ENGINEERING COLLEGE

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

Answers

Answer 1

Answer:

The solution is in the attachment

Answer 2

Answer:

Answer:

please find attached.

Explanation:

Related Questions

At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.
three string are attached to a small metal ring, two of the strings make and angle of 35° with the vertical and each is pulled with a force of 7Newton. What force must be applied to the third string to keep the rings stationary?
B. Suppose R1 is a fuse which burns out due to a sudden surge of current, thus, it essentially becomes an open switch. How do the currents change after this?
The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)and is placed in the clamp when the temperature is T1 = 15°C. If the two 304-stainless-steel (E =193GPa, a = 17x10^-6/°C) carriage bolts of the clamp each have a diameter of 10mm, and they holdthe cylinder snug with a negligible force against the rigid jaws, determine the temperature at whichthe average normal stress in either the magnesium or steel becomes 12 MPa.
One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.

In the hydrodynamic entrance region of a pipe with a steady flow of an incompressible liquid A. The average velocity increases with distance from the entrance.
B. The average velocity stays the same with distance from the entrance.
C. The maximum velocity increases with distance from the entrance.
D. The maximum velocity decreases with distance from the entrance.
E. B and C
F. B and D

Answers

Answer:

D. The maximum velocity decreases with distance from the entrance.

Explanation:

This is because over time, the pressure with with the incompressible liquid enters decreases with distance from the entrance

Answer:

C. The maximum velocity increases with distance from the entrance

Explanation:

As the fluid particles moves into the pipe, the layer of fluid in contact with the surface of the pipe come to a complete stop. This layer also causes the fluid

particles in the adjacent layers to gradually slow down as a result of friction between fluid molecules, leaving the fluid at the center of the pipe with the maximum velocity.

Since the fluid is incompressible, to make up for this velocity reduction, the velocity of the fluid at the mid-

section of the pipe has to increase to keep the mass flow rate through the

pipe constant. As a result, a velocity gradient develops along the pipe and the maximum velocity which is at the center of the pipe increases with distance from entrance.

With the aid of a labbled diagram describe the operation of a core type single phase transformer

Answers

Answer:

A simple single-phase transformer has each winding being wound cylindrically on a soft iron limb separately to provide a necessary magnetic circuit, which is commonly referred to as “transformer core”. It offers a path for the flow of the magnetic field to induce voltage between two windings.

A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the vertex. The speed of the liquid leaving the funnel is V= √ 2gy where y is the height of the liquid free surface above the hole. The funnel initially is filled to height y0 = 300 mm. Obtain an expression for the time, t, for the funnel to completely drain, and evaluate. Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm), and from 150 mm to completely empty (also a change in depth of 150 mm). Can you explain the discrepancy in these times?

Answers

Answer:

$3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

t = 12.03

t = 81.473

velocity of the fluid decreases with level of fluid. Hence, in later stages the time taken is greater than in earlier stages

Explanation:

Given:

- The half angle θ = 30°

- The diameter of the small hole d = 6.25 mm

- The flow rate out of the funnel Q = A*√ 2gy

- The volume of frustum of cone is given by:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Where,

D: Is the larger diameter of the frustum

d: Is the smaller diameter of the frustum

y: The height of the liquid free surface from small diameter d base.

Find:

- Obtain an expression for the time, t, for the funnel to completely drain, and evaluate.

- Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm) and from 150 mm to completely empty (also a change in depth of 150 mm)

- Can you explain the discrepancy in these times?

Solution:

- We will use rate of change analysis by considering the rate of change of volume, and then apply the chain rule.

- The Volume of the frustum is a function of d , D , y. V = f ( d , D , y ). In our case the diameter of base d remains constant. Then we are left with:

V = f ( D , y )

- Determine a relationship between y and d. We will use half angle θ to determine the relationship between D and y by applying trigonometric ratios:

tan ( θ ) = D / 2*y

D = 2*y*tan ( θ )

- We developed a relationship between D and y in terms of half angle which remains constant. So the Volume is now only a function of one variable y:

V = f ( y )

- The volume of frustum of the cone can be written as:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Substituting the relationship for D in terms of y we have:

$V = (\pi )/(12)*y*(4*y^2*tan(Q) ^2 + d^2 + 2*d*y*tan(Q)})$

- Now by rate of change of Volume analysis we have:

dV / dt = [dV / dy] * [dy / dt]

- Computing dV / dy, where V = f(y) only:

$V = (\pi )/(12)*(4*y^3*tan(Q) ^2 + y*d^2 + 2*d*y^2*tan(Q)})\n\n(dV)/(dy) = (\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)})\n\n$

- Where, dV/dt = Volume flow rate:

$(dV)/(dt) = - Q\n(dV)/(dt) = - A*V\n(dV)/(dt) = - (\pi*d^2 )/(4) *√(2*g*y)$

- Then from Chain rule we have:

[dy / dt] = [dV / dt] / [dV / dy]

$(dy)/(dt) = (- (\pi*d^2 )/(4) *√(2*g*y))/((\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n} \n(dy)/(dt) = (-d^2 *√(2*g*y))/((4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n}$

- Separate variables:

$\frac {(4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)} {√(2*g*y)} .dy = {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n$

- Integrate both sides:

$\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*((8)/(5)*y^2^.^5*tan(Q) ^2 + 2*d^2*y^0^.^5 + (8)/(3)*d*y^1^.^5*tan(Q)) = -d^2*t + C\n\n\frac { 1 } { √(2*y*g) }*((8)/(5)*y^3*tan(Q) ^2 + 2*d^2*y + (8)/(3)*d*y^2*tan(Q)) = -d^2*t + C\n\n0.1204*y^3 + 0.000017638*y + 0.00217*y^2 = -0.0000390625*t + C\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = -t + C$

- Evaluate @ t = 0 , y = 0.3 m

$3175.424*(0.15)^3 + 0.45153*(0.15) + 55.552*(0.15)^2 = 0 + C\n\nC = 12.0347055\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

- Time taken from y = 300 to 150 mm:

$3175.424*0.15^3 + 0.45153*0.15 + 55.552*0.15^2 = -t + 90.871587\n\nt = 0 - -12.0347055\n\nt = 12.0347055 s\n$

- Time taken from y = 150 to 0 mm:

t = t_(0-300) - ( t_(0-150) = 90.871587 - 12.0347055 = 81.4721 s

- The discrepancy is time can be explained by the velocity of fluid coming out of bottom is function of y. The velocity of the fluid decreases as the level of fluid decreases hence the time taken from y =150 mm to 0 mm is larger than y = 300 mm to y = 150 mm.

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 ksi (1400 MPa).True/False

Answers

Answer:

True

Explanation:

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 kpsi (1400 MPa).

Also, It is a simplistic rule of thumb that, for steels having a UTS less than 160 kpsi, the endurance limit for the material will be approximately 45 to 50% of the UTS.

A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa

Answers

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto (1)/(V)$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $(v)/(3)L$

$150* v=P_2* (v)/(3)$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.