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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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When you see a street with white markings only, what kind of street is it?
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4 years ago
A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei
miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

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4 0
4 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
3 years ago
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