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labwork [276]
3 years ago
13

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

d 5.0 cm, find the volume flow rate and the average velocity in each pipe section.
Engineering
1 answer:
ser-zykov [4K]3 years ago
7 0

Answer:

volumetric flow rate = 0.0251 m^3/s

Velocity in pipe section 1 = 6.513m/s

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = 0.0251 m^3/s

volumetric flow rate = 0.0251 m^3/s

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s

<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s

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A piston-cylinder device contains 0.58 kg of steam at 300°C and 0.5 MPa. Steam is cooled at constant pressure until one-half of
Mumz [18]

Answer:

a) Tբ = 151.8°C

b) ΔV = - 0.194 m³

c) The T-V diagram is sketched in the image attached.

Explanation:

Using steam tables,

At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.

Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C

b) The volume change

Using data from A-5 and A-6 of the steam tables,

The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume

(αբ) (which is calculated from the final quality and the consituents of the specific volumes).

ΔV = m(αբ - αᵢ)

αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)

q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg

αբ = 0.00109 + 0.5(0.3748 - 0.00109)

αբ = 0.187945 m³/kg

αᵢ = 0.5226 m³/kg

ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³

c) The T-V diagram is sketched in the image attached

3 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are he
aleksandr82 [10.1K]

Answer:

do the wam wam

Explanation:

6 0
3 years ago
Find the speed of a transverse wave on a 75-cm length of a cord when the tension in the cord is 320 N. The mass of the cord is 1
tino4ka555 [31]

Answer:

The speed of transverse wave will be 28.2842 m/sec  

Explanation:

We have given length of the card = 75 cm = 0.75 m

Tension on the card = 320 N

Mass of the card = 120 gram = 0.12 kg

So linear density =\frac{mass}{length}=\frac{0.12}{0.75}=0.4kg/m

We have to find the speed of the transverse wave

Speed is given by v=\sqrt{\frac{T}{linear\ density}}

v=\sqrt{\frac{320}{0.4}}=28.2842m/sec

So the speed of transverse wave will be 28.2842 m/sec

8 0
3 years ago
A corn ethanol production plant receives 500,000.0 kg/day corn feedstock at a moisture content of 15.5% (wet basis). If all of t
Alenkasestr [34]

Answer:

207 m³/day

Explanation:

Dry corn feed stock = 500000 × ( 100 - 15.5%) = 500000 × 84.5% = 500000× 0.845 = 422500

Starch yield = 68.5% × 422500 = 289412.5

Glucose yield = 1.11 × 289412.5 = 321247.875 where 1.11 is the scarification factor of starch to glucose

Ethanol yield = 0.51 × 321247.875 = 163836.416 where 0.51 is theoretical yield of ethanol from one mole of glucose

density = mass / volume

volume = mass / density = 163836.416 / 789 = 207 m³ / day

4 0
3 years ago
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