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iVinArrow [24]
2 years ago
9

A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly Problems 49 CH001.qx

d 2/24/11 12:03 PM Page 49 insulated. What is the minimum thickness of styrofoam insulation (k 0.030 W/m K) that must be applied to the top and side walls to ensure a heat load of less than 500 W, when the inner and outer surfaces are 10 and 35 C
Engineering
1 answer:
marusya05 [52]2 years ago
3 0

Answer:

30 mm is the minimum thickness that must be applied.

Explanation:

Given the data in the question;

Using Fourier's equation. the heat rate is  

q = kA(ΔT/Δx)

where

A is the surface area, we must consider all surfaces through which the heat can dissipate through

i.e 2×2 for one wall gives you 4m²,

there are 5 walls, so we will  have 20m² for surface area.

k is thermal conductivity of the styrofoam ( 0.030 W/m K)    

q is the heat loss (500 W  )

ΔT is the Temperature difference ( 35 - 10) = 25°C

Δx  = ?

So we substitute

500 = (0.030)(20)(25/Δx)

500 = 0.6 (25/Δx)

500 = 15 / Δx

Δx = 15 / 500

Δx = 0.03 m = 30 mm

Therefore, 30 mm is the minimum thickness that must be applied.

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Fill in the blank to correctly complete the statement below.
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3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
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Answer:

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Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

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