1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
maria [59]
2 years ago
10

A sphere is assumed to have the properties of water and has an initial heat generation 46480 W/m^3 How much should the heat gene

ration term be lowered, if the maximum temperature at the center of the spherical geometry is to be limited to 360 Kelvin? The radius of the sphere is 0.1m ambient air temperature is 25 C. Convert the resulting heat generation to Food Calories per hour
Engineering
1 answer:
Verdich [7]2 years ago
3 0

Answer:

Resulting heat generation, Q = 77.638 kcal/h

Given:

Initial heat generation of the sphere, Q_{Gi} = 46480 W/m^{3}

Maximum temperature, T_{m} = 360 K

Radius of the sphere, r = 0.1 m

Ambient air temperature, T = 25^{\circ}C = 298 K

Solution:

Now, maximum heat generation, Q_{m} is given by:

T_{m} = \frac{Q_{m}r^{2}}{6K} + T                     (1)

where

K = Thermal conductivity of water at T_{m} = 360 K = 0.67 W/m^{\circ}C

Now, using eqn (1):

360 = \frac{Q_{m}\times 0.1^{2}}{6\times 0.67} + 298

Q_{m} = 24924 W/m^{3}

max. heat generation at maintained max. temperature of 360 K is 24924W/m^{3}

For excess heat generation, Q:

Q = (Q_{Gi} - Q_{m})\times volume of sphere, V

where

V = \frac{4}{3}\pi r^{3}

Q = (46480 - 24924)\times \frac{4}{3}\pi\0.1^{3} = 21556\times \frac{4}{3}\pi\0.1^{3} W/m^{3}

Q = 90.294 W

Now, 1 kcal/h = 1.163 W

Therefore,

Q = \frac{90.294}{1.163} = 77.638 kcal/h

You might be interested in
A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu
Ksju [112]

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

7 0
3 years ago
1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

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

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0
2 years ago
64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump eff
grin007 [14]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

5 0
2 years ago
Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the
Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen         σ_{max}                      σ_{min}

A                       +450                      -150

B                       +300                      -300

C                       +500                      -200

Solution:

Compute the mean stress

σ_{m} =  (σ_{max}  +  σ_{min})/2

σ_{mA} =  (450 + (-150)) / 2

       =  (450 - 150) / 2  

       = 300/2

σ_{mA} = 150 MPa

σ_{mB}  = (300 + (-300))/2

        = (300 - 300) / 2

        = 0/2  

σ_{mB}  = 0 MPa

 

σ_{mC}  = (500 + (-200))/2

        = (500 - 200) / 2

        = 300/2

σ_{mC}  = 150 MPa  

Compute stress amplitude:

σ_{a} =  (σ_{max}  -  σ_{min})/2    

σ_{aA} =  (450 - (-150)) / 2

       =  (450 + 150) / 2

       = 600/2

σ_{aA} = 300 MPa

σ_{aB} =  (300- (-300)) / 2

       =  (300 + 300) / 2

       = 600/2

σ_{aB}  = 300 MPa

σ_{aC}  = (500 - (-200))/2

        = (500 + 200) / 2

        = 700 / 2

σ_{aC}   = 350 MPa

From the above results it is concluded that the longest  fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0
3 years ago
In python/Spyder
lianna [129]

Answer:

Answer to both the parts A and B are provided in the word document attached.

Explanation:

The explanation of the programs is provided in the attached file along with the screenshot of ATM application output.

Download docx
8 0
3 years ago
Other questions:
  • A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap
    11·1 answer
  • What does basic levels of competence involves??​
    13·2 answers
  • What is the capacity of the machine in batches?
    10·1 answer
  • What is Differential Analysis in fluid mechanics?
    13·1 answer
  • Design an Armstrong indirect FM modulator to generate an FM signal with a carrier frequency 98.1 MHz and a frequency deviation △
    15·1 answer
  • Realize the function f(a, b, c, d, e) = Σ m(6, 7, 9, 11, 12, 13, 16, 17, 18, 20, 21, 23, 25, 28)using a 16-to-1 MUX with control
    13·1 answer
  • Sam constructs a circuit, connects a lead acid battery of 2 V to a lamp of resistance 3 Ω and places an ammeter across it. What
    8·2 answers
  • Documentation of a flow chart?
    7·1 answer
  • What two factors are changing when the current is changed on an electric generator
    7·1 answer
  • You may wonder who the rest goes
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!