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Given that h = 760mm of Hg for P_atm 

P_atm = ρgh = 13.6 × 10³ × 9.81 × 760/1000 = 101396.16

 N/m² = 101396.16Pa = 101.39KPa                                                              ...(i)                                                       

(a) 90cm Hg gauge 

P_gauge = ρgh = 13.6 × 10³ × 9.81 × 90/100 = 120.07KPa                             ...(ii)

P_abs = P_atm + P_gauge  = 101.39 + 120.07 

Pabs = 221.46KPa .......ANS

 (b) 40cm Hg vacuum

 P_vacc = ρgh = 13.6 × 10³ × 9.81 × 40/100 = 53.366KPa                              ...(iii)

Pabs = Patm – Pvacc 

= 101.39 – 53.366

 Pabs = 48.02KPa .......ANS

 (c)1.2m Water gauge

 P_gauge = ρgh = 1000 × 9.81 × 1.2 = 11.772KPa                                          ...(iv)

P_abs = P_atm + P_gauge

 = 101.39 + 11.772

 P_abs = 113.162KPa .......ANS

The refrigerator must pump the heat up to 20°C to reject it to the ambient. The refrigerator must then have a work input that will exactly offset the increased work output of the power plant, if they are both ideal. As we can not build ideal devices the actual refrigerator will require more work than the power plant will produce extra.

This means that there are temperature differences between the source of energy and the working substance so T_H is smaller than the source temperature. This lowers the maximum possible efficiency. 

As heat is rejected the working substance must have a higher temperature T_L than the ambient receiving the Q_L, which lowers the efficiency further.

For a refrigerator the high temperature must be higher than the ambient to 

which the Q_H is moved. Likewise the low temperature must be lower than the cold space temperature in order to have heat transfer from the cold space to the cycle substance. So the net effect is the cycle temperature difference is larger than the reservoir temperature difference and thus the COP is lower than that estimated from the cold space and ambient temperatures.

T_MAX = 150°C = 423.2 K = T_H ; T_Min = 20°C = 293.2 K = T_L 

 η_{TH MAX} = T_H - T_L/T_H = 130/423.2 = 0.307

Yes. Saturated vapor source at 150°C would remain at 150°C as it condenses to liquid, providing a large energy supply at that temperature. 

F_d = 0.225 A ρV² Units = m² × ( kg/m³ ) × ( m²/ s² )

= kg m / s² = N

Given that PVacc = 260mm of Hg

 P_Vacc = ρgh = 13.6 × 10³ × 9.81 × 260/1000

 = 34.688 × 103 N/m²           .......ANS                                                               ...(i)

 P_atm = ρgh = 13.6 × 10³ × 9.81 × 760/1000 = 101396.16

 N/m² = 101.39 × 10³ N/m²                                                                              ...(ii)

 P_abs = P_atm – P_Vacc

 = 101.39 × 10³ – 34.688 × 10³

 = 66.61 × 10³ N/m²            .......ANS

 Now if   

 P_gauge = 260mm of Hg =

 P_gauge = 260mm of Hg = 13.6 × 10³ × 9.81 × 260/1000 = 34.688 × 10³ N/m² Pabs = Patm +P gauge

 = 101.39 × 10³ + 34.688 × 10³ = 136.07 × 103 N/m2 .......ANS

 ANS: P_vacc = 34.7×10³ N/m²(vacuum), P^_abs = 66.6kPa, 136kpa

Difference is because vacuum pressure is always Negative gauge pressure.

 Or 

vacuum in a gauge pressure below atmospheric pressure and gauge pressure is above atmospheric pressure.

m = 2Kg

T₁ = 125°C

T₂ = 30°C

W = 152KJ

H = 212.8KJ

C_P = ?, C_V = ?, R = ?

We know that during adiabatic process is:

W.D. = P₁V₁ – P₂V₂/γ–1 = mR(T₁ – T₂)/ γ–1

152 × 10³ = 2 × R (125 – 30)/(1.4 – 1)

R = 320J/Kg°K = 0.32 KJ/Kg°K

H = mcp dT

212.8 = 2.C_P.(125 – 30)

C_P = 1.12 KJ/Kg°K

C_P – C_V = R

CV = 0.8 KJ/Kg°K

Given that:

Hydrogen temperature = 300C = 303K

Load lifting = 400Kg

Atmospheric pressure = 13.6 × 10³ × 0.75 × 9.81 = 1.00 × 10⁵ N/m² = 1.00 bar

Atmospheric Temperature = 27⁰C = 300K

The mass that can be lifted due to buoyancy force, 

So the mass of air displaced by balloon(ma ) = Mass of balloon hydrogen gas (mb) + load lifted ...(i)

