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3 years ago

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process

Engineering

1 answer:

mihalych1998 [28]3 years ago

5 0

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

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Which of the following have the capacity to display formatted data?

sukhopar [10]

Answer:

D. A and B

Explanation:

1. The method Console.Write() is an overloaded method in a language like C#.

One of its variations could be as follows;

Console.Write(String format, Object a, Object b).

This contains three parameters and will write the text representation of the specified objects to the standard output stream using the information specified by the format specifier. Parameter 1 is format which is a composite format string representing the format specifier. Parameter 2 is a, which is the first object to be written using format. Parameter 3 is b, which is the second object to be written using format.

2. The method Console.WriteLine() has the same characteristics as Console.Write() above, except that it writes the text representation of the specified objects, followed by current line terminator then to the standard output stream using the information specified by the format specifier.

3. Console.WriteFormat() does not exist, at least not in C# or .NET

Therefore, Console.Write() and Console.WriteLine() have the capacity to display formatted data.

Hope this helps!

6 0

3 years ago

While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic

makvit [3.9K]

Answer:

$Given data:\\While walking across campus one windy day\\Frontal area, $A=0.3 m ^{2}$\\Wind speed $V=24 Km / hr$\\The drag coefficient $C_{D, b}=1.2$\\The combined mass $m=75 kg$\\Umbrella diameter, $D=1.22 m$\\Velocity of wind $V=24 \frac{ km }{ hr }$\\The rolling resistance $C_{R}=0.75 \%$$

Solution:

Note: Refer the diagram

$Basic equation:\\'s law of motion: $\sum F_{x}=m a_{x}$\\Lift coefficient, $C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}$\\Drag coefficient, $C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}$$

$From force balance equation:\\$\sum F_{x}=F_{D}-F_{R}=0$\\But $F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}$$F_{R}=C_{R} m g$\\Area of the Umbrella $A_{u}=\frac{\pi D_{u}^{2}}{4}$$A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}$$A_{v}=1.17 m ^{2}$$

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), $C_{D, x}=1.42$

Substituting all parameters,

$\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}$

Then,

$\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}$

And the equation becomes,

$\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}$

Thus the floyds travels at $68.3^{\circ}$ wind speed.

7 0

4 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0

3 years ago

Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96

Viktor [21]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

5 0

3 years ago

The toughness of a material does what, when it's been heated?

Sindrei [870]

It melt making it easily malleable

6 0

3 years ago

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process​

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process