Need help I’m tryna get a good grade

Air enters a turbine with a stagnation pressure of 900 kPa and a stagnation temperature of 658K, and it is expanded to a stagnat

bezimeni [28]

Answer:

12.332 KW

The positive sign indicates work done by the system ( Turbine )

Explanation:

Stagnation pressure( P1 ) = 900 kPa

Stagnation temperature ( T1 ) = 658K

Expanded stagnation pressure ( P2 ) = 100 kPa

Expansion process is Isentropic, also assume steady state condition

mass flow rate ( m ) = 0.04 kg/s

<u>Calculate the Turbine power </u>

Assuming a steady state condition

( p1 / p2 )^(r-1/r) = ( T1 / T2 )

= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )

= ( 9 )^0.285 = 658 / T2

∴ T2 = 351.22 K

Finally Turbine Power / power developed can be calculated as

Wt = mCp ( T1 - T2 )

= 0.04 * 1.005 ( 658 - 351.22 )

= 12.332 KW

The positive sign indicates work done by the system ( Turbine )

6 0

2 years ago

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C

Qa= 5 MW

Lets heat supplied at 150°C = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

$COP=\dfrac{Qr}{W}$

$2.5=\dfrac{5}{W}$

W=2 MW

For Carnot heat pump

$COP=\dfrac{T_2}{T_2-T_1}$

$2.5=\dfrac{T_2}{T_2-(273+85)}$

2.5 T₂ - 895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0

2 years ago

I need a thesis statement about Engineers as Leaders.

algol [13]

Answer:

Engineers are a very beneficial contribution in which offers great solutions to national problems.

5 0

2 years ago

A two-dimensional flow field described by

Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

5 0

2 years ago

A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi

hodyreva [135]

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 ) = Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

6 0

2 years ago