# Air is entering a 4200-kW turbine that is operating at its steady state. The mass flow rate is 20 kg/s at 807 C, 5 bar and a velocity of 100 m/s. This air then expands adiabatically, through the turbine and exits at a velocity of 125 m/s. Afterwards the air then enters a diffuser where it decelerates isentropically to a velocity of 15 m/s and a pressure of 1 bar. Using the ideal gas model, determine, (a) pressure and temperature of the air at the turbine exit, in units of bar and Kelvin. (b) Entropy production rate in the turbine in units of kW/k, and (c) draw the process on a T-s Diagram.

2 months ago

## Solution 1

Guest #9791770
2 months ago

a)$$T_2=868.24 K$$ ,$$P_2=2.32 bar$$

b) $$s_2-s_1=0.0206$$KW/K

Explanation:

P=4200  KW ,mass flow rate=20 kg/s.

Inlet of turbine

$$T_1$$=807°C,$$P_1=5 bar,V_1=100 m/s$$

Exits of turbine

$$V_2=125 m/s$$

Inlet of diffuser

$$P_3=1 bar,V_3=15 m/s$$

Given that ,use air as ideal gas

R=0.287 KJ/kg-k,$$C_p$$=1.005 KJ/kg-k

Now from first law of thermodynamics for open system at steady state

$$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$$

Here given that turbine is adiabatic so Q=0

Air treat ideal gas   PV=mRT, Δh=$$C_p(T_2-T_1)$$

$$w=\dfrac{P}{mass \ flow\ rate}$$

$$w=\dfrac{4200}{20}$$

w=210 KJ/kg

Now putting the values

$$1.005\times (273+807)+\dfrac{100^2}{2000}=1.005T_2+\dfrac{125^2}{2000}+210$$

$$T_2=868.24 K$$

Now to find pressure

We know that for adiabatic $$PV^\gamma =C$$ and for ideal gas Pv=mRT

⇒$$\left (\dfrac{T_2}{T_1}\right )^\gamma=\left (\dfrac{P_2}{P_1}\right )^{\gamma -1}$$

$$\left(\dfrac{868.24}{1080}\right )^{1.4}=\left (\dfrac{P_2}{5}\right )^{1.4-1}$$

$$P_2=2.32 bar$$

For entropy generation

$$s_2-s_1=1.005\ln\dfrac{868.24}{1080}-0.287\ln\dfrac{2.32}{5}$$

$$s_2-s_1=0.00103$$KJ/kg_k

$$s_2-s_1=0.00103\times 20$$KW/K

$$s_2-s_1=0.0206$$ KW/K

## 📚 Related Questions

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A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.
Solution 1

w =  -28.8 kJ/kg

q = 723.13 kJ/kg

Explanation:

Given :

Initial properties of piston  cylinder assemblies

Temperature, $$T_{1}$$ = -20°C

Quality, x = 70%

= 0.7

Final properties of piston  cylinder assemblies

Temperature, $$T_{2}$$ = 180°C

Pressure, $$P_{2}$$ = 6 bar

From saturated ammonia tables at $$T_{1}$$ = -20°C  we get

$$P_{1}$$ = $$P_{sat}$$ = 1.9019 bar

$$v_{f}$$ = 0.001504 $$m^{3}$$ / kg

$$v_{g}$$ = 0.62334 $$m^{3}$$ / kg

$$u_{f}$$ = 88.76 kJ/kg

$$u_{g}$$ = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

$$v_{1}$$ = $$v_{f}$$+x ($$v_{g}$$-$$v_{f}$$

= 0.001504+0.7(0.62334-0.001504)

= 0.43678 $$m^{3}$$ / kg

$$u_{1}$$ = $$u_{f}$$+x ($$u_{g}$$-$$u_{f}$$

= 88.76+0.7(1299.5-88.76)

