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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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Explanation:

Given data in question

mean stress  = 50 MPa

amplitude stress  = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first  maximum stress  and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

\sigma _{m} =  \sigma _{maximum} + \sigma _{minimum}  / 2

50 =  \sigma _{maximum} + \sigma _{minimum}  / 2    ...........1

and

225 =  \sigma _{maximum} + \sigma _{minimum}  / 2      ...........2

from eqution 1 and 2 we get maximum and minimum stress

\sigma _{maximum} = 275 MPa        ............3

and \sigma _{minimum} = -175 MPa     ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio =  \sigma _{minimum} / \sigma _{maximum}

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Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur
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Answer:

(a) BP = 11.99 KPa

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(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

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BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

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(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

<u>h = 2 m</u>

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