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3 years ago

8

You are an engineer in an electric-generation station. You know that the flames in the boiler reach a temperature of 1200 K and

that cooling water at 300 K is available from a nearby river. What is the maximum efficiency your plant will ever achieve?

1 answer:

olga nikolaevna [1]3 years ago

5 0

Answer:

<em>The maximum efficiency the plant will ever achieve is 75%</em>

<em>Explanation:</em>

From the question given, we recall the following:

<em>Th flames in the boiler reaches a temperature of = 1200K</em>

<em>the cooling water is = 300K</em>

<em>The maximum efficiency the plant will achieve is defined as:</em>

Let nmax = 1 - Tmin /Tmax

Where,

Tmin = Minimum Temperature in plants

Tmax = Maximum Temperature in plants

The temperature of the cooling water = Tmin = 300K

The temperature of the flames in boiler = Tmax = 1200k=K

The maximum efficiency becomes:

nmax = 1 - Tmin /Tmax

nmax = 1 - 300 /1200

nmax = 1-1/4 =0.75

nmax = 75%

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Define predictive analysis

svet-max [94.6K]

Answer:

Predictive analytics encompasses a variety of statistical techniques from data mining, predictive modelling, and machine learning, that analyze current and historical facts to make predictions about future or otherwise unknown events.

Explanation:

4 0

3 years ago

How many gallons of 25% concentrated due will Lucy need to add to the 15% concentrated dye to make a batch with a concentration

Volgvan

Complete question is;

Lucy works for a company that manufactures liquid dyes for fabric. She currently wants to mix up some 23%-concentrated teal dye. She has 16 gallons of 15%-concentrated teal dye, as well as plenty of 25%-concentrated teal dye. How many gallons of the 25%-concentrated teal dye will Lucy need to add to the 15%-concentrated teal dye to make a batch with a concentration of 23%?

Answer:

64 gallons

Explanation:

We are given;

Quantity of 15%-concentrated teal dye = 16 gallons

Let the number of gallons of 25%-concentrated teal dye needed to add to make the batch with a concentration of 23% be represented as x.

Thus;

(15% × 16) + (25%)x = 23%(16 + x)

Multiply each term by 100 to give;

(15 × 16) + 25x = 23(16 + x)

240 + 25x = 368 + 23x

Rearranging, we have;

25x - 23x = 368 - 240

2x = 128

x = 128/2

x = 64

8 0

3 years ago

Piessunzed water (pin-10 bar, Tin= 1 10°C) enters the hottom of an L 12-m-long vertical tube of diameter D110 mm at a mass flow

Westkost [7]

Answer:

(1) $\Delta E_{tw}=4845.43 kW$

(2) $\Delta E_m=6.329 kW$

(3) $\Delta E_t= 4851.759 kW$

(4) $q= 4851.759 kW$

Explanation:

At the saturation temperature, water starts boiling, and before that heat is added at constant pressure as latent heat.

From the saturated water-pressure table, at the pressure $P=10$ bar, we have

The saturated temperature of the water, $T_{sw}=179.88^{\circ} C$

The specific volume of water, $v_{ws}=v_f=0.00127 m^3/kg$

Specific enthalpy of water, $h_{ws}=h_f=762.50 kJ/kg$

The given inlet temperature of the water, $T_i=110^{\circ}$ , so, latent heat added to the water to reach the saturation temperature is

$h_l=C_P(T_{sw}-T_i)$

$\Rightarrow h_l=4.187(179.88^{\circ} -110^{\circ}$

$\Rightarrow h_l=292.587 kJ/kg$

Now, specific enthalpy of the water at the inlet $=$ (specific enthalpy of the water at the saturation temperature) $-$ (Latent heat capacity).

$\Rightarrow h_i=h_{sw}-h_l$

$\Rightarrow h_i=762.50-292.587=469.912kJ/kg$

The specific volume of the water at intel is the same as the specific volume at the saturation temperature as volume remains unchanged on the addition of latent heat.

So, $v_i=v_{ws}=0.00127 m^3/kg$ .

The outlet temperature, $T_o=600^{\circ} C$ and pressure, $P_o=7$ bar. From the superheated water table, we have

The specific volume of water, $v_o=0.5738 m^3/kg$

The specific enthalpy of water, $h_{wo}=3700.9 kJ/kg$

The given mass flow rate, $\dot{m} =1.5 kg/s$ .

The inlet radius and outlet diameter are the same, i.e

$d_i=d_o=110 mm=0.11m$ .

So, Inlet and outlet areas, $A_i=A_f=9.5033\times 10^{-3} m^2$ .

Let the inlet and outlet velocities be $V_i$ and $V_o$ respectively.

For the given specific volume, $v$ , and mass flow rate, $\dot{m}$ , the velocity, $V$ , at any cross-section having an area $A$ is

$V=\frac{v\dot{m}}{A}$ .

So, the inlet velocity,

$V_i=\frac{v_i \dot{m}}{A_i}$

$\Rightarrow V_i=\frac{0.00127\times 1.5}{9.5033\times 10^{-3}}$

$\Rightarrow V_i=0.20 m/s$ .

Similarly, the outlet velocity,

$V_o=\frac{v_o \dot{m}}{A_o}$

$\Rightarrow V_o=\frac{0.5738\times 1.5}{9.5033\times 10^{-3}}$

$\Rightarrow V_0=90.57 m/s$ .

(1) The change in combined thermal energy and work flow $=$ Change in the thermal energy $+$ Change in the flow work

$\Delta E_{tw}= \dot{m}(u_f-u_i)+\dot{m} (P_fv_f-P_iv_i)$

$\Rightarrow \Delta E_{tw}=\dot{m}[(u_f+P_fv_f) - (u_i+P_iv_i)$

$\Rightarrow \Delta E_{tw}=\dot{m} (h_f-h_i)$

$\Rightarrow \Delta E_{tw}=1.5(3700,20-469.912)=4845.43 kW$

(2)The change in mechanical energy

$\Delta E_m=$ Change in kinetic energy + change in potential energy

$\Rightarrow \Delta E_m=\left(\frac 1 2 \dot{m} V_f^2-\frac 1 2 \dot{m} V_i^2\right)+(\dot{m} g z_f-\dot{m} g z_i)$

$\Rightarrow \Delta E_m=\frac 1 2 \dot{m}(V_f^2-V_i^2)+\dot{m} g (z_f-z_i)$

$\Rightarrow \Delta E_m=\frac 1 2\times 1.5((90.57)^2-(0.2)^2)+1.5 \times 9.81 \times12$

$\Rightarrow \Delta E_m=6328.74 J/s=6.329 kW$

(3) The change in the total energy of water,

$\Delta E_t=$ chnge in the thernal energy + change in the flow work + change in the mechanical energy

$\Rightarrow \Delta E_t=4845.43+6.329= 4851.759 kW$ [from part (1) and (2)]

(4) Now, as there is no work done by the water, so, the heat input only caused the change in the total energy.

Hence, the rate of heat transfer, $q= 4851.759 kW$ [from part (3).

7 0

3 years ago

A batch of parts is produced on a semi-automated production machine in a sequential batch production operation. Batch quantity i

vredina [299]

Answer:

a) 4.21 min

b) 21.9666 hrs

c) 1.3657 Pc/hr

Explanation:

Given that;

Batch quantity = 300 units

time setup = 55min

unload/loading time = 0.75

processing time = 3.46

a) Average cycle time;

Average Cycle Time TC = loading time + processing time

TC = 0.75 + 3.46

Tc = 4.21 min

b) Time to complete the batch

Time to complete the batch Tb = setup time + process time + non operation time

Tb = (55min * 1hr/60min) + (300 * 3.46 * 1hr/60min) + (300 * 0.75 * 1hr/60min)

Tb = 0.9166 + 17.3 + 3.75

Tb = 21.9666 hrs

c) Average production rate

Average production rate Rp = 1 / ( Tb / batch size)

we substitute

Rp = 1 / ( 21.9666 / 300 )

Rp = 1 / 0.7322

Rp = 1.3657 Pc/hr

4 0

3 years ago

Turpentine flows through a 12-nominal schedule 40 pipe. What is the flow rate that corresponds to a Reynolds number of 2000?

r-ruslan [8.4K]

Answer:

flow rate is 8.0385 × $x^{-4}$ m³/s or 12.741 gpm

Explanation:

given data

12-nominal schedule 40 pipe

Reynolds number = 2000

to find out

What is the flow rate

solution

we know the diameter of 12-nominal schedule 40 pipe is

Diameter = 12.75 inch

D = 0.32385 m

and

dynamic viscosity of Turpentine is = 0.001375 Pa-s

and Density of Turpentine is 870 kg/m³

so

Reynolds number is express as

Re = $\frac{\rho*V*D}{\mu}$

here ρ is density and D is diameter and V is velocity and µ is viscosity

so put here all value

2000 = $\frac{870*V*0.3238}{0.001375}$

V = 9.7619 × $x^{-3}$ m/s

and

flow rate is

Q = V × A

here A is area and Q is flow rate

Q = 9.7619 × $x^{-3}$ × $\frac{\pi }{4} * 0.3238^2$

Q = 8.0385 × $x^{-4}$ m³/s

so flow rate is 8.0385 × $x^{-4}$ m³/s or 12.741 gpm

4 0

4 years ago