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andreev551 [17]

3 years ago

6

Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an ins

trument used to determine density) indicates that the stream has a density of 0.810 g/mL. Using specific tractors from Table B.1, estimate the mass and volumetric feed rates of the two hydrocarbons to the mixing vessel (in U.S. customary units). State at least two assumptions required to obtain the estimate from the recommended date.

1 answer:

Taya2010 [7]3 years ago

5 0

Let me think of that

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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

3 years ago

A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat

ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n ..........1

and

total cost of capital intensive = $1,244,000 + $92.5 n ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n < $33800 + $143 n

solve we get

n > $\frac{906000}{50.5}$

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > $\frac{1244000}{104.5}$

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0

3 years ago

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

<u>FOR SUBSTANCE A:</u>

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

<u>FOR SUBSTANCE B:</u>

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

<u>C = 1.16 J/g</u>

5 0

3 years ago

Describe the three primary duties for engineer

Wewaii [24]

Answer:Prepare plans with detailed drawings that include project specifications and cost estimates.

Design and execute engineering experiments to create workable solutions.

Develop engineering calculations, diagrams and technical reports.

Explanation:

4 0

3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

2 years ago