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
lilavasa [31]
3 years ago
14

During a long run a very well-trained dog can use up to 1000 ‘cal’/hour (Note: Food calories differ by a factor of one thousand

from the scientific calorie). If it is assumed that the dog has a mass of twenty five kilograms and the physical properties of water, determine the uniform heat generation per unit volume under the assumption that the heat generation is uniform in the dog’s body(Note: You must determine the dog’s volume as part of the calculation.).
Engineering
1 answer:
cricket20 [7]3 years ago
5 0

Answer:

46488.8 W / m³

Explanation:

Given:

calories used by the dog = 1000 cal/ hour

1 Food calories = 1000 scientific calories

also,

1 scientific calorie = 4.184 J

thus,

total Power used by the dog per second = ( 1000 × 1000 ×  4.184 ) / 3600

or

total power used by the dog per second =  1162.22 W

Mass of the dog = 25 kg

since, the physical properties of the dog is similar to water

thus,

density of the dog = 1000 kg/m³

thus,

the volume of the dog = Mass / Density = 25 / 1000 = 0.025 m³

Now,

we have the relation

Heat supplied (Q) = volume × Heat generated per unit volume (q)

or

1162.22 = 0.025 × q

or

q = 46488.8 W / m³

You might be interested in
Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes
loris [4]

Answer:

a). 8.67 x 10^{-3} m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L  = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X 10^{-5} Pa-s

We know density of air is ρ = 1.21 kg /m^{3}

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = \frac{\rho .U.x}{\mu }

Re = \frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = \frac{5.x}{\sqrt{Re}}

   = \frac{5\times 1}{\sqrt{332052.6}}

   = 8.67 x 10^{-3} m

a). Boundary layer thickness at x = 1 is δ = 8.67 X 10^{-3} m

b). Given Re = 100000

    Therefore the critical distance from the leading edge can be found by,

     Re = \frac{\rho .U.x}{\mu }

     100000 = \frac{1.21\times5\times x}{1.822 \times10^{-5}}

     x = 0.3011 m

c). Given x = 3 m from the leading edge

    The Reyonld no at x = 3 m from the leading edge

     Re = \frac{\rho .U.x}{\mu }

     Re = \frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }

     Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

    δ = \frac{0.38\times x}{Re^{\frac{1}{5}}}

       = \frac{0.38\times 3}{996158.06^{\frac{1}{5}}}

       = 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

  Re = \frac{\rho \times U\times x}{\mu }

5 X 10^{5} = \frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}

 x = 1.505 m

We know that the force acting on the plate is

F_{D} = \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

and C_{D} at x= 1.505 for a laminar flow is = \frac{1.328}{\sqrt{Re}}

                                                                         = \frac{1.328}{\sqrt{5\times10 ^{5}}}

                                                                       = 1.878 x 10^{-3}

Therefore, F_{D} =  \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

                                          = \frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}

                                         = 0.2137 N

e). The flow is turbulent at the end of the plate.

  Re = \frac{\rho \times U\times x}{\mu }

       = \frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }

       = 1992316

Therefore C_{D} = \frac{0.072}{Re^{\frac{1}{5}}}

                                           = \frac{0.072}{1992316^{\frac{1}{5}}}

                                           = 3.95 x 10^{-3}

Therefore F_{D} = \frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}

                                           = \frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}

                                          = 1.792 N

3 0
3 years ago
Which investigative process is most helpful for learning about past societies?
tatuchka [14]

Answer: think it A

Explanation: makes

6 0
3 years ago
Wastewater flows into a _________ once it is released into a floor drain.
vodomira [7]

Answer:

a

Explanation:

5 0
3 years ago
Which part of a machine control unit interacts with the machine tools through electric signals?=]
FrozenT [24]

Answer:

control loop unit

Explanation:

Edmentum/Plato

4 0
2 years ago
Draw the internal connections of motor generator set​
Akimi4 [234]

Answer:

tyjtgfjhgk vgjyg7igjccxfb  rt5bshe dgrty5rm nry5ghbhjyrdegbtyr45bh4 cnbfgcb xdftjrnn hdftytr s  jhbgfhtyujt ntj

Explanation:

3 0
3 years ago
Other questions:
  • Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!
    13·1 answer
  • HELP PLEASE<br> this is for drivers ed btw
    5·1 answer
  • A flat-plate solar collector is 2 m long and 1 m wide and is inclined 60o to the horizontal. The cover plate is separated from t
    13·1 answer
  • Are routers better for internet connection rather than a WiFi modem?
    6·2 answers
  • Module 42 Review and Assessment
    7·1 answer
  • The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect
    10·2 answers
  • list out main types of material used in design and Manufacture of product give one example for each in engineering application ?
    10·1 answer
  • A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a
    14·1 answer
  • In-------process the hot drawn bar or rod is pulled through the die.
    7·1 answer
  • when discussing valve train components, technician a says stamped rocker arms are very strong and may be used in high-horsepower
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!