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3 years ago

9

2.44mW of incident 520 nm light is directed through a1 cm sample cuvette and 0.68 mW of Plight exits the sample what is the abso

rbance of the sample?

1 answer:

Flura [38]3 years ago

7 0

Answer:

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Explanation:??

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g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70%

Bond [772]

Answer:

<em>Heat rejected to cold body = 3.81 kJ</em>

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

<em>efficiency of the Carnot's engine = 1 - </em> $\frac{Tc}{Th}$ <em> </em>

eff. of the Carnot's engine = 1 - $\frac{400}{600}$

eff = 1 - 0.25 = 0.75

<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>

work done by heat engine = 2 kJ

<em>eff. of heat engine is gotten as = W/Q</em>

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>

6 0

3 years ago

Identify the true statements about the relationship between the distillate and the sample mixture in a simple distillation.

fenix001 [56]

Answer:

The answer is "Choice b,c, and d".

Explanation:

The distillation would be the separation phase of working liquids that boil close to each other. Whenever the formulation is long-term, a much more fraction of a compound collected will have a lower boiling point than it would otherwise be. In other phrases, it'll be pulverized with both the lower level, also with the boiling point. Its structure of both the diluted volatile component may therefore include another cell's contaminants.

8 0

3 years ago

Finally you will implement the full Pegasos algorithm. You will be given the same feature matrix and labels array as you were gi

Diano4ka-milaya [45]

Answer:

In[7] def pegasos(feature_matrix, labels, T, L):

"""

.

let learning rate = 1/sqrt(t),

where t is a counter for the number of updates performed so far (between 1 and nT inclusive).

Args:

feature_matrix - A numpy matrix describing the given data. Each row

represents a single data point.

labels - A numpy array where the kth element of the array is the

correct classification of the kth row of the feature matrix.

T - the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

L - The lamba valueto update the pegasos

Returns: Is defined as a tuple in which the first element is the final value of θ and the second element is the value of θ0

"""

(nsamples, nfeatures) = feature_matrix.shape

theta = np.zeros(nfeatures)

theta_0 = 0

count = 0

for t in range(T):

for i in get_order(nsamples):

count += 1

eta = 1.0 / np.sqrt(count)

(theta, theta_0) = pegasos_single_step_update(

feature_matrix[i], labels[i], L, eta, theta, theta_0)

return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

labels = np.array([1, 1])

T = 1

L = 1

exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

Explanation:

In[7] def pegasos(feature_matrix, labels, T, L):

"""

.

let learning rate = 1/sqrt(t),

where t is a counter for the number of updates performed so far (between 1 and nT inclusive).

Args:

feature_matrix - A numpy matrix describing the given data. Each row

represents a single data point.

labels - A numpy array where the kth element of the array is the

correct classification of the kth row of the feature matrix.

T - the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

L - The lamba valueto update the pegasos

Returns: Is defined as a tuple in which the first element is the final value of θ and the second element is the value of θ0

"""

(nsamples, nfeatures) = feature_matrix.shape

theta = np.zeros(nfeatures)

theta_0 = 0

count = 0

for t in range(T):

for i in get_order(nsamples):

count += 1

eta = 1.0 / np.sqrt(count)

(theta, theta_0) = pegasos_single_step_update(

feature_matrix[i], labels[i], L, eta, theta, theta_0)

return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

labels = np.array([1, 1])

T = 1

L = 1

exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

6 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

4 years ago

Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as a

Goshia [24]

Answer:

a. Wa = 73.14 Btu/lbm

b. Sgen = 0.05042 Btu/lbm °R

c. Isentropic efficiency is 70.76%

d. Minimum specific work for compressor W = -146.2698 Btu/lbm [It is negative because work is being done on the compressor]

Explanation:

Complete question is as follows;

Air initially at 120 psia and 500oF is expanded by an adiabatic turbine to 15 psia and 200oF. Assuming air can be treated as an ideal gas and has variable specific heat.

a) Determine the specific work output of the actual turbine (Btu/lbm).

b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R).

c) Determine the isentropic efficiency of this turbine (%).

d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 oF, and the ideal exit state is 120 psia. What is the minimum specific work the compressor will be required to operate (Btu/lbm)?

solution;

Please check attachment for complete solution and step by step explanation

8 0

3 years ago