=936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

$$v_{2}$$ = 0.3639 $$m^{3}$$ / kg

$$u_{2}$$ = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

$$w = \left (\frac{P_{1}+P_{2}}{2} \right )\left ( v_{2}-v_{1} \right )$$

$$w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )$$

$$w = -28794.52$$ J/kg

= -28.8 kJ/kg

Now for heat transfer

$$q = (u_{2}-u_{1})+w$$

$$q = (1688.2-936.27)-28.8$$

= 723.13 kJ/kg

Question
A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the piston until the air reaches to 0.1 m3 and 1,000 °C, in which the air undergoes a polytropic process (PV" const). Assume that heat loss from the cylinder, friction of piston, kinetic and potential effects are negligible. 1) Determine the polytropic constant n. 2) Determine the work transfer in ki for this process, and diseuss its direction. 3) sketch the process in T-V (temperature-volume) diagram.
Solution 1

n=2.32

w= -213.9 KW

Explanation:

$$V_1=0.3m^3,T_1=298 K$$

$$V_2=0.1m^3,T_1=1273 K$$

Mass of air=1 kg

For polytropic process  $$pv^n=C$$ ,n is the polytropic constant.

$$Tv^{n-1}=C$$

$$T_1v^{n-1}_1=T_2v^{n-1}_2$$

$$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$$

n=2.32

Work in polytropic process given as

w=$$\dfrac{P_1V_1-P_2V_2}{n-1}$$

w=$$mR\dfrac{T_1-T_2}{n-1}$$

Now by putting the values

w=$$1\times 0.287\dfrac{289-1273}{2.32-1}$$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

Question
A flat rectangular door in a mine is submerged froa one side in vater. The door dimensions are 2 n high, 1 n vide and the vater level is 1,5 m higher than the top of the door. The door has two hinges on the vertical edge, 160 mm from each corner and a sliding bolt on the other side in the niddle. Calculate the forces on the hinges and sliding bolt. Hint: Consider the door from a side view and from a plaa vies respectively and take moments about a point each time.)
Solution 1

Force on the bolt = 24.525 kN

Force on the 1st hinge = 8.35 kN

Force on the 2nd hinge = 16.17 kN

Explanation:

Given:

height = 2 m

width =1 m

depth of the door from the water surface = 1.5 m

Therefore,

$$\bar{y}$$ =1.5+1 = 2.5 m

Hydrostatic force acting on the door is

$$F= \rho \times g\times \bar{y}\times A$$

$$F= 1000 \times 9.81\times 2.5\times 2\times 1$$

= 49050 N

= 49.05 kN

Now finding the Moment of inertia of the door about x axis

$$I_{xx}=\frac{1}{12}\times b\times h^{3}$$

$$I_{xx}=\frac{1}{12}\times1\times 2^{3}$$

= 0.67

Now location of force, $$y^{*}$$

$$y^{*}=\bar{y}+\frac{I_{xx}}{A\times \bar{y}}$$

$$y^{*}=2.5+\frac{0.67}{2\times 1\times 2.5}$$

= 2.634

Therefore, calculating the unknown forces

$$F=F_{A}+R_{B}+R_{C} = 49.05$$  ------------------(1)

Now since $$\sum M_{R_{A}}=0$$

∴ $$R_{B}\times L+R_{C}\times L-F\times \frac{1}{2}=0$$

$$R_{B}+R_{C}-F\times \frac{1}{2}=0$$

$$R_{B}+R_{C}=\frac{F}{2}$$

$$R_{B}+R_{C}=24.525$$        -----------------------(2)

From (1) and (2), we get

$$R_{A} = 49.05-24.525$$

= 24.525 kN

This is the force on the Sliding bolt

Taking $$\sum M_{R_{C}}=0$$

$$F\times 0.706-R_{A}\times 0.84-R_{B}\times 1.68 = 0$$

$$49.05\times 0.706-24.525\times 0.84-R_{B}\times 1.68 = 0$$

$$R_{B}$$ =8.35 kN

This is the reaction force on the 1st hinge.

Now from (1), we get

$$R_{C}$$ =16.17 kN

This is the force on the 2nd hinge.

