What is the force in kN of work done is 1.2 ms moves through 120m

Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo

deff fn [24]

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

8 0

3 years ago

Read 2 more answers

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago

Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz

butalik [34]

Answer:

Exit velocity $V_2=1472.2$ m/s.

Explanation:

Given:

At inlet:

$P_1=100 bar,T_=600°C,V_1=35m/s$

Properties of steam at 100 bar and 600°C

$h_1=3624.7\frac{KJ}{Kg}$

At exit:Lets take exit velocity $V_2$

We know that if we know only one property inside the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

$h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}$

$h_2=2541.62\frac{KJ}{Kg}$

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

In the case of adiabatic nozzle Q=0,W=0

$3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0$

$V_2=1472.2$ m/s

So Exit velocity $V_2=1472.2$ m/s.

4 0

3 years ago

Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water

olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate = 2 L/s

water temperature 30 degree celcius

$Q = A\times V$

$2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity$

$V = \frac{20\times 4}{9\times \pi} = 2.83 m/s$

$Re = \frac{\rho\times V\times D}{\mu}$

at 30 degree celcius $= \mu = 0.798\times 10^{-3}Pa-s , K = 0.6154$

$Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}$

Re = 106390

So ,this is turbulent flow

$Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}$

$Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419$

$\frac{h\times 0.03}{0.615} = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}$

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0

3 years ago

Imagine yourself as an Engineer. How would you design your own Wind Turbine to generate electricity (at least 40 watts) at your

topjm [15]

Answer: You need metal and other stuff

Explanation:

6 0

2 years ago

What is the force in kN of work done is 1.2 ms moves through 120m​

What is the force in kN of work done is 1.2 ms moves through 120m