Applied Thermodynamic solved problem manual PDF

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Thermodynamics Assignment Numericals QNO1: In a non-flow process there is a heat transfer loss of 1055kJ and an internal energy increase of 210kJ. Determine the work transfer state whether the process is an expansion or a compression. ( Basic Engineering Thermodynamics Author: Rayner Joel; Fifth Edition; Numerical no 1; Page No: 47.) Given Data: Q = -1055 ( Heat is rejected by system so negative) E = 210KJ Required: Work done = W = ? State =? Sol: As we know from first law of thermodynamics Q=E+W W=Q–E W = -1055-210 W = -1265Kj As work is negative, so work is done on the system and hence compression occure. QNO2: Air enters a gas turbine system with a velocity of 105m/s and has a specific volume of 0.8m3/kg. The inlet area of the gas turbine system is 0.05m2. At exit the air has a velocity of 135m/s and has a specific volume of 1.5m3/kg. In its passage through the turbine system, the specific enthalpy of air is reduced by 145kJ/kg and the air also has a heat transfer loss of 27Kj/kg. Determine (a) The mass flow rate of the air through the turbine system in kg/s (b) The exit area of the turbine system in m2 (c) The power developed by the turbine system in Kw (Basic Engineering Thermodynamics Rayner Joel; Fifth Edition; Numerical no 3; Page No: 47.) 2017-ME-405

Given Data: C1 = 105m/s Specific volume at inlet = V1 = 0.8m3/kg A1 = 0.05m2 C2 = 135m/s Specific volume at exit = V2 = 1.5m3/kg ΔH = H1-H2 = 145 kJ/kg 𝑄󰇗 = -27kJ/kg Required: (a) Mass flow rate = m’ = ? (b) A2 = ? (c) Power developed = W = ? Sol: (a) As we know mass flow rate is as follow 𝑚󰇗 = 𝑚󰇗 =

𝐶𝐴 𝑉 (105)(0.05) 0.8

𝑚󰇗 = 6.56 kg/s (b) From the formula we know 𝐶1 𝐴1 m1 = m2 =

𝑉1 𝐶2 𝐴2 𝑉2

By comparing these two equation, we have 𝐶1 𝐴1 𝐶2 𝐴2 = 𝑉1 𝑉2 (1.5)(105)(0.05) (135)(0.8)

= A2

A2 = 0.073m2 (c) From steady flow equation, we have H1 + 1/2mc12 + mgz1 + Q = W + 1/2mc22 + mgz2 + H2 By taking common mass we have, h1 + 1/2c12 + gz1 + 𝑄󰇗 = h2 + 1/2c22 + gz2 + 𝑊󰇗 1 (h1-h2) + (c12 – c22) + 𝑄󰇗 = 𝑊󰇗 2

1 𝑊󰇗 = ΔH + (c12 – c22) + 𝑄󰇗 2

1 𝑊󰇗 = 145 + 2×1000 [ (105)2 – (135)2 ] – 27 𝑊󰇗 = 747.46kW P = 𝑊󰇗 = 747.46kW

2017-ME-405

C = Velocity 𝑚󰇗 = mass flow rate v = specific volume h= specific enthalpy z = height

QNO3: A nozzle is a device for increasing the velocity of a steady flow stream. At the inlet of certain nozzle, the enthalpy of the fluid passing is 3000kJ/kg and the velocity is 60m/s. At the discharge end, the enthalpy is 2762kJ/kg. The nozzle is horizontal and there is negligible heat loss from it. (a) Find the velocity at exit from the nozzle. (b) If the inlet area is o.1m2 and the specific volume at inlet is 0.187m3/kg, find the mass flow rate. (c) If the specific volume at the nozzle exit is 0.498m3/kg, find the exit area of the nozzle. Given data: h1 = 3000kJ/kg c1 = 60m/s h2 = 2762kJ/kg Required: (a) c2 = ? (b) mass flow rate = m’ = ? (c) A2 = ? Sol: (a) From steady flow equation, we have H1 + 1/2mc12 + mgz1 + Q = W + 1/2mc22 + mgz2 + H2 By taking common mass we have, h1 + 1/2c12 + gz1 + 𝑄󰇗 = h2 + 1/2c22 + gz2 + 𝑊󰇗 As nozzle is horizontal and heat is negligible, so height “z” and heat “Q” is zero. And as we know that in thermodynamics, work is only done if turbine rotate and piston move. So there is work done also zero because no piston move and no turbine rotate. So, now steady flow equation becomes, 1

1

(h1-h2) + 2c12 = c22 2 1

c22 = 2[ (h1-h2) + c22 ] 2

1

c22 = 2[ (3000- 2762) + (60)2 ] 2

By solving, we have c2 = 692.5m/s (b) inlet area = 0.1m2 specific volume at inlet = 0.187m3/kg From the formula of flow rate 𝐶1𝐴1 𝑚󰇗 = 𝑚󰇗 =

𝑉1 (60)(0.1) 0.187

𝑚󰇗 = 32.08kg/s 2017-ME-405

(c) specific volume at exit = 0.498m3/kg velocity at exit = c2 = 692.5m/s From the formula of flow rate, we have 𝐶2𝐴2 𝑚󰇗 = A2 =

𝑉2 (32.08)(0.498) 692.5

A2 = 0.023m

2017-ME-405...