Since PV = mRT; m_a = P_a V_a /RT_a ; R = 8314/29 = 287 KJ/Kgk For Air; 29 = Mol. wt of air

= 1.00 × 10⁵ × V/ 287 × 300 = 1.162V Kg ...(ii)

Mass of balloon with hydrogen

mb = PV/RT = 1.00 × 10⁵ × V/ (8314/2 × 300) = 0.08V Kg ...(iii)

Putting the values of (ii) and (iii) in equation (i)

1.162V = 0.08V + 400

V = 369.67 m³

But we know that the volume of a balloon (sphere) = 4/3Πr³

322 = 4/3Πr³ 

r = 4.45 m

P_abs = P_atm + P_gauge

P_abs = P_atm + ñ.g.h

= 101 × 10³ + 13.6 × 10³ × 9.81 × 250 × 10^–2

= 434.2 × 10³ N/m² 

= 0.4342 MPa

Battery thermally insulated ⇒ Q = 0

For constant voltage E and current i,

Power = E i = 12.3 × 6 = 73.8 W 

[Units V × A = W

W = ∫power dt = power ∆t

= 73.8 × 4 × 60 × 60

= 1 062 720 J = 1062.7 kJ

The point where all three phases are shown of water is known as triple point of water. 

Triple point of water T = 273.16°K

Let t represent the Celsius temperature then 

t = T – 273.15°C 

Where t is Celsius temperature °C and Kelvin temperature T(°K) 

T°_F = 9/5T°C + 32 = 9/5 × 0.01 + 32 = 32.018°F

T°_R = 9/5T°K = 9/5 × 273.16 = 491.7 R

T°_C = 9/5(T°_K–32)

T_K = ^t°_C + 273.16

T_R = t°_F + 459.67

T_R/T_K = 9/5

Given data:

m = 1.5kg

P₁ = 0.1MPa = 0.1 × 10⁶ Pa = 10⁵ N/m²

P₂ = 0.7MPa = 0.7 × 10⁶ Pa = 7 × 10⁵ N/m²

PV = c or Temp is constant

ρ = 1.16 Kg/m³ 

W.D. by the piston = ?

For PV = C; 

WD = P₁V₁log V₂/V₁ or P₁V₁logP₁/P

ρ = m/V; i.e.; V₁ = m/ρ = 1.5/1.16 = 1.293m³ ...(i)

W_1–2 = P₁V₁logP₁/P₂ = 10⁵ × 1.293 loge (10⁵/7 × 10⁵)

= 10⁵ × 1.293 × (–1.9459)

= –251606.18J = –251.6 KJ (–ive means WD on the system)

– 251.6K

The constant v line has a higher slope than the constant P line also at positive slope. Thus both the constant P and v processes have an increase in T. As T goes up the change in s is smaller.

The constant T (isothermal) process therefore changes s the most. 

Given that

P_atm = 101.325 × 10³ N/m²

Revolution = 10000

Torque = 1.275 × 10⁶ N 

Dia = 0.6m

Distance moved = 0.8m

Work transfer = ?

W.D by stirring device W₁ = 2Π × 10000 × 1.275 J = 80.11 KJ ...(i)

This work is done on the system hence it is –ive.

Work done by the system upon surrounding

W₂ = F.dx = P.A.d×

= 101.32 × Π/4 × (0.6)² × 0.8

= 22.92 KJ ...(ii)

Net work done = W₁ + W₂

= –80.11 + 22.92 = –57.21KJ (–ive sign indicates that work is done on the system)

W_net = 57.21KJ

Given that : 

P =10bar

x = 0.95 

 t_sup = 250°C

From Saturated steam table

t_sat = 179.9

Now, entropy of steam at the entry of the superheater

s₁ = s_f1 + x₁s_fg1

= 2.1386 + 0.95 × 4.4478 = 6.3640 kJ/kg K

entropy of the steam at exit of superheater 

s₂ = s_gf + C_ps ln(T_sup/T_sat)

= 6.5864 + 2.25 ln (250+ 273/179.9 +273)

= 6.9102 kJ/kg K

Change in entropy = s₂ – s₁ = 6.9102 – 6.3640 

= 0.5462 kJ/kg K

Most electric appliances such as TV, VCR, stereo and clocks dissipate power in electrical circuits into internal energy (they get warm) some power goes into light and some power into mechanical energy. The light is absorbed by the room walls, furniture etc. and the mechanical energy is dissipated by friction so all the power eventually ends up as internal energy in the room mass of air and other substances.

