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Archy [21]
3 years ago
7

A 350 m^3 retention pond that holds rainwater from a shopping mall is empty at the beginning of a rainstorm. The flow rate out o

f the retention pond must be restricted to 320 L/min to prevent downstream flooding from a 6-hour storm. What is the maximum flow rate (in L/min) into the pond from a 6-hr storm that will not flood it? a. 5,860 L/min b. 321 L/min c. 1.290 L/min d. 7.750 L/min
Engineering
1 answer:
White raven [17]3 years ago
6 0

Answer:

correct option is c. 1.290 L/min

Explanation:

given data

retention volume = 350 m³

flow rate out = 320 L/min

time = 6 hour

to find out

maximum flow rate (x)

solution

we apply here Mass rate of accumulation formula that is  

Mass rate of accumulation = mass rate of input - mass rate of output ................1

350 × 1000 L/m³ = x × ( 6hr × 60min ) -  ( 320 L × 6 hr × 60 min )

3.5 × 10³ + 4.65 × 10³ = x × 360

x = 1290 L/min

so correct option is c. 1.290 L/min

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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}

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}

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