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Oksanka [162]
3 years ago
6

Ok there..............................................

Engineering
1 answer:
aleksley [76]3 years ago
7 0

Answer:

kidney, uterus, bladder  

the second one might be a cell has received instruction to close its voltage gated salt channels

Explanation:

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Short-term memoryA) has a larger storage capacity than long-term memory.B) takes longer to retrieve than long-term memory.C) inv
zaharov [31]

Answer:D

Explanation:

Take longer time to retrieve than long term memory, involves transient modifications in the function of pre existing synapses, such as channel modifications.

6 0
3 years ago
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
Anastaziya [24]

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}

5 0
3 years ago
Water is being heated in a closed pan on top of a range whilebeing stirred by a paddle wheel. During the process 30kJ of heat is
My name is Ann [436]

Answer:

38 kJ

Explanation:

The solution is obtained using the energy balance:  

ΔE=E_in-E_out

U_2-U_1=Q_in+W_in-Q_out

U_2=U_1+Q_in+W_in-Q_out

      =38 kJ

4 0
3 years ago
Why is sssniperwolf so fine
RSB [31]
Bc she’s just a baddie
5 0
3 years ago
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Mr. Blue lives in a blue house, Mrs. Pink lives in a pink house and Mr. Red lives in a red house. Who lives in the White House?
Cerrena [4.2K]

Answer:

the president and mr.white my history teacher lol

6 0
2 years ago
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