These are not heat engines, just the opposite happens, namely electrical power is turned into internal energy and redistributed by heat transfer. These are irreversible processes. 

Energy Eq.: W = Q_H − Q_L = change in energy of air. OK 

2^nd law: Exchange energy with only one reservoir. NOT OK.

This is a violation of the statement of Kelvin-Planck.

Remark: You cannot create and maintain your own energy reservoir.

As the polytropic process is Pvⁿ = C, then n = 0 is a constant pressure process. This can be a pipe flow, a heat exchanger flow (heater or cooler) or a boiler. 

We would expect the lowest possible exit temperature when the maximum amount of work is taken out. This happens in a reversible process so if we assume it is adiabatic this becomes an isentropic process.

Exit: 200 kPa, s = sin = 7.6758 kJ/kg K ⇒ T = 241.9°C

The efficiency from measures the turbine relative to an isentropic turbine, so the efficiency will be 100%. 

No. Since the ocean is at the ambient T (it is the ambient) it is not possible to extract any work from it. You can extract wave energy (wind generated kinetic energy) or run turbines from the tide flow of water (moon generated kinetic energy). However, since the ocean temperature is not uniform there are a few locations where cold and warmer water flows close to each other like at different depths. In that case a heat engine can operate due to the temperature difference. 

Energy Eq.:  W = Q_H − Q_L = change in energy of air. OK 

2nd law:  Exchange energy with only one reservoir. NOT OK.

This is a violation of the statement of Kelvin-Planck.

Remark: You cannot create and maintain your own energy reservoir. 

Yes. There can be heat transfer involved and that has an availability associated with it, which then equals the change of availability of the substance. 

No. It depends on the definition of ideal work. The ideal device does not necessarily have the same exit state as the actual device. An ideal turbine is approximated as a reversible adiabatic device so the ideal work is the isentropic work. The reversible work is between the inlet state and the actual exit state that do not necessarily have the same entropy

That happens when the inlet and exit states (or beginning and end states) have the same entropy. 

Equilibrium is that state of a system in which the state does not undergo any change in itself with passage of time without the aid of any external agent. Equilibrium state of a system can be examined by observing whether the change in state of the system occurs or not. If no change in state of system occurs then the system can be said in equilibrium.

 Let us consider a steel glass full of hot milk kept in open atmosphere. It is quite obvious that the heat from the milk shall be continuously transferred to atmosphere till the temperature of milk, glass and atmosphere are not alike. During the transfer of heat from milk the temperature of milk could be seen to decrease continually. Temperature attains some final value and does not change any more. This is the equilibrium state at which the properties stop showing any change in themselves. 

Generally, ensuring the mechanical, thermal, chemical and electrical equilibriums of the system may ensure thermodynamic equilibrium of a system. 

1. Mechanical Equilibrium: When there is no unbalanced force within the system and nor at its boundaries then the system is said to be in mechanical equilibrium. For a system to be in mechanical equilibrium there should be no pressure gradient within the system i.e., equality of pressure for the entire system.

 2. Chemical Equilibrium: When there is no chemical reaction taking place in the system it is said to be in chemical equilibrium. 

3. Thermal equilibrium: When there is no temperature gradient within the system, the system is said to be in thermal equilibrium.

 4. Electrical Equilibrium: When there is no electrical potential gradient within a system, the system is said to be in electrical equilibrium.

 When all the conditions of mechanical, chemical thermal, electrical equilibrium are satisfied, the system is said to be in thermodynamic equilibrium

Correct option (2) BaCl₂·2 H₂O(s) → BaCl₂(s) + 2 H₂O(g)

Explanation:

We can predict the sign and magnitude of ∆S by noting the relative order of entropy: solids (lowest) < liquids < solutions < gases (highest) and the number of moles of each type. For the reactions given we have NH₃(g) + HCl(g) → NH₄Cl(s 2 mol gas → 1 mol solid; 

∆S < 0 2 H₂(l) + O₂(l) → 2 H₂O(g)

3 mol liquid → 2 mol gas; ∆S > 0

N₂(g) + 3 H₂(g) → 2 NH₃(g) 4 mol gas → 2 mol gas; ∆S < 0

K(s) + O₂(g) → KO₂(s) 1 mol solid + 1 mol gas → 1 mol solid; ∆S < 0

BaCl₂ · 2 H₂O(s) → BaCl₂(s) + 2 H₂O(g)

1 mol solid → 1 mol solid + 1 mol gas; ∆S > 0

The greatest increase in S would be for the reaction

BaCl₂· 2 H₂O(s) → BaCl₂(s) + 2 H₂O(g)

Correct option 1. None of these is

Explanation:

For spontaneous changes, the entropy of the universe increases.

