A liquid (r 1.65 g/cm3) fl ows through two horizontal

Ответ:

julie1624

21.01.2022

Mathematics

The question is incomplete: This is the complete Question

A liquid (ρ = 1.65 g/cm3) flows through two horizontal sections of

tubing joined end to end. In the first section the cross-sectional area is

10.0 cm2 , the flow speed is 275 cm/s, and the pressure is 1.20 x 105

Pa. In the second section the cross-sectional area is 2.50 cm2. Calculate

the smaller section’s (1) flow speed and (2) pressure.

(1) Smaller's section flow speed = 1100cm/s

(2) Smaller's section pressure = $2.64*10^{5}Pa$

Step-by-step explanation:

Question 1

Using the continuity equation, we have that:

$A_{1}v_{1}=A_{2}v_{2}$

$v_{2}=\frac{A_{1}v_{1}}{A_{2}} \\\\v_{2}=\frac{10cm^{2} *275cm/s}{2.5cm^{2}}\\\\v_{2}=4*275=1100cm/s$

Question 2

Next, we apply Bernoulli's Equation

$P_{2} + \frac{1}{2}rv_{2} ^{2} = P_{1} + \frac{1}{2}rv_{1} ^{2}$

Let r = ρ(the density)

$P_{2} = P_{1} + \frac{1}{2}rv_{1} ^{2} -\frac{1}{2}rv_{2} ^{2}\\\\P_{2} = P_{1} + \frac{1}{2}r(v_{1} ^{2} -v_{2} ^{2})$

The pressure is given to us in its fundamental unit, Pa. So, we need to convert the speeds and density to their fundamental units.

$v_{1} = 275cm/s=\frac{275}{100}m/s\\ v_{1}=2.75m/s\\\\v_{2} = 1100cm/s=\frac{1100}{100}m/s\\ v_{2}=11m/s\\\\r=1.65g/cm^{3} = 1.65*1000(kg/m^{3})\\ r=1650kg/m^{3}$

$P_{2} = 1.2*10^{5} + \frac{1}{2}(1650) (2.75 ^{2} -11 ^{2})\\\\P_{2} = 1.2*10^{5} + 825 ( 7.5625-121) = 1.2*10^{5} + 825 (-113.4375)\\\\P_{2} = 120000 - 93585.9375 = 26414.0625\\\\P_{2} = 2.6414*10^{5}$

$P_{2}$ ≅ $2.64*10^{5}Pa$

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question