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Orlov [11]
3 years ago
7

Flow at the inlet and outlets can be assumed One-dimensional for the sake of simplicity. (a) Explain how this assumption simplif

ies the Reynolds Transport Theorem equation (provide complete, detailed answer) (b) This assumption introduces errors into the calculations, as it assumes flow properties do not vary across the inlet/outlet area. How this error changes with the area (does it increase or decrease)?
Engineering
1 answer:
Anna [14]3 years ago
5 0

Answer:

a) reduces the need to determine complicated flow properties.

b) The error decreases with the change in the area.

Explanation:

a) The mechanical properties of a fluid can sometimes be complicated. This results in an a need to calculate all the forces acting on the surface of the fluid particle. This then introduces the concept of the resolving the various forces. To reduce the error, we use the control volume. The  Reynolds Transport Theorem states that the rate of change of an extensive property N, for a system is equal to the time rate of change of N in the control volume and the net rate of flux of the property N through the control surface.

b) By making the assumptions, it decreases the likelihood of errors occurring in the equation.

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State the two Carnot corollaries (principles)
Salsk061 [2.6K]

Answer  and Explanation:

The two principles or corollaries of Carnot Theorem are listed below:

1). The efficiencies of all the reversible heat engines between any two thermal reservoirs working between the same temperatures will be equal to each other.

2). For every Carnot engine working between any two thermal reservoirs will have the same efficiency independent of the operating conditions and the nature of working substance. It only depends on the temperature of the thermal energy reservoirs.

8 0
4 years ago
Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel f
Morgarella [4.7K]

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

T_{ \infty} = 27^0C= 273+27 = 300K

T_s= 127^0C = (127+273)= 400K

P = 1atm

Film temperature

T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K

dynamic viscosity =

\mu =20.9096\times 10^{-6} m^2/sec

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}

=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043

we have,

Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k

we have,

heat transfer rate from top plate

\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w

7 0
3 years ago
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr
maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0
3 years ago
a torque of 24 ft pounds is the result when a force of ______ pounds is applied to a wrench that is 18 inches long.​
andrew-mc [135]

Answer:

  16

Explanation:

The product of feet and pounds is the torque. 18 inches is 1.5 feet.

  (1.5 ft)(x lbs) = 24 ft·lbs

  x lbs = (24 ft·lbs)/(1.5 ft) = 16 lbs

The force will be 16 pounds applied to an 18-inch wrench to obtain 24 ft·lbs.

5 0
3 years ago
Read 2 more answers
4-6. A vertical cylindrical storage vessel is 10 m high and 2 m in diameter. The vessel contains liquid cyclohexane currently at
Alex73 [517]

Answer:

See attachment for step by step procedure into getting answer.

Explanation:

Given that;

Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

4-6. A vertical cylindrical storage vessel is 10 m high and 2 m in diameter. The vessel contains liquid cyclohexane currently at a liquid level of 8 m. The vessel is vented to the atmosphere. A 1-cm-diameter hole develops in the bottom of the vessel. Calculate and plot the discharge rate of cyclohexane as a function of time until the vessel is completely empty. Show that the time to empty is the same as the time calculated using Equation 4-21.

See attachment for completed steps.

5 0
4 years ago
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