Question
A roller support acts like a contact boundary condition as it can produce a reaction force as a push response to a body but will not produce a pull force to hold a body from moving away. a)True b)- False
Solution 1

a) True

Explanation:

Roller can provide reaction for push support but it can not provide reaction for pull support.

From the free body diagram of roller and hinge support we can easily find that ,Roller providing vertical reaction and can not provide horizontal reaction.

On the other hand hinge support can provide reaction in both the direction.

So we can say that roller can not proved reaction for pull support.

Question
A pendulum has an oscillation frequency (T) which is assumed to depend upon its length (L), load mass (m) and the acceleration of gravity (g). Determine the relationship between oscillation frequency, length, load mass and acceleration of gravity. Differentiate as well which variable does not affect the oscillation frequency.
Solution 1

Mass does not affect oscillation frequency.

Explanation:

Let the bob of the pendulum makes a small angular displacement θ. When the pendulum is displaced from the equilibrium position, a restoring force tries to act upon it and it tries to bring the pendulum back to its equilibrium position. Let this restoring force be F.

Therefore, F = -mgsinθ

Now for pendulum, for small angle of θ,

sinθ$$\simeq$$θ

Therefore, F = -mgθ

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F = m.a = -mgθ

$$\Rightarrow m.\frac{d^{2}x}{dt^{2}} = - mg\Theta$$

Now since, x = θ.L

$$\Rightarrow L.\frac{d^{2}\Theta }{dt^{2}}= -g\Theta$$

$$\Rightarrow \frac{d^{2}\Theta }{dt^{2}}= -\frac{g}{L}.\Theta$$

$$\Rightarrow \frac{d^{2}\Theta }{dt^{2}}+\frac{g}{L}.\Theta =0$$

Therefore, angular frequency

$$\omega ^{2}$$ = $$\frac{g}{L}$$

ω = $$\sqrt{\frac{g}{L}}$$

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Solution 1

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$$\frac{C(s)}{R(s)}$$=$$\frac{G(s)}{1+G(s)H(s)}$$       (1)

Explanation:

with reference to fig1:

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$$\frac{5}{s(s+1)}$$

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(b)     Irreversible cycle.

Explanation:

Given;

$$T_2=540R ,Q_2= 500 Btu/s ,T_1=240 R ,Q_1= 200 Btu/s$$

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If it is greater than zero ,then cycle will be impossible .

Now find

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Question
Describe the operational principle of a unitary type air conditioning equipment with a suitable sketch.
Solution 1

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coolant, a heat exchanger (outdoor) for heat exchange, an expander attached to the heat exchanger for expansion of coolant and a duct.

It continuously removes heat and moisture from inside an occupied space and cools it with the help of heat exchanger and condensor in the condensing unit and discharges back into the same occupied indoor space that is supposed to be cooled.

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refer to fig 1

Question
What is a truss? What separates a truss from a frame and other forms of rigid bodies?
Solution 1

The application of force is the main difference between truss and frame.

Explanation:

Truss:

Truss is a collection of beams,which use to handle the tensile and   compression loads . That collection of beams creates rigid structure.

The load on the truss will be acting always at the  at the hinge. Truss is     widely used in the construction areas.

Frame:

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Truss and frame both forms a rigid structure and is used in the construction areas.

Question
Water flows at the rate of 200 I/s upwards through a tapered vertical pipe. The diameter at Marks(3) CLO5) the bottom is 240 mm and at the top 200 mm and the length is 5m. The pressure at the bottom is 8 bar, and the pressure at the topside is 7.3 bar. Determine the head loss through the pipe. Express it as a function of exit velocity head.
Solution 1

Explanation:

By using energy equation:

$$\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g} =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}$$

$$\gamma=specific weight$$

$$v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s$$

$$h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}$$

$$h_{L}=\frac{(8-7.3)\times 100 }{9.81} +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}$$

$$h_L=7.135+3.927\\h_L=11.062m$$

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head loss as a function of exit velocity head is=$$\frac{11.062}{2.068}$$

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