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NikAS [45]
3 years ago
15

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

ic and potential energies, determine the mass ow rate required for a power output of 5MW.
Engineering
1 answer:
Amanda [17]3 years ago
3 0

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

        P_1 = 10 MPa

        T_1 = 500 C

- Outlet of Turbine

        P_2 = 10 KPa

        x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

          P_1 = 10 MPa

          T_1 = 500 C            ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

          P_2 = 10 KPa       -------------> h_f = 191.81 KJ/kg

          x = 0.9                  -------------> h_fg = 2392.1 KJ/kg

          h_2 = h_f + x*h_fg

          h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

          W_out = flow(m) * ( h_1 - h_2 )

          flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

          flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

          flow ( m ) = 4.852 kg/s

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Cavitation usually occurs because:
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2 years ago
An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in
Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, T_{a} = 140^{\circ}C = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, T_{i} = 60^{\circ}C = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

At room temperature

Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)

Thus

T_{e} = 109.71^{\circ}C = 273 + 109.71 = 382.71 K

where

T_{e} = Equilibrium temperature

Now,

To calculate the changer in entropy:

\Delta s = \Delta s_{a} + \Delta s_{i}

Now,

For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K

For Iron:

\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

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3 years ago
(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i
kirill115 [55]

Answer:

The equation used to solve a diode is

i_d = I_se^\frac{V_d}{V_T}-1

  • i_d is the current going through the diode
  • I_s is your saturation current
  • V_D is the voltage across your diode
  • V_T is the voltage of the diode at a certain room temperature. by default, you always use V_T=25.9mV for room temperature.

If you look at the equation, i_d = I_se^\frac{V_d}{V_T}-1, you'd notice that the e^\frac{V_d}{V_T} grow exponentially fast, so we can ignore the -1 in the equation because it's so small compared to the exponential.

i_d = I_se^\frac{V_d}{V_T}-1

i_d\approx I_se^\frac{V_d}{V_T}

Therefore, use i_d= I_se^\frac{V_d}{V_T} to solve your equation.

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a.)

i.)

You're given I_s=10^{-11}A

at i_d=10\mu A,     V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{10\cdot10^{-6}}{10\cdot10^{-11}})=.298V

at i_d=100\mu A,   V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{100\cdot10^{-6}}{10\cdot10^{-11}})=.358V

at i_d=1mA,      V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{1\cdot10^{-3}}{10\cdot10^{-11}})=.417V

<em>note: always use</em>  V_T=25.9mV

ii.)

Just repeat part (i) but change to I_s=-5\cdot10^{-12}A

b.)

same process as part A. You do the rest of the problem by yourself.

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Answer:

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Hello,

In this case, considering the given information and that the feed is a saturated liquid, the solution is shown on the attached pictures with the proper information and procedure.

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