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Nastasia [14]
3 years ago
15

thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat

e saturated steam. what is the temperature of the 75% quality steam.?
Engineering
1 answer:
eduard3 years ago
7 0

Answer:

q_{in} = 528.6\,\frac{kJ}{kg}

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

q_{in} = h_{g} - h_{mix}

The specific enthalpies are:

Liquid-Vapor Mixture:

h_{mix} = 2217.2\,\frac{kJ}{kg}

Saturated Vapor:

h_{g} = 2745.8\,\frac{kJ}{kg}

The thermal energy per unit mass required to heat the steam is:

q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

q_{in} = 528.6\,\frac{kJ}{kg}

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Ammonia enters the expansion valve of a refrigeration system a temperature of 24oC and pressure of10 bar and exits at 1 bar. If
miskamm [114]

Answer:

X=0.194

T=-33.6C

Explanation:

Hello!

To solve this problem use the following steps!

1. We will call the expansion valve inlet 1 and exit 2

2.Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

3. Find the enthalpy of state 1 using pressure and temperature using thermodynamic tables

h1=Enthalpy(Ammonia;T=24C;P=1000kPa)=312.9KJ/kg

4. An expansion valve is a device which does not have heat or work exchange which means that the enthalpy of state one is equal to that of state 2, so using thermodynamic tables uses the pressure of state 2 and enthalpy of state 1 to find quality and temperature

x2=Quality(Ammonia;P=100kPa;h=h1=312.9KJ/kg) =0.194

T2=Temperature(Ammonia;P=100kPa;h=h1=312.9KJ/kg)=-33.6C

6 0
4 years ago
What is the smallest variable type I can use to represent the number 27?
oksano4ka [1.4K]

Answer:3

Explanation:

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3 0
3 years ago
As car production became more efficient, cars became more affordable for american families. One result of this increase in autom
tankabanditka [31]

Answer:

IT IS SIMPLE LIKE HENRY FORD MADE THIS OPTION POSSIBLE

Explanation:

5 0
2 years ago
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.
Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

E_{in}- E_{out}=E_{system,final}- E_{system,initial}

Where

E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

E_{in}- E_{out}=-Q_{out}+W_{in}

Regarding the energy system we have the following:

E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}

By doing the calculations we have:

E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0
3 years ago
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