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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

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A tank of final volume 10 m3 contains compressed air at 15◦C. The gage pressure in the tank is 4.50 MPa.
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Answer:

The answers to the question are as follows

(a) W = -175.6 MJ

(b) W = -329.256 MJ

The peak temperature of the isentropic compression process is 886.974 K

Explanation:

(a) We are given the initial conditions as

v₂ = 10 m³

T₂ = 15 °C

p₂ (gauge) = 4.5 MPa gauge  → 4.5 MPa + 1 atm = 4.5 MPa + 101325 Pa = 4.601 MPa

p₁ = 1 atm

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W_{12} = p_{2}  v_{2}ln(\frac{v_{1} }{v_{2} } ) per unit mass of the given gas, hence

From the relation

p₁·v₁ =p₂·v₂  therefore v₁ = p₂·v₂/p₁  = 4.6 MPa× 10 m³/(1 atm) = 4.6 MPa× 10 m³/(‪101325‬ Pa) = 454.115 m³

Therefore W₁₂ = 101325 Pa × 454.11 m³× ㏑((10 m³)/(454.115 m³)) = 46013250×(-3.82) = -175575813.855 J = -175.6 MJ

W = -175.6 MJ

(b) For isentropic compression we have

W = m×cv×(T₂ -T₁)

\frac{p_{1} }{p_{2} }  = (\frac{v_{2} }{v_{1} } )^{K} =(\frac{T_{1} }{T_{2} } )^{\frac{K}{K-1} }

for air we put K = 1.4

therefore we have \frac{101325 }{4601325 }  = (\frac{10 }{v_{1} } )^{1.4} from which

v₁ = 152.65 m³

We also have \frac{T_{1} }{T_{2} }  =(\frac{p_{1} }{p_{2} } )^{\frac{K-1}{K} }  or \frac{T_{2} }{T_{1} }  =(\frac{p_{2} }{p_{1} } )^{\frac{K-1}{K} }  from which we find the value of T₂ as {T_{2} }  =298.15( \frac{4601325 }{101325 })^{\frac{0.4}{1.4} }  = 886.974 K (peak temperature)

Therefore from pv = RT and R  =cp -cv = 1.005 -0.718 = 0.287 kJ/kg·K

Therefore number of moles = pv/(RT) = (4601325×10)(287×288.15) = 556.394 kg

m = 556.394 kg

Therefore work done at constant pressure = m·cp·(T₂-T₁)  gives

556.394 kg × 1.005 kJ/kg⋅K×(298.15 K-886.974 K ) = -329256.19 kJ or -329.256 MJ

The peak temperature of the isentropic process = 886.974 K

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