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4 years ago

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thermo A large container ship is propelled by a heat engine that uses fuel oil as a heat source and sea water as a heat sink. Th

e engine produces 75000 W while consuming 13600 Litres per hour of fuel oil. The fuel oil releases 43000 kJ per litre burned. Calculate the rate of heat transfer to the ocean and the thermal efficiency of the ships heat engine. (5 points)

1 answer:

TEA [102]4 years ago

3 0

Answer:

$\eta = 4.617\times 10^{-4}\,(0.046\,\%)$ , $\dot Q_{out} = 162369.444\,kW$

Explanation:

The definition of thermal efficiency follows to this expression:

$\eta = \frac{\dot W}{\dot Q_{in}}$

$\eta = \frac{75\,kW}{\left(43000\,\frac{kJ}{L} \right)\cdot \left(13600\,\frac{L}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}$

$\eta = 4.617\times 10^{-4}\,(0.046\,\%)$

The rate of heat transfer to the ocean is:

$\dot Q_{out} = \dot Q_{in}-\dot W$

$\dot Q_{out} = \left(43000\,\frac{kJ}{L} \right)\cdot \left(13600\,\frac{L}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)-75\,kW$

$\dot Q_{out} = 162369.444\,kW$

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Which statement best explains the underlined portion of the quote from the declaration of independence

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Missing part of question

("We hold these truths to be self-evident, that all men are created equal....")

Answer:

Both citizens and political leaders have equal rights

Explanation:

The statement tries to remind those elected and in power as our political leaders that we are all equal with those who elected us. Therefore, both citizens and political leaders have equal rights. The statement was common among the founding fathers of several nations in Europe and America.

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3 years ago

The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Cel

Oksi-84 [34.3K]

Answer:

the rate of heat transfer into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

the rate of heat output is $\mathbf{q_{out} =182 \ W/m^2}$

the rate of change of energy stored by the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

the convection coefficient is h = 4.26 W/m².K

Explanation:

From the question:

The temperature distribution across the wall is given by :

$T(x) = ax+bx+cx^2$

where;

T = temperature in ° C

and a, b, & c are constants.

replacing 200° C for a, - 200° C/m for b and 30° C/m² for c ; we have :

$T(x) = 200x-200x+30x^2$

According to the application of Fourier's Law of heat conduction.

$q_x = -k \dfrac{dT}{dx}$

where the rate of heat input $q_{in} = q_k$ ; Then x= 0

So:

$q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}$

$q_{in}= -1 (-200+60x)_{x=0}$

$\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

Thus , the rate of heat transfer into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The rate of heat output is:

$q_{out} = q_{x=L}$ ; where x = 0.3

$q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}$

replacing T with $200x-200x+30x^2$ and k with 1 W/m.K

$q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}$

$q_{out} = -1 (-200+60x)_{x=0.3}$

$q_{out} = 200-60*0.3$

$\mathbf{q_{out} =182 \ W/m^2}$

Therefore , the rate of heat output is $\mathbf{q_{out} =182 \ W/m^2}$

Using energy balance to determine the change of energy(internal energy) stored by the wall.

$\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\$

$\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

Thus; the rate of change of energy stored by the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

We all know that for a steady state, the heat conducted to the end of the plate must be convected to the surrounding fluid.

So:

$q_{x=L} = q_{convected}$

$q_{x=L} = h(T(L) - T _ \infty)$

where;

h is the convective heat transfer coefficient.

Then:

$Replacing \ 182 W/m^2 \ for \ q_{x=L} , (200-200x +30x \ for \ T(x) \ , 0.3 m \ for \ x \ and \ 100^0 C for \ T$ We have:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient is h = 4.26 W/m².K

6 0

4 years ago

Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i

Brilliant_brown [7]

Answer:

$\dot Q_{in} = 372.239\,MW$

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

$w_{in} + h_{in}- h_{out} = 0$

$h_{out} = w_{in}+h_{in}$

$h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}$

$h_{out} = 203.81\,\frac{kJ}{kg}$

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

$h_{out} = 2685.4\,\frac{kJ}{kg}$

The rate of heat transfer in the boiler is:

$\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)$

$\dot Q_{in} = 372.239\,MW$

3 0

3 years ago

Read 2 more answers

An insulated 40 ft^3 rigid tank contains air at 50 psia and 580°R. A valve connected to the tank is now opened, and air is allow

Pachacha [2.7K]

Answer:

The answer is " $\bold{W_{in} = 645.434573 \ Btu}$ ".

Explanation:

Its air enthalpy is obtained only at a given temperature by A-17E.

The solution from the carbon cycle is acquired:

$\to \Delta U= W_{in}-m_{out}h_{out}=0\\\\\to W_{in} = (m_1 - m_2)h$

$=\frac{Vh}{RT}(P_1- P_2)\\\\= \frac{40 \times 138.66}{0.3704\times 580}(50-25) Btu\\\\= \frac{5,546.4}{214.832}(25) Btu\\\\= 25.8173829(25) Btu\\\\ =645.434573 Btu$

$W_{in} = 645.434573 \ Btu$

8 0

3 years ago

A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40

tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/ $mm^{2}$

c) 127.307 N/ $mm^{2}$

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi* $r^{2}$ = 78.55 $mm^{2}$

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give, A2 = 0.1 A1 = 7.855 $mm^{2}$

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area = F / A1 = 1000 / 78.55

Engineering stress = 12.730 $N / mm^{2}$

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final area = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 $N / mm^{2}$

e) strain = change in length / original length

strain = 10 / 40 = 0.25

8 0

3 years ago