Answer:
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
