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2 years ago

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A vertical cylinder (Fig. P3.227) has a 61.18-kg piston locked with a pin, trapping 10 L of R-410a at 10◦C with 90% quality insi

de. Atmospheric pressure is 100 kPa, and the cylinder cross-sectional area is 0.006 m2. The pin is removed, allowing the piston to move and come to rest with a final temperature of 10◦C for the R-410a. Find the final pressure, the work done, and the heat transfer for the R-410a.

1 answer:

Whitepunk [10]2 years ago

4 0

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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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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:

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

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

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

The capacity ratio is:

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

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

$c = 0.541$

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:

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

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

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

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

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

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

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

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

$L = 220\,m$

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.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

3 years ago