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2 years ago

The PPC is bowed outward because:

Engineering

1 answer:

9966 [12]2 years ago

6 0

PPC is following me

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Based on these statements:

Evgen [1.6K]

Answer:

the third statement is true

Explanation:

given data

Lenovos cost more than Dells

Lenovos cost less than Apples

solution

we have given 1st statement that is express as

cost (Lenovo) > cost (Dell) ..................1

and

2nd statement that is express as

cost (Lenovo) < cost (Apple)

so we can say it as

cost (Apple) > cost (Lenovo) ......................2

and

now above Both equation 1 and 2 can be written as

cost (Apple) > cost (Lenovo) > cost (Dell) .........................3

so we can say cost of Apples is more than the cost of Lenovos and the cost of Dells

so as that given 3rd statement is true

7 0

3 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

7 0

3 years ago

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

zvonat [6]

Answer:

R= 1.25

Explanation:

As given the local heat transfer,

$Nu_x = 0.035 Re^{0.8}_x Pr^{1/3}$

But we know as well that,

$Nu=\frac{hx}{k}\\h=\frac{Nuk}{x}$

Replacing the values

$h_x=Nu_x \frac{k}{x}\\h_x= 0.035Re^{0.8}_xPr^{1/3} \frac{k}{x}$

Reynolds number is define as,

$Re_x = \frac{Vx}{\upsilon}$

Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity

Then replacing we have

$h_x=0.035(\frac{Vx}{\upsilon})^{0.8}Pr^{1/3}kx^{-1}$

$h_x=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}kx^{0.8-1}$

$h_x=Ax^{-0.2}$

*Note that A is just a 'summary' of all of that constat there.

That is $A=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}k$

Therefore at x=L the local convection heat transfer coefficient is

$h_{x=L}=AL^{-0.2}$

Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so

$h=\frac{1}{L}\int\limit^L_0 h_x dx\\h=\frac{1}{L}\int\limit^L_0 AL^{-0.2}dx\\h=\frac{A}{0.8L}L^{0.8}\\h=1.25AL^{-0.2}$

The ratio of the average heat transfer coefficient over the entire plate to the local convection heat transfer coefficient is

$R = \frac{h}{h_L}\\R= \frac{1.25Al^{-0.2}}{AL^{-0.2}}\\R= 1.25$

3 0

2 years ago

Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3

Neko [114]

Answer:

$\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

Explanation:

Diffusion flux of a gas, J is given by

$J=P_m\frac {\triangle P}{\triangle x}$ where $P_m$ is permeability coefficient, $\triangle$ P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is $38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)$

Since the thickness of sheet is given as 1mm= 0.1 cm then

$J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}$

Therefore, the diffusion flux is $\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

7 0

3 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago