1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
snow_lady [41]
3 years ago
6

Superheated steam at 20 bar and 450oC flows at a rate of 200 kg/min to an adiabatic turbine, where it expands to 10 bar. The tur

bine produces 1500 kW of work. The steam then flows to a heat exchanger where it is heated isobarically to its initial temperature. Neglect kinetic energy.
1) Using the energy balance on the turbine, solve for the outlet temperature (in oC).

2) Using the energy balance on the heater, solve for the required heat input (in kW) to the steam.
Chemistry
1 answer:
sergejj [24]3 years ago
8 0

Explanation:

The given data is as follows.

Pressure of steam at inlet of turbine, P_{1} = 20 bar

Temperature at inlet of turbine, T = 450^{o}C

Pressure at outlet of turbine, P_{2} = 10 bar  

Mass flow rate of steam, m = 200 kg/min

Work produced by the turbine, W_{s} = 1500 kW

Steam is heated at constant pressure to its initial temperature, i.e., temperature at outlet of heat exchanger, T_{3} = 450^{o}C.

(1)   For an adiabatic turbine, the energy balance is  as follows.

           -W_{s} = m({H}_{2} - {H}_{1})

where W_{s} = work done by the turbine

                      m = mass flow rate of steam

H_{1} and H_{2} are the specific enthalpy of steam at inlet and outlet conditions of turbine.

Obtain the specific enthalpy of steam from Properties of Superheated Steam table

At 20 bar and 450^{o}C, H_{1} = 3358 kJ/kg

           H_{2} = H_{1} - \frac{W_{s}}{m}

      H_{2} = 3358 kJ/kg - \frac{1500kJ/s \times 60 s/min}{200 kg/min}

           H_{2} = 2908 kJ/kg

For P = 10 bar, H =2875 kJ/kg for T= 200^{0}C and H = 2975 kJ/kg for T=250^{o}C. Interpolate the values.

The temperature corresponding to P = 10 bar and H_{2} = 2908 kJ/kg is T = 216.5^{o}C

Therefore, the outlet temperature is T_{2} = 216.5^{o}C.

(2)     Energy balance on the heater is  as follows.

         Q = \Delta H = m(H_{3} - H_{2})

where,        Q = heat input required by the steam

                \Delta H = specific enthalpy change

           H_{3} = specific enthalpy of steam at the outlet conditions of heat exchanger

At P = 10 bar and T_{3} = 450^{o}C,  H_{3} = 3371 kJ/kg.

          Q = \frac{200 kg/min}{60 s/min} \times (3371 - 2908)kJ/kg

[/tex]

          Q = 1543.33 kJ/s

or,    Q = 1543.33 kW

Therefore, the heat input required is Q = 1543.33 kW.

You might be interested in
I need help.... How does the kinetic energy of molecules explain both Boyle’s and Charles’ Laws?
Aleks [24]

Answer:

Kinetic Molecular Theory states that gas particles are in constant motion and exhibit perfectly elastic collisions. Kinetic Molecular Theory can be used to explain both Charles' and Boyle's Laws. The average kinetic energy of a collection of gas particles is directly proportional to absolute temperature only. Hope this helps!!

Explanation:

4 0
2 years ago
Read 2 more answers
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!
e-lub [12.9K]
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol


Find the molar mass
2Na = 23 * 2 = 46 grams
1S   = 32 * 1 = 32 grams
O4   = 16 * 4 = 64 grams
Total =            142 grams / mol

Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???

given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams. 

85.2 are in a 2 L solution that has a concentration of 0.6 mol/L


4 0
3 years ago
Read 2 more answers
What force is the type of intermolecular force that would be present between molecules of Cl2 ?
KatRina [158]

Cl2 is nonpolar so it has to be only London dispersion force (LDF)

7 0
3 years ago
Read 2 more answers
Eeeeeeeeeeeeeeeeeeeeeeeeeee
Varvara68 [4.7K]

Answer:

dolo HAHAHAAHAHHAHAHAHA

6 0
2 years ago
Other questions:
  • Why there is not reaction in nac2h3o2(aq)+pb(no3)2(aq)→?
    5·1 answer
  • 1.The diagram below shows the stages of a land feature being formed over
    7·1 answer
  • Consider the reaction:C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l); DH = -1.37 x 103 kJConsider the following propositions:I. The reacti
    11·1 answer
  • The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
    6·2 answers
  • PLS HELPPP!!!!
    13·1 answer
  • What is geography and importance of geography<br><br>​
    15·1 answer
  • C+ O2 --&gt; CO2 +60kJ <br> is the reaction exothermic or<br> endothermic?
    5·1 answer
  • 46.0 g of barium nitrate is dissolved in 360 g of water. What is the percent concentration of the barium nitrate solute in the s
    12·1 answer
  • HELP PLEASE!!!!<br> Can someone answer it as a paragraph please I should submit it today
    8·1 answer
  • How did humans create cow breeds that survive well in cold weather?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!