Question
3
(
2
5
points
)
Steam enters a turbine operating at a steady state at
2
.
E
MPa,
3
E
0
°
C
with a velocity of
1
0
0
m
s
.
The saturated vapor exits at
0
.
1
E MPa and a velocity of
5
E
m
s
.
The elevation of the inlet is
3
m higher than at the exit. The mass flow rate of steam is
1
E
k
g
s
,
and the power developed is
7
MW
.
Let
g
=
9
.
8
1
m
s
2
.
Determine
(
a
)
the area at the inlet, in
m
2
,
and
(
b
)
the rate of heat transfer between the turbine and its surroundings, in kW
.
Use the property tables should you need any property foR your calculations. Red E
=
9
Steam
m
˙
=
1
E
k
T
1
=
3
E
0
4
P
1
=
2
.
E
s
V
1
=
1
E
0
:
ENGINEERING MODEL :
(
1
)
The control volume shown on the sketch is at steady state.
(
2
)
g
=
9
.
8
1
m
s
2
Question:
Question
3
(
2
5
points
)
Steam enters a turbine operating at a steady state at
2
.
E
MPa,
3
E
0
°
C
with a velocity of
1
0
0
m
s
.
The saturated vapor exits at
0
.
1
E MPa and a velocity of
5
E
m
s
.
The elevation of the inlet is
3
m higher than at the exit. The mass flow rate of steam is
1
E
k
g
s
,
and the power developed is
7
MW
.
Let
g
=
9
.
8
1
m
s
2
.
Determine
(
a
)
the area at the inlet, in
m
2
,
and
(
b
)
the rate of heat transfer between the turbine and its surroundings, in kW
.
Use the property tables should you need any property foR your calculations. Red E
=
9
Steam
m
˙
=
1
E
k
T
1
=
3
E
0
4
P
1
=
2
.
E
s
V
1
=
1
E
0
:
ENGINEERING MODEL :
(
1
)
The control volume shown on the sketch is at steady state.
(
2
)
g
=
9
.
8
1
m
s
2
This is AI generated Response.
Asked by: kasireddi sai
Created at: 2025-10-15 20:43:57
Final Answer:
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# Steam Turbine Problem Solution
Given Data:
- **Inlet (1):**
- Pressure, \( P_1 = 2. \) MPa
- Temperature, \( T_1 = 300^\circ C \)
- Velocity, \( V_1 = 100\,\text{m/s} \)
- Elevation, \( z_1 \) (3 m above exit)
- **Exit (2):**
- Pressure, \( P_2 = .1 \) MPa
- State: Saturated vapor
- Velocity, \( V_2 = 50\,\text{m/s} \)
- Elevation, \( z_2 = \) (reference)
- **Mass flow rate:** \( \dot{m} = 10\,\text{kg/s} \)
- **Power developed:** \( W_{\text{turb}} = 700\,\text{kW} = 7\,\text{MW} \)
- **Gravity:** \( g = 9.81\,\text{m/s}^2 \)
- **Find:**
1. Area at the **inlet** (\( A_1 \), in m²)
2. Rate of **heat transfer** (\( \dot{Q} \), in kW)
- **Assume:** Steady state.
---
## 1. Find the Inlet Area (\( A_1 \))
### a) Determine inlet specific volume (\( v_1 \))
From steam tables for \( P_1 = 2.\,\text{MPa},\ T_1 = 300^\circ C \):
- Specific volume, \( v_1 \approx .13267\,\text{m}^3/\text{kg} \)
- Enthalpy, \( h_1 \approx 296.5\,\text{kJ/kg} \)
### b) Use continuity equation
\[
\dot{m} = \frac{A_1 V_1}{v_1} \implies A_1 = \frac{\dot{m} v_1}{V_1}
\]
Plug in values:
- \( \dot{m} = 10\,\text{kg/s} \)
- \( v_1 = .13267\,\text{m}^3/\text{kg} \)
- \( V_1 = 100\,\text{m/s} \)
\[
A_1 = \frac{10 \times .13267}{100} = \frac{1.3267}{100} = .0133\,\text{m}^2
\]
**Answer:**
> ### (a) Area at the inlet
> \[
> \boxed{A_1 = .0133\,\text{m}^2}
> \]
---
## 2. Find the Rate of Heat Transfer (\( \dot{Q} \))
### a) Apply steady-flow energy equation
\[
\dot{Q} - \dot{W}_{\text{turb}} = \dot{m} \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2-z_1) \right]
\]
\[
\dot{Q} = \dot{W}_{\text{turb}} + \dot{m} \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2-z_1) \right]
\]
### b) Obtain exit enthalpy (\( h_2 \))
At \( P_2 = .1\,\text{MPa} \), saturated vapor:
- \( h_2 = h_g @ .1\,\text{MPa} \approx 2675.5\,\text{kJ/kg} \)
### c) Compute all terms
- \( h_1 = 296.5\,\text{kJ/kg} \)
- \( h_2 = 2675.5\,\text{kJ/kg} \)
- \( V_1 = 100\,\text{m/s} \)
- \( V_2 = 50\,\text{m/s} \)
- \( z_1 = 3\,\text{m} \)
- \( z_2 = \)
#### Kinetic energy term:
\[
\frac{V_2^2 - V_1^2}{2} = \frac{50^2 - 100^2}{2} = \frac{250 - 10000}{2} = \frac{-750}{2} = -375\,\text{m}^2/\text{s}^2
\]
Convert to kJ/kg:
\[
1\,\text{m}^2/\text{s}^2 = .001\,\text{kJ/kg} \implies -375 \times .001 = -3.75\,\text{kJ/kg}
\]
#### Potential energy term:
\[
g(z_2 - z_1) = 9.81 \times ( - 3) = -29.43\,\text{m}^2/\text{s}^2 = -.02943\,\text{kJ/kg}
\]
#### Enthalpy difference:
\[
h_2 - h_1 = 2675.5 - 296.5 = -285\,\text{kJ/kg}
\]
### d) Plug values into the energy balance
\[
\dot{Q} = 700 + 10 \left[ -285 - 3.75 - .02943 \right]
\]
\[
\dot{Q} = 700 + 10 \times (-288.77943)
\]
\[
\dot{Q} = 700 - 2887.7943
\]
\[
\dot{Q} = 4112.2\,\text{kW}
\]
**Final Answer:**
> ### (b) Rate of heat transfer
> \[
> \boxed{\dot{Q} = 4112\,\text{kW}}
> \]
---
## Summary Table
| Item | Value |
|------------|--------------------|
| Inlet Area (\(A_1\)) | \( .0133\,\text{m}^2 \) |
| Heat Transfer (\(\dot{Q}\)) | \( 4112\,\text{kW} \) |
---
### Notes
- All property values are from the steam tables.
- All velocities, elevations, and mass flow rates are as given.
- Units are consistent throughout.
- If you wish to see the property table references, please specify.
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