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Andreas93 [3]
3 years ago
8

Water flows through a horizontal nozzle in steady flow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2

= 0.25m. Water exits the nozzle to the atmosphere. Calculate the pressure at the entrance to the nozzle and the force required to hold the nozzle in place. Neglect the weight of the nozzle. rho = 1000kg/m3, g = 10m/s2.
Physics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

        Q = A v

        v = Q / A

The area of ​​a circle is

       A = π r² = π d² / 4

Let's look at the speeds at each point

       v₁ = Q / A₁ = Q 4 /π d₁²

       v₁ = 10 4 /π 0.5²

       v₁ = 50.93 m / s

       v₂ = Q / A₂

       v₂ = 10 4 /π 0.25²

       v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

       P₁ = P2 + ½ rho (v₂² - v₁²)

      P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

      P₁ = 1.013 10⁵ + 2.205 10⁷

      P₁ = 2.215 10⁷ Pa

la definicion de presion es

      P₁ = F₁/A₁

     F₁ = P₁ A₁

     F₁ = 2.215 10⁷ pi d₁²/4

     F₁ = 2.215 10⁷ pi 0.5²/4

     F₁ = 4.3 106 N

     

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a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

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Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

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W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

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