$

$\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

$\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right$

$5^{0.33} = \left \dfrac{T_1}{T_2} \right$

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3× $c_v$ ×(T₂ - T₁)

$c_v$ = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86 kJ/(kg·K)

We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3× $c_v$ ×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = $U_g = h_g -pV_g$

$h_g$ = 2777.12 kJ/kg

$V_g$ = 0.194349 m³/kg

p = 1000 kPa

$U_{g1}$ = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, $U_{g1}$ + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0

3 years ago

Consider the control volume form of the basic laws. For the conservation of mass form for a control volume with mass flow into a

aleksandrvk [35]

Answer:

C) Dependent on the mass flows in and out.

Explanation:

Lets take control volume(CV)

Take

$m_i$ =inlet mass flow rate

$m_e$ =exit mass flow rate

If we take unsteady flow process then inlet mass can not be equal to exit mass.Some mass can store if inlet mass flow rate is high and exit mass flow rate is low.

So mass of control volume

$m_{cv}=m_i-m_e$ .

so above we can say that mass of control volume dependent on inlet and exit mass.

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3 years ago

You don't have to notify employees that a lockout/tagout is about to begin?

Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

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3 years ago