Correct option 1. either, positive

Explanation:

Only the sign of ∆S°_rxn determines how ∆G°_rxn will be effected by changes in temperature. When ∆S°_rxn is positive, ∆G°_rxn will always decrease in value as you raise the temperature. This can be intuited from the Maxwell equation, ∆G = ∆H − T∆S. Note that only ∆S is multiplied by T, not ∆H.

Correct option 2.  All responses are

Explanation:

Entropy (S) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of freedom, disorder or randomness.

S(g) > S(l) > S(s).

A reaction is spontaneous only when ∆G is negative. ∆H is positive for this reaction and ∆S is positive.

∆G = ∆H − T ∆S

= (+) − T (+) 

= (+) − T

∆G will be negative (spontaneous reactions) only at high values of T.

Correct option 6. I, II, III    

Explanation:

Statement I is true; the second law equation states that the change in entropy of the universe is the sum of the change in entropy of the system and the surroundings. Statement II is true; this is a mathematical identity derived from the second law equation. Statement III is true; the third law states that as the temperature of a system approaches absolute zero, the entropy of the system approaches its minimum, which in the case of a perfect crystalline solid, is zero.

Correct option 3. II and III only

Explanation:

Statement I is false; the first law states that the energy of the universe is conserved, in other words a constant value. Statement II and III are true; internal energy in the universe is conserved, and thus energy lost by the system is always gained by the surroundings. The universe is the most obvious example of an isolated system in that energy and matter are conserved in the universe

Correct option 5. IV only

Explanation:

Sublimation results in a significant increase in the volume of the system, allowing it to do work on its surroundings, i.e. the pressure-volume work function is negative. Sublimation is also an endothermic process, making heat positive. Since the process described ”happens” as a given in the problem, the change in free energy must be negative. Change in entropy must be positive since a solid is becoming a gas.

Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (ΔH), the sum of all potential and kinetic energies. ΔH is determined by the system, not the surrounding environment in a reaction. A system that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system.

A system of reactants that absorbs heat from the surroundings in an endothermic reaction has a positive  ΔH , because the enthalpy of the products is higher than the enthalpy of the reactants of the system.

We know that Gibbs Free Energy is used to determine whether a reaction is favored or disfavored. It is given by the equation: 

ΔG = ΔH - TΔS

Where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature.

Entropy change (ΔS) - Change in entropy is the change in degree of randomness or disorder.

A system will tend towards maximum entropy.

If ΔH and ΔS are either both negative, or both positive, the thermodynamic favorability of the reaction can depend on the temperature.

Reactions can be 'driven by enthalpy' (where a very exothermic reaction (negative ΔH) overcomes a decrease in entropy) or 'driven by entropy' where an endothermic reaction occurs because of a highly positive ΔS.

Favorable Conditions - ΔH < 0 & ΔS > 0

Unfavorable Conditions - ΔH > 0 & ΔS < 0

We see hot high pressure steam flowing in at state 1 from the steam drum through a flow control (not shown). The steam leaves at a lower pressure to the condenser (heat exchanger) at state 2. A rotating shaft gives a rate of energy (power) to the electric generator set. 

The electrical power also leaves the C.V. to be used for lights, instruments and to charge the batteries. 

ΔE or E₂ – E₁ is the change from state 1 to state 2 and depends only on states 1 and 2 not upon the process between 1 and 2.

₁Q₂ and ₁W₂ are amounts of energy transferred during the process between 1 and 2 and depend on the process path.

Thermal engineering is a very important associate branch of mechanical, chemical, metallurgical, aerospace, marine, automobile, environmental, textile engineering, energy technology, process engineering of pharmaceutical, refinery, fertilizer, organic and inorganic chemical plants. Wherever there is combustion, heating or cooling, exchange of heat for carrying out chemical reactions, conversion of heat into work for producing mechanical or electrical power; propulsion of rockets, railway engines, ships, etc., application of thermal engineering is required. Thermodynamics is the basic science of thermal engineering.

Normal Temperature and Pressure (N.T.P.): The conditions of temperature and pressure at 0°C (273K) and 760 mm of Hg respectively are called normal temperature and pressure (N.T.P.).

Standard Temperature and Pressure (S.T.P.): The temperature and pressure of any gas, under standard atmospheric conditions are taken as 150C(288K) and 760 mm of Hg respectively. Some countries take 250C(298K) as temperature.

