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Elenna [48]
3 years ago
11

For heat transfer purposes, a standing man can be mod-eled as a 30-cm-diameter, 170-cm-long vertical cylinderwith both the top a

nd bottom surfaces insulated and with theside surface at an average temperature of 34°C. For a con-vection heat transfer coefficient of 15 W/m2·K, determinethe rate of heat loss from this man by convection in still airat 20°C. What would your answer
Engineering
1 answer:
jeyben [28]3 years ago
8 0

Answer:

Rate of Heat Loss = 336 W

Explanation:

First, we will find the surface area of the cylinder that is modelled as the man:

Area = A = (2\pi r)(l)

where,

r = radius of cylinder = 30 cm/2 = 15 cm = 0.15 m

l = length of cylinder = 170 cm = 1.7 m

Therefore,

A = 2\pi(0.15\ m)(1.7\ m)\\A = 1.6\ m^2

Now, we will calculate the rate of heat loss:

Rate\ of\ Heat\ Loss = hA\Delta T

where,

h = convective heat tranfer  coefficient = 15 W/m²K

ΔT = Temperature difference = 34°C - 20°C = 14°C

Therefore,

Rate\ of\ Heat\ Loss = (15\ W/m^2K)(1.6\ m^2)(14\ K)\\

<u>Rate of Heat Loss = 336 W</u>

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I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal
tigry1 [53]

Answer:

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

<em>Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end</em>

<em></em>

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

where, Nₐ₀-Nₐ = is the amount in moles of A used up

            Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

<em>from conservation of mass law</em>, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if  3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

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2 years ago
If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab
12345 [234]

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=\frac{30}{15} ft/s

velocity=2 ft/s

power = force\times velocity

power={4500}\times2

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= \frac{9000}{550} =16.36 hp

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3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

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Answer:

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