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3 years ago

How many times has the ITU-R revised the CCIR 601 international standard? A. four B. five C. six D. seven

Engineering

1 answer:

Alex787 [66]3 years ago

8 0

Answer:

b or c I think it's c

Explanation:

..............

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Why do we study National studies

Gre4nikov [31]

Answer:

It’s so we can then understand other cultures and other ways that people do things or we can also be prepared for what a country that we have not been to yet is going to relatively be like. This can also help with vacation planning.

Explanation:

it can also help us understand different people from different cultures.

3 0

2 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

Energy transfer in mechanical systems: During steady-state operation, a mechanical gearbox receives 70 KW of input power through

Degger [83]

Answer:

Heat transfer rate(Q)= 1.197kW

Power output(W)=68.803kW

3 0

3 years ago

A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract

Hitman42 [59]

Answer:

radius = 9.1 × $10^{-3}$ m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support = 45 mm

solution

we apply here minimum radius formula that is

radius = $\sqrt[3]{\frac{FL}{\sigma \pi}}$ .................1

here F is applied load and is length

put here value and we get

radius = $\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}$

solve it we get

radius = 9.1 × $10^{-3}$ m

8 0

3 years ago

To do the right thing or to have moral principles is called _____ . a code

crimeas [40]

Answer:

Ethics

Explanation:

To do the right thing or to have moral principles is called ethics. Ethics affect how people make decisions and lead their lives. Some words people associate with ethics includes:

- Honesty

- Integrity

- Promise keeping and Trustworthiness

- Loyalty

- Fairness

- Concern for others

- Respect for others

- Law abiding

etc.

7 0

3 years ago