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Nata [24]
3 years ago
7

In a simple ideal Rankine cycle, water is used as the working fluid. The cycle operates with pressures of 2000 psi in the boiler

and 4 psi in the condenser. What is the minimum temperature required at the turbine inlet, so that the quality of the steam at the turbine outlet is not less than 85%. What would be the thermal efficiency of the cycle?
Engineering
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible,  therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is  not a given, so it cannot be calculated.

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6 0
2 years ago
A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed
yan [13]

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

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3 0
3 years ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
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