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Eva8 [605]
3 years ago
5

Steam flows steadily through a turbine at a rate of 420 kg/min. The enthalpy of the steam decreases by 600 kJ/kg as it flows ste

adily through the turbine. If the power generated by the turbine is 4 MW, determine the rate of heat loss from the steam turbine [kW).
Engineering
1 answer:
Ghella [55]3 years ago
4 0

Answer:

the rate of heat loss from the steam turbine  is Q = 200 kW

Explanation:

From the first law of thermodynamics applied to open systems

Q-W₀ = F*(ΔH + ΔK + ΔV)

where

Q= heat loss

W₀= power generated by the turbine

F= mass flow

ΔH = enthalpy change

ΔK = kinetic energy change

ΔV = potencial energy change

If we neglect the changes in potential and kinetic energy compared with the change in enthalpy , then

Q-W₀ = F*ΔH

Q =  F*ΔH+ W₀

replacing values

Q =  F*ΔH+ W₀ = 420 kg/min * (-600 kJ/kg) * 1 min/60 s * 1 MW/1000 kW + 4 MW = -0.2 MW = -200 kW (negative sign comes from outflow of energy)

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A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft
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Answer with Explanation:

By the equation or Torque we have

\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}

where

T is the torque applied on the shaft

I_{p} is the polar moment of inertia of the shaft

\tau is the shear stress developed at a distance 'r' from the center of the shaft

\theta is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft I_{p}=\frac{\pi R^4}{2}

For a hollow shaft I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

\frac{2T}{\pi R^4}\times r=\tau _{solid}

2) For hollow shaft we have

\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}

Comparing the above 2 relations we see

\frac{\tau _{solid}}{\tau _{hollow}}=0.76

Similarly for angle of twist we can see

\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316

Part b)

Strength of solid shaft = \tau _{max}=\frac{T\times R}{I_{solid}}

Weight of solid shaft =\rho \times \pi R^2\times L

Strength per unit weight of solid shaft = \frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}

Strength of hollow shaft = \tau '_{max}=\frac{T\times R}{I_{hollow}}

Weight of hollow shaft =\rho \times \pi (R^2-0.7R^2)\times L

Strength per unit weight of hollow shaft = \frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}

Thus \frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16

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A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i
lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s

h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

m_1+m_2+m_3=m

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

m_1h_1+m_2h_2+m_3h_3=mh

By putting the values

1\times 100+1.5\times 120+2\times 500=4.5\times h

So h=284.44 KJ

4 0
3 years ago
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