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9 days ago

Consider a convergent-divergent duct with exit and throat areas of 0.5 m2 and 0.25 m2, respectively. The inlet reservoir pressur

e is 1 atm and the exit static pressure is 0.6 atm. For this pressure ratio, the flow will be supersonic in a portion of the nozzle, terminating with a normal shock inside of the nozzle. Calculate the local area ratio (A/A*) at which the shock is located inside the nozzle.

Engineering

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An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

Mrrafil [318]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

Final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

Substituting all values yields

0.0130 = $\frac{\pi }{4} *2^2*dl$

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0

2 months ago

The molecular weight of a 10g rubber band

Viktor [391]

The response to this query is 1 * 10 g/mole = 10.

8 0

2 months ago

The uniform dresser has a weight of 90 lb and rests on a tile floor for which the coefficient of static friction is 0.25. If the

Kisachek [356]

Answer:

a) $F = 736.065\,lbf$ , b) $\mu_{k} = 0.15$

Explanation:

a) The uniform dresser can be modeled using specific equilibrium equations:

$\Sigma F_{x} = F - \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

Following some algebraic manipulations, the formulated equation is derived:

$F = \mu_{k}\cdot m \cdot g$

$F = (0.25)\cdot (90\,lbm)\cdot (32.714\,\frac{ft}{s^{2}} )$

$F = 22.5\,lbf$

b) Similarly, the man can be represented by a set of equilibrium equations:

$\Sigma F_{x} = -F + \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

After some algebraic changes, the expression for the coefficient of static friction comes out as:

$\mu_{k} = \frac{F}{m\cdot g}$

$\mu_{k} = \frac{22.5\,lbf}{150\,lbf}$

$\mu_{k} = 0.15$

3 0

2 months ago

6.15. In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20

Viktor [391]

Explanation:

At a temperature of $33^{\circ} C$ and relative humidity of 86%, the humidity ratio stands at 0.0223 with a specific volume of 14.289.

At a temperature of $33^{\circ} C$ and relative humidity of 40%, the humidity ratio is 0.0066 while the specific volume is 13.535.

To determine the mass of air, the following formula can be used:

$\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}$

Now, we will calculate the volume

$\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}$

The duration required to fill the cistern can be determined with the equation:

$Time $=\frac{\text { cistern volume }}{\text { removal water perminute volume }}$$

By substituting the values into the preceding formula, we find:

$$\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}$\\$=279.09$ minutes\\$=4.65$ hours.$

Thus, the hours necessary to fill the cistern amount to 4.65 hours.

3 0

2 months ago

(a) For BCC iron, calculate the diameter of the minimum space available in an octahedral site at the center of the (010) plane,

choli [298]

Response:

a) The diameter available is 0.0384 nm

b)This space is less than the size of a carbon atom, which has a radius of 0.077 nm, indicating that the carbon atom won't occupy these sites.

Clarification:

For BCC iron

According to the information in Appendix B, the lattice parameter (a) is determined to be 0.2866 nm

BCC iron encompasses 4 atomic radii, thus the body diagonal length = $a(3)^\frac{1}{2}$

which represents the atomic radius of BCC iron

4r = $a(3)^\frac{1}{2}$

Substituting the value of (a) from Appendix B, set as 0.2866 nm

4r = $0.2866 nm (3)^\frac{1}{2}$

leading to r = 0.4964 nm / 4 = 0.1241 nm

Refer back to Appendix C, where the atomic radius of BCC iron is stated as 0.1241 nm, assuming the atomic sizes for iron remain consistent.

Thus, the radius ratio = 0.62

According to Figure 3.2, the space necessary for an interstitial at the BCC site exists between atoms located at the FCC site, containing two atoms, each equal to a radius of 0.2482 nm

The diameter of the minimum space available

$d_{a} = a - r_{a}$

$r_{a} = atomic radii$ = 0.2482 nm

With a = 0.2666 nm

therefore

d $_{a}$ = 0.2866 nm - 0.2482 nm = 0.0384 nm

When comparing with the diameter of a carbon atom

This space is smaller than that of a carbon atom which has a radius of 0.077 nm, confirming that the carbon atom will not be able to occupy these positions.

7 0

2 months ago