Question

An ideal turbojet engine is analyzed using the cold air standard method. Given specific operating conditions determine the temperature, pressure, and enthalpy at each state, and the exit velocity.

--Given Values--
T1 (K) = 249
P1 (kPa) = 61
V1 (m/s) = 209
rp = 10.7
rc = 1.8

Required:
a. Determine the temperature (K) at state 2.
b. Determine the pressure (kPa) at state 2.
c. Determine the specific enthalpy (kJ/kg) at state 2.
d. Determine the temperature (K) at state 3.

Answer 1

a. the temperature (K) at state 2 is
`\mathbf{T_2 =270.76 \ K}}`

b. the pressure (kPa) at state 2 is
`\mathtt{ \mathbf{ p_2 = 81.79 \ kPa }}`

c. the specific enthalpy (kJ/kg) at state 2 is
`\mathbf{h_2 = 271.84 \ kJ/kg}}`

d. the temperature (K) at state 3 is
`\mathbf{ T_3 = 532.959 \ K}`

Step-by-step explanation:

From the given information:

T1 (K) = 249

P1 (kPa) = 61

V1 (m/s) = 209

rp = 10.7

rc = 1.8

The objective is to determine the following:

a. Determine the temperature (K) at state 2.

b. Determine the pressure (kPa) at state 2.

c. Determine the specific enthalpy (kJ/kg) at state 2.

d. Determine the temperature (K) at state 3.

To start with the specific enthalpy (kJ/kg) at state 2.

By the relation of steady -flow energy balance equation for diffuser (isentropic)

`h_1 + (V_1^2)/(2)=h_2+(V^2_2)/(2)`

`h_1 + (V_1^2)/(2)=h_2+0`

`h_2=h_1 + (V_1^2)/(2)`

For ideal gas;enthalpy is only a function of temperature, hence
`c_p`T = h

where;

`h_1` is the specific enthalpy at inlet =
`c_pT_1`

`h_2` is the specific enthalpy at outlet =
`c_pT_2`

`c_p` = 1.004 kJ/kg.K or 1004 J/kg.K

Given that:

`T_1` (K) = 249

`V_1` (m/s) = 209

∴

`h_2=C_pT_1+ (V_1^2)/(2)`

`h_2=1004 * 249+ (209^2)/(2)`

`h_2 = 249996+21840.5`

`\mathbf{\mathtt{h_2 = 271836.5 \ J/kg}}`

`\mathbf{h_2 = 271.84 \ kJ/kg}}`

Determine the temperature (K) at state 2.

SInce;
`\mathtt{h_2 = c_pT_2 = 271.84 \ kJ/kg}`

`\mathtt{ c_pT_2 = 271.84 \ kJ/kg}`

`\mathtt{T_2 = (271.84 \ kJ/kg)/( c_p)}`

`\mathtt{T_2 = (271.84 \ kJ/kg)/(1.004 \ kJ/kg.K)}`

`\mathbf{T_2 =270.76 \ K}}`

Determine the pressure (kPa) at state 2.

For isentropic condition,

`\mathtt{ (T_2)/(T_1)= \begin {pmatrix} (p_2)/(p_1) \end {pmatrix} ^(k-1)/(k)}`

where ;

k = specific heat ratio = 1.4

`\mathtt{ (270.76)/(249)= \begin {pmatrix} (p_2)/(61) \end {pmatrix} ^(1.4-1)/(1.4)}`

`\mathtt{ 1.087389558= \begin {pmatrix} (p_2)/(61) \end {pmatrix} ^(0.4)/(1.4)}`

`\mathtt{ 1.087389558 * 61 ^ {^ (0.4)/(1.4) }}=p_2} ^(0.4)/(1.4)}`

`\mathtt{ 3.519487255=p_2} ^(0.4)/(1.4)}`

`\mathtt{ \mathbf{ p_2 = \sqrt[0.4]{3.519487255^(1.4)} }}`

`\mathtt{ \mathbf{ p_2 = 81.79 \ kPa }}`

d. Determine the temperature (K) at state 3.

For the isentropic process

`\mathtt{(T_3)/(T_2) = \begin {pmatrix} (p_3)/(p_2) \end {pmatrix}^{(k-1)/(k)}}`

where;

`\mathtt{(p_3)/(p_2) }` is the compressor ratio
`\mathtt{r_p}`

Given that ; the compressor ratio
`\mathtt{r_p}` = 10.7

`\mathtt{(T_3)/(T_2) = \begin {pmatrix} r_p \end {pmatrix}^{(k-1)/(k)}}`

`\mathtt{(T_3)/(270.76) = \begin {pmatrix} 10.7 \end {pmatrix}^{(1.4-1)/(1.4)}}`

`\mathtt{(T_3)/(270.76) = \begin {pmatrix} 10.7 \end {pmatrix}^{^ (0.4)/(1.4)}}`

`\mathtt{{T_3}{} =270.76 *\begin {pmatrix} 10.7 \end {pmatrix}^{^ (0.4)/(1.4)}}`

`\mathbf{ T_3 = 532.959 \ K}`