Multiple Select

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

4 0

3 years ago

According to the eNotes, a program that eliminates sales and promotions in an effort to minimize the bullwhip effect would be ca

Ilia_Sergeevich [38]

Options: True or false

Answer:True, it will be called CRMWRONGEDLP.

Explanation:eNotes was founded in 1998 by Brad Satoris and Alexander Bloomingdale, to enable and enhance the ability of students to solve assignments as given and other home works,eNote is also used by students to prepare effectively and efficiently for examinations.

Since its launch eNote has been a great source of educational materials for students who have gained a lot in the use of its materials.

bullwhip effect is a concept in supply chain, where forcasts causes the supply chain to be inefficient, it usually starts from retailers who raise concern of high demands.

CRMWRONGEDLP, is the program that minimizes the bullwhip effects according to eNote.

7 0

3 years ago

Assume you are an observer standing at a point

Arlecino [84]

Answer:

3.1m

Explanation:

6 0

3 years ago

Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe

Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

$\frac{p1}{p*}$ = 0.3636

$\frac{T1}{T*}$ = 0.5289

$\frac{T01}{T0*}$ = 0.7934

Isentropic Flow Chart: M1 = 2.0 , $\frac{T01}{T1}$ = 1.8

T1 = $\frac{1}{0.7934}$ (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= $\frac{T02}{T1}T1$ = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = $\frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K$ = 135.7* $10^{3}$ J/Kg.

5 0

3 years ago

Read 2 more answers

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

4 0

3 years ago