Superheated steam at 40 bar absolute and 500°C flows at a rate of 250 kg/min to an adiabatic turbine, where it expands to 5 bar. The turbine develops 1500 kW. From the turbine the steam flows to a heater, where it is reheated isobarically to its initial temperature. Neglect kinetic energy changes. Write an energy balance on the turbine and use it to determine the outlet stream temperature. Write an energy balance on the heater and use it to determine the required input (kW) to the steam. Verify that an overall energy balance on the two-unit process is satisfied. Suppose the turbine inlet and outlet pipes both have diameters of 0.5 meter. Show that it is reasonable to neglect the change in kinetic energy for this unit.

a. The outlet stream temperature = 310°C

b. Required Input = 1662.5kW

c. Energy is Balanced at 1662.5 kW

d. Neglecting the change in kinetic energy for this unit is reasonable

Explanation:

Given

T1 = 500°C

Rate=250kg/min

Power = 1500kW

a. From the steam table, we'll determine the following

H1(v, 40 bar, 500°C) = 3445kg/kj

H3(v,5 bar, 500°C) = 3484 kg/kj

The energy in energy, ∆E = 0 on the turbine.

So,the energy Balance is given as

∆H = -Ws = m(H2-H1)

H2 = H1 - Ws/m

H2 = 3445 - (1500)(1 min/250kg)(60s/1min)

H2 = 3085Kg/kj

From the steam table, we calculate T where H = 3085Kg/Kj and P = 5 bars

T = 310°C

The outlet stream temperature = 310°C

b. Write an energy balance on the heater and use it to determine the required input (kW) to the steam

On the heat exchanger, ∆E = 0; Ws = 0 and ∆E,k = 0.

So, we have

Q = ∆H

Q = m(H3 - H2)

Q = 250(3484 - 3085) * (1min/60s)

Q = 1662.5kW

Required Input = 1662.5kW

c. Verify that an overall energy balance on the two-unit process is satisfied

In the overall energy balance, ∆E,p = 0 and ∆E,k = 0

So,

∆H = Q - Ws

Where ∆H = m(H3 - H1)

m(H3 - H1) = Q - Ws

250(3484-3445)*(1min/60s) = Q - Ws

Q = 250(3484-3445)*(1min/60s) + Ws

Q = 250(3484-3445)*(1min/60s) + 1500kW

Q = 1662.5kW

d. Suppose the turbine inlet and outlet pipes both have diameters of 0.5 meter. Show that it is reasonable to neglect the change in kinetic energy for this unit.

First, we need to calculate the velocities from the specific volume and diameter.

V1(v,40 bar, 500°C) = 0.0864m³/kg

V2(v,5 bar, 310°C) = 0.5318m³/kg

Velocity= Ratio of volumetric flow rate and tube cross-sectional area

u1 = V1/A1 = (250 * 1/60 * 0.0864) / (0.25π * 0.5²) = 1.83m/s

u2 = V2/A2= (250 * 1/60 * 0.5318) / (0.25π * 0.5²) = 11.3m/s

∆E = ½m(u2² - u1²)

= ½ * 250 * 1/60 (11.3² - 1.83²)

= 260W

Since 260W < 1500kW then neglecting the change in kinetic energy for this unit is reasonable

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