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2 years ago

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A jet of water 75m in diameter,issues with a velocity of 30m/s and impinge on a stationary plate which distort its forward motio

n find the forces exerted by the jet on the plate

1 answer:

pashok25 [27]2 years ago

5 0

Forces exerted by the jet on the plate is=3976N

<h3>How to calculate forces exerted by the jet on the plate?</h3>

A force is an effect that can change the motion of an object. A force can generate an object with mass to change its velocity, i.e., to accelerate. Force can also be explained intuitively as a push or a pull. A force has both volume and direction, making it a vector amount.

A jet of water 75m in diameter

velocity = 30m/s

The forces exerted by the jet on the plate is

F=1000×44178×10^-3×30²

=3976N

the jet on the plate work done by Is zero .

To learn more about Force, refer

brainly.com/question/12970081

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Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

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Answer:

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Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

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$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

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Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

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b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

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Viefleur [7K]

Answer:

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Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

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Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05

convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$ ------------- equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$ ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

Replacing the given data; we have:

$\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}$ $= \pi (0.1)*2000(276-273)$

$\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96$

$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

$\frac {0.05}{r} =e^{0.0970}$

$\frac {0.05}{r} =1.102$

$r=\frac{0.05}{1.102}$

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

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t = (0.05 - 0.0454)

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3 years ago

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timurjin [86]

Answer:

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Explanation:

Given:

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- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

A = pi*D*L

A = pi*(0.0005)*(0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

P = I*dV = 0.1*5 = 0.5 W

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h = P /A*(T_s - T_inf)

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V = 6.25*10^-5 * h^2

V = 6.25*10^-5 * (318.27)^2

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3 years ago