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3 years ago

Read Shakespeare's "Sonnet 100.”

Engineering

1 answer:

liraira [26]3 years ago

7 0

Answer: quatrain

Explanation:

Reading the Shakespeare's "Sonnet 100", we can infer that the underlined section is referred to as a quatrain.

The quatrain simply refers to a type of stanza that is made up of four lines. For example, based on the information given, we can deduce that the rhyme scheme for the second quatrain is given as cdcd.

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Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow

melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

$m_{in}h_{in}=W_{out}+m_{out}h_{out}$

Whereas the mass flow is just the same, which means we have:

$W_{out}=m_{out}(h_{out}-h_{in})$

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

$h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R$

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

$s_{out}= 2.2096BTU/lbm-R$

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

$h_{out}=1659.15BUT/lbm$

Then, the isentropic work (negative due to convention) is:

$W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU$

And the real produced work is:

$W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU$

Finally, in horsepower:

$W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP$

Regards!

6 0

3 years ago

The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are

IrinaK [193]

Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum = $9.93*10^{-8}$

Number of vacant atoms = $5.854 * 10^{15} \ vacancies/cm^3$

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = $6.022*10^{23$

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

$5.854*10^{15}$ = $9.93*10^{-8}$ × Total number of sites(N)

Total number of sites (N) = $\dfrac{5.854*10^{15}}{9.93*10^{-8}}$

Total number of sites (N) = $5.895*10^{22}$

From the expression of the total number of sites; we can determine the density of the electrum;

$N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}$

where ;

$N_A$ = Avogadro's Number

$\rho_{electrum} =$ density of the electrum

$A_{electrum}$ = Atomic mass

$5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}$

$5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}$

$8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}$

$\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}$

$\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}$

Thus; the density of the electrum is 14.30 g/cm³

7 0

3 years ago

why did my dirtbike just stop when i was riding it, and now when i start it it sounds like it has a knock, and wont stay running

chubhunter [2.5K]

Answer: check the engines i swear if ur talking about an actual bike im gonna be so embarrassed lma0

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3 years ago

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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di

SSSSS [86.1K]

Answer:

The solution is attached in the attachment.

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3 years ago

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Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro

Liula [17]

Answer:

The answer is "0.0728"

Explanation:

Given value:

$P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\$