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Olin [163]
2 years ago
9

Problem 1: Experts on hypothermia tell us that heat loss from the body is much greater in 10°C water than in air at the same tem

perature. To test this, assume that heat transfer from a person can be approximated as that from a vertical cylinder of diameter D = 0.3 m and length L = 1.8 m, with a constant surface temperature of 37°C. Assuming that heat losses from the cylinder occur only at the lateral surfaces, calculate the ratio of the heat transfer rate from the cylinder to water to that for the cylinder to air. Data for the heat transfer properties for water and air can be found in the Appendix in Tables A.2-11 and A.3-3, respectively, and appropriate correlations can be found in eqns (4.7-4), and Tables 4.7-1 and 4.7-2. What do you conclude from this about your odds of survival if you fall into a cold water lake/ocean? (Hint: neglect the cylinder height restriction in Table 4.7-1).

Engineering
1 answer:
pochemuha2 years ago
4 0

Answer:

See explaination

Explanation:

Please see attachment for the step by step solution of the given problem .

It is very detailed in the file attached.

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Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a
AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2)  + x(2) * h(fg2)

so

exit quality  = \frac{h1 - h(f2)}{h(fg2)}

exit quality  = \frac{332.17- 9.04}{1382.73)}

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0
3 years ago
Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

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n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0
3 years ago
A right triangle has a base of 12 inches and a height of 30 inches, what is the centroid of the triangle?​
aliina [53]

Answer:

the correct answer is 42

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