If a 110-volt appliance requires 20 amps, what is the total power consumed?

H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

2 years ago

A long, horizontal, pressurized hot water pipe of 15cm diameter passes through a room where the air temperature is 24degree C. T

solmaris [256]

Answer:

Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m

Explanation:

Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.

We know that heat transfer by convection is given by

$\dot{Q}=hA(T-T_{\infty })$

where,

h = heat transfer coefficient = 10.45 $W/m^{2}K$ (free convection in air)

A = Surface Area of the pipe

Applying the given values in the above formula we get

$\dot{Q}=10.45\times \pi DL\times (130+273-(24+273))\\\\\frac{\dot{Q}}{L}=10.45\times 0.15\times \pi \times (130-24)\\\\\frac{\dot{Q}}{L}=521.99W/m$

5 0

3 years ago

You have removed a very large, thick plate of steel (AISI 1010) from your heat treat oven and have placed it on a large insulate

Andreas93 [3]

Answer:

for got sorry

Explanation:

6 0

3 years ago

What is the specific volume of oxygen at 40 psia and 80°F?

lisabon 2012 [21]

This is the answer
Ur welcome

6 0

3 years ago

A screw extruder is 50 mm in diameter, 1 m long, has a 50mm lead, a channel 5 mm deep and a flight 3 mm wide. The circular die t

Margarita [4]

Answer:

A) 105.7 rpm

B) 11.32 kw

C) 20.85 NPa

Explanation:

Number of solid rods to be manufactured = 3600

a) Determine the RPM of the screw

we will apply the relation below

discharge rate ( Qd ) = 0.5 π^2 * D^2 * N di * sinA * cos A ------- ( 1 )

where : D = 50 mm , di = 5 mm , N = ?

Tan A = p / πD = 50 / π*50 ∴ A = 17.65°

Insert values into equation ( 1 )

Qd = 17.83 * 10^-6 * N

required discharge rate ( Q ) = $\frac{\frac{\pi D^2}{4}*L*N }{Time}$ ------ ( 2 )

where : D = 0.01 , L = 25 * 10^-2 , N = 3600 , time = 10 * 3600

input value into 2

Q = 31.415 * 10^-6 m^3/s

Hence the RPM of the screw ( N )

= Q / Qd = 31.415 * 10^-6 / 17.83 * 10^-6 = 1.76 rev/s = 105.7 rpm

b) Determine the power requirements of the extruder

max power requirement = Pm * A * πDN / 60

= ( 20.85 * π * ( 50 )^2 / 4 ) * π * 150 *1.76

max power requirement = 11.32 kw

c) What is the pressure buildup within the extruder

Pressure buildup within the extruder = ( 6π*D*N*L* η * cot A ) / di^2

= ( 6π * 0.05 * 1.76 * 1 * 100 * cot17.65 ) / ( 5 * 10^-3 )^2

therefore ; Pm = 20.85 NPa

3 0

2 years ago