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valkas [14]
3 years ago
5

Water enters the turbine of an ideal Rankine cycle as a superheated vapor at 18 MPa and 550°C. If the condenser pressure is 9 kP

a, find: (a) the rate of heat addition into the cycle, in kJ/kg of water flowing, (b) the thermal efficiency of the system, (c) the rate of heat rejection from the cycle, in kJ/kg of water flowing, (d) the back work ratio.

Physics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

For this ideal Rankine cycle:

A) The rate of heat addition into the cycle Qh is 3214.59KJ/Kg.

B) The thermal efficiency of the system εt=0.4354.

C) The rate of heat rejection from the cycle Qc is 1833.32KJ/Kg.

D) The back work ratio is 0.013

Explanation:

To solve the ideal Rankine cycle we have to determinate the thermodynamic information of each point of the cycle. We will use a water thermodynamic properties table. In an ideal Rankine cycle, the process in the turbine and the pump must be isentropic. Therefore S₃=S₄ and S₁=S₂.

We will start with the point 3:

P₃=18000KPa  T₃=550ºc ⇒ h₃=3416.12 KJ/Kg  S₃=6.40690 KJ/Kg

Point 4:

P₄=9KPa S₄=S₃ ⇒ h₄=2016.60 (x₄=0.7645 is wet steam)

Point 1:

P₁=P₄ in the endpoint of the steam curve. ⇒ h₁= 193.28KJ/Kg  S₁=0.62235KJ/Kg  x₁=0

Point 2:

P₂=P₃ and S₂=S₁ ⇒ h₂=201.53KJ/Kg

With this information we can obtain the heat rates, the turbine, and the pump work:

Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

W_{turb}=W_{3-4}=h_3-h_4=1399.52KJ/Kg

W_{pump}=W_{1-2}=h_2-h_1=18.25KJ/Kg

We can answer the questions with this data:

A) Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

B) \epsilon_{ther}=\displaystyle\frac{W_{turb}}{Q_h}=0.4354

C)Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

D) r_{bW}=\displaystyle\frac{W_{pump}}{W_{turb}}=0.013

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onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
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Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA =  1.00 * 10⁻⁴ * 1

Number of moles of HA =  1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of  A⁻ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into  H⁺  and  A⁻, 84% of HA is left

Number of mols of HA left = 0.84 *  1.00 * 10⁻⁴

Number of mols of HA left =  8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ =  1.6 * 10⁻⁵/V

To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

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