Ask question

Ask question

All categories

soldi70 [24.7K]

3 years ago

11

Determine the minimum necessary clearance height of the overpass and the resultant elevation of the bottom of the overpass over

the PVI. Ignore the cross-sectional width of the overpass.

1 answer:

Nastasia [14]3 years ago

5 0

Answer:

attached below

Explanation:

You might be interested in

A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False

ozzi

Answer:

True

Explanation:

For point in xz plane the stress tensor is given by $\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]$

where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )

thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane

8 0

3 years ago

Drag the tiles to the boxes to form correct pairs. Identify the designations of the three employees in an automobile company fro

aniked [119]

Answer:

is the fare of our responsibility towards

7 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

If the price of the car is less than or equal to your available cash, display "no". If the price of the car is more than your av

Ede4ka [16]

Answer:

function decision(car_price, available_cash) {

if(car_price <= available_cash) {

console.log("no");

}

else {

console.log("yes");

}

}

decision(car_price, available_cash); or decision(available_cash, car_price);

Explanation:

using functions in Javascript:

functions; this refers to dividing codes into reusable parts.

e.g function function_name() {

console.log("How are you?");

}

you can call or invoke this function by using its name followed by parenthesis, like this: function_name(). each time the function is called it will print out "How are you?".

Parameters: these are variables that act as placeholders for the values that are to be input into a function when it is called

Arguments: The actual values that input or passed into a function when it is called.

e.g

function function_name(parameter1, parameter2) {

console.log(parameter1, parameter2);

}

then we call function_name: function_name("please", "leave"):we have passed two arguments, "please" and "leave". Inside the function parameter1 equals "please" while parameter2 equals "leave".

Hence, from the question given the two parameters "car_price" and "available_cash" respectively, we write the function with name function_name:

function decision(car_price, available_cash) {

if(car_price <= available_cash) {

console.log("no");

}

else {

console.log("yes");

}

}

decision(car_price, available_cash); or decision(available_cash, car_price);

7 0

3 years ago

Consider the velocity boundary layer profile for flow over u flat plate to be of the form u = C_1 + C_2 y. Applying appropriate

ra1l [238]

Answer:

The result in terms of the local Reynolds number ⇒ Re = [μ_∞ · x] / v

Explanation:

See below my full workings so you can compare the results with those obtained from the exact solution.

4 0

3 years ago