Why dose bob not let humans touch him one and only Ivan

Air at 2.5 bar, 400 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchange

Shkiper50 [21]

Answer:

a) Power developed by the turbine = 132.89 kW

b) magnitude of the rate of heat transfer from the air to the ambient, in kw = 251.25 kW

Explanation:

b) The process is a constant pressure process (Isobaric process)

The constant pressure specific heat of air, $c_{p} = 1.005 kJ/kg -K$

Specific heat ratio for air, $\gamma = 1.4$

The mass flow rate of air, $\dot{m} = 2.5 kg/s$

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

Using the steady flow energy equation:

$Q_{1-2} = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2} = 2.5 * 1.005 * (300 - 400)\\Q_{1-2} = -251.25 kW$

Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, $Q_{1-2} = 251.25 kW$

a) For the isentropic process:

Power developed by the turbine is given by the relation $\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})$

Isentropic efficiency, $\eta_{t} = 80%$

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, $T_{3s}$ = ? where $T_{3s}$ is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

$\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }$

$T_{3s} = 230.9 K$

To get the temperature at 80% efficiency, we will use the relation:

$\eta_{t} = \frac{T_{2} - T_{3} }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3} }{300 - 230.9 }$

T₃ = 244.72 K

Power developed by the turbine is given by the relation:

$\dot{W} = \dot{M} c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW$

4 0

3 years ago

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What is the purpose of a heater core

aliina [53]

A heater core is a radiator-like device used in heating the cabin of a vehicle. Hot coolant from the vehicle's engine is passed through a winding tube of the core, a heat exchanger between coolant and cabin air.

Explanation:

4 0

3 years ago

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5) A 80-kg man has a total foot imprint area of 480 cm2. Determine the pressure this man exerts on the ground if (a) he stands o

attashe74 [19]

Answer:

The pressure exerted by this man on ground

(a) if he stands on both feet is 8.17 KPa

(b) if he stands on one foot is 16.33 KPa

Explanation:

(a)

When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:

Area = A = 2 x 480 cm²

A = 960 cm²

A = 0.096 m²

The force exerted by man on his area will be equal to his weight.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.096 m²

(b)

When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:

Area = A = 480 cm²

A = 0.048 m²

The force exerted by man on his area will be equal to his weight, in this case, as well.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.048 m²

4 0

3 years ago

). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially

Delicious77 [7]

Answer:

$\frac{e'_z}{e_z} = 0.87142$

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

εx = - [ σx - ν( σy + σz ) ] / E

εy = - [ σy - ν( σx + σz ) ] / E

εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

εx = [ ν*σz ] / E

εy = [ ν*σz ] / E

εz = - [ σz ] / E .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E

εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note: σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2

σx' ( 1 - v ) = [ ν*σz ] / 2

σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

εz' = - [ σz - ν( σy' + σx' ) ] / E

εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

$\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142$ ... Answer

5 0

3 years ago

saan nag tungo si Aguinaldo at ilang pinuno ng kilusan pagkatapos mapairal ang kasunduan na pansamantalang nag dulot ng kapayapa

avanturin [10]

Answer:

sa china po

Explanation:

sana makatulong ako

6 0

2 years ago