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Korolek [52]
3 years ago
15

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

d a velocity of 10 m/s. The steam expands to the turbine exit where the pressure is 0.07 MPa, specific enthalpy is 2431.7 kJ/kg, and the velocity is 90 m/s. The mass flow rate is 11.95 kg/s. Neglecting potential energy effects, determine the power developed by the turbine, in kW.
Engineering
1 answer:
Anarel [89]3 years ago
7 0

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

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2 years ago
Anne works with other engineers studying physical processes that involve the flow of particles. Which field of engineering would
slega [8]

Answer:

Option A

Chemical engineering

Explanation:

Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.

3 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA
weqwewe [10]
OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.


It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
5 0
2 years ago
The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
2 years ago
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