T_r = ( 250 + 460 ) / 549.7 = 1.29 and P_r = 440/708 = 0.621 

 ⇒ Z = 0.9

m = PV/ZRT = 440 × 144 × 25 / (51.38 × 710 × 0.9)

= 48.25 lbm

Ideal gas Z = 1

⇒ m = 43.21 lbm 10% error

For the compressed liquid and the solid phases the specific volume and thus density is nearly constant. These surfaces are very steep nearly constant v and there is then no reason to fill up a table with the same value of v for different P and T.

Given that P = 1bar = 10⁵N/m²

 P_atm = ρgh , for water equivalent

10⁵ = 1000 × 9.81 × h 

h = 10.19m                                                .......ANS 

P_atm = ρgh , for Hg

10⁵ = 13.6 × 10³ × 9.81 × h

 h = 0.749m                                              .......ANS

 ANS: 10.19m, 0.75m

Given that:

Capacity of drum V₁ = 0.24m³

Pressure of steam P₁ = 11bar

Temperature of steam T₁ = 200ºC

At pressure 11bar from super heated steam table

At 10 bar and T = 200ºC; Å =0.2060 m³/kg

At 12 bar and T = 200ºC; Å =0.1693 m³/kg

Using linear interpolation: 

(Å – 0.2060)/(0.1693 – 0.206) = (11 – 10)/(12 – 10)

Å = 0.18765 m³/kg

Quantity of steam = V/Å = 0.24/0.18765 = 1.2789 kg

From Saturated steam table

At 11bar; T_sat = 184.09ºC

2000C > 184.09ºC

i.e. steam is superheated 

If the vessel is cooled until the steam becomes dry saturated, its volume will remain the same but its pressure will change. 

From Saturated steam table; corresponding to Å_g = 0.18765, the pressure is 1122.7KPa

Given that

Mass of steam m = 1kg

Pressure of steam P₁ = 1.5MPa = 15bar

Temperature of steam T₁ = 400ºC

From superheated steam table

At P₁ = 15bar, T₁ = 400ºC

Å₁ = 0.1324 m³/kg

Å₂ = (2 x 0.1324)/3 = 0.0882 m³/kg

Change in volume “Å = Å₁ - Å₂ = 0.1324 – 0.0882 = 0.0441 m³/kg

The steam is wet at 15bar, therefore, the temperature will be 198.32ºC.

 P_gauge = 1.5 × 10⁶ N/m²

 P_vacc = 710 mm of Hg

 P_atm = 76 cm of Hg = 101.3 × 10³ N/m²

 P_inlet = ? 

P_inlet = P_abs = P_gauge(inlet)+ P_atm

 = 1.5 × 10⁶ + 101.3 × 10³

 P_inlet = 1.601 × 10⁶ Pa           .......ANS

 Since discharge is at vacuum i.e.;

 P_exhaust = P_abs = P_atm – P_vacc

 = 101.3 × 10³ – 13.6 × 10³ × 9.81 × 710/1000

 P_exhaust = 6.66 × 10⁶ Pa        .......ANS

 ANS: P_inlet = 1.6×10⁶Pa, P_exhaust = 6.66×10³Pa 

Density of steel in 

ρ = 7820 kg/m³

Volume of steel: V = m/ρ = 2 kg/7820 kg/m3   = 0.000 256 m³

Density of water in ρ = 997 kg/m³

Mass of water: m

 = ρV = 997 kg/m³ ×0.004m³ 

= 3.988 kg

Total mass: m = m_steel + m_water 

= 2 + 3.988 = 5.988 kg

Total volume: V = V_steel + V_water 

= 0.000 256 + 0.004

= 0.004 256 m³ = 4.26 L

Q = hΑ ΔT; Q = .QΔt = hA ΔT Δt

Q = 3 (Btu/h ft² R) × 10 ft² × (95 –70) F × (15/60) h

= 187.5 Btu

(a) The space heater.

Electrical work (power) input, and equal (after system warm up) Q out to the room.

(b) Room

Q input from the heater balances Q loss to the outside, for steady (no temperature change) operation.

(c) The space heater and the room together 

Electrical work input balances Q loss to the outside, for steady operation. 

No.

The statement for a cycle involves an integral of dQ/T so T can vary, which it does during most processes in actual devices. This just means that you cannot that easily get a closed expression for the integral.

From the definition of entropy

ds =  dq/T

for a reversible process. Thus only heat transfer gives a change in s, expansion/compression involving work does not give such a contribution.

Yes

During various parts of the cycle work and heat transfer may be transferred. That happens at different P and T. The net work out equals the net heat transfer in (energy conservation) so dependent upon the sign it is a heat engine or a heat pump (refrigerator). The net effect is thus a conversion of energy from one storage location to another and it may also change nature (some Q got changed to W or the opposite)