What was the main

Steam enters an adiabatic turbine at 8 MPa and 500C with a mass flow rate of 3

Vinil7 [7]

Answer:

a)temperature=69.1C

b)3054Kw

Explanation:

Hello!

To solve this problem follow the steps below, the complete procedure is in the attached image

1. draw a complete outline of the problem

2. to find the temperature at the turbine exit use termodinamic tables to find the saturation temperature at 30kPa

note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa

4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.

Note: Remember that for a turbine with a single input and output, the power is calculated as the product of the mass flow and the difference in enthalpies.

5. Find the real power of the turbine

3 0

3 years ago

For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a domin

Ad libitum [116K]

Answer:The awnser is 5

Explanation:Just divide all of it

4 0

3 years ago

The B-pillar may also be called the:

slega [8]

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

looked it up

6 0

3 years ago

Saturated liquid water flows through 2 cm ID stainless steel tubes at 200 g/s. The water is at 80oC and the inside surface of th

EleoNora [17]

Answer:

steel

Explanation:

8 0

3 years ago

A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De

andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes = {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes = {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes = {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0

3 years ago