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Anit [1.1K]
3 years ago
12

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

m ( 0.573 m) and the flow speed of the petroleum is 13.5 m/s. At the refinery, the petroleum flows at 5.83 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?
Physics
1 answer:
Mashcka [7]3 years ago
4 0

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

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Calculate the current through a 3.0 ω resistor with a voltage of 9.0 v across it.
gulaghasi [49]

Answer: 3 A

Explanation:

According to<u> Ohm's law</u>:  

V=R.I

Where:

V=9 V is the voltage

R=3\Omega is the resistance of the resistor

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Isolating I:

I=\frac{V}{R}

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A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
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Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

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