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fiasKO [112]
3 years ago
12

The densities of several materials are given in SI units. Convert these to densities in U.S. customary units (slug/ft3), and als

o compute the specific weights of these materials in U.S. customary units (lb/ft3).The density of ceramic (alumina Al2O3), rho = 3.9 Mg/cm3.

Engineering
1 answer:
Nuetrik [128]3 years ago
5 0

Answer:

a) Density of Lead = 22.0029 slug/ft3

b) Density of Ceramic Alumina = 7567163.99 slug/ft3

c) Density of polyethylene = 1.8626slug/ft3

d) Density of Balsa wood = 0.388slug/ft3

Explanation:

The question is incomplete as other information's are missing, here is the complete question ;

The densities of several materials are given in SI units . Convert these to densities in U.S . Customary units (slug/ft3), and also compute the specific weights of these materials in U.S . Customary units (lb/ft3). (a) Lead (pure), p = 11.34g/cm3. (b) Ceramic (alumina Al2O3), p = 3.90 Mg/cm3. (c) Polyethylene (high density), p = 960 kg/m3. (d) Balsa wood, p = 0.2 Mg/m3.

The specific weight is the weight per unit volume, i.e W = mg

but mass = density x volume

W = rho x v x g = (rho)vg

W/v = rho x g

where W/v = is the specific weight and g = is the acceleration due to gravity depending on the units, but in this case as told, g is in the US Customary units (lb/slug) which is 32.2 from conversion factor, and since we are told to calculate the specific weight of each in lb/ft3.

All you have to do is to multiply each density by the acceleration due to gravity and

repeat the same step for each of the materials.

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2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin
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Answer:

a) L = 220\,m, b) U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}

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\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}

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c = \frac{C_{min}}{C_{max}}

c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}

c = 0.541

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that NTU = 2.5. The efectiveness of the heat exchanger is:

\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}

\epsilon \approx 0.824

The real heat transfer rate is:

\dot Q = \epsilon \cdot \dot Q_{max}

\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})

\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)

\dot Q = 489.795\,kW

The exit temperature of the hot fluid is:

\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})

T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}

T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}

T_{h,out} = 96.719^{\textdegree}C

The log mean temperature difference is determined herein:

\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }

\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }

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A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}

A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }

A_{i} = 9.676\,m^{2}

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L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}

L = 220\,m

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

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