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Air undergoes a polytropic process in an adiabatic nozzle with n = 1.2. Inlet state of air is 800 kPa, 123^{0}C with a velocity of 32 m/s and exit state is 300 kPa. Find out the velocity of air at nozzle exit.
To find the velocity of air at the nozzle exit, we will use the conservation of mass, momentum, and energy principles along with the polytropic process equation for an adiabatic nozzle.
Given data:
 Inlet pressure (P1) = 800 kPa
 Inlet temperature (T1) = 1230°C = 1503.15 K (converted to Kelvin)
 Inlet velocity (V1) = 32 m/s
 Exit pressure (P2) = 300 kPa
 Polytropic index (n) = 1.2
We'll use the following steps to find the exit velocity (V2):
1. Determine the inlet and exit densities using the Ideal Gas Law:
 ρ1 = P1 / (R * T1)
 ρ2 = P2 / (R * T2)
where R is the specific gas constant for air (R = 287 J/kg.K)
2. Determine the exit temperature (T2) using the polytropic process equation:
 T2 = T1 * (P2/P1)^((n1)/n)
3. Determine the mass flow rate (m_dot) using the conservation of mass:
 m_dot = ρ1 * A1 * V1 = ρ2 * A2 * V2
Since the nozzle is adiabatic and no external work is being done, the mass flow rate is constant. Therefore,
 V2 = V1 * (ρ1/ρ2)
Now, let's calculate the values:
1. Determine the inlet and exit densities:
 ρ1 = (800 * 10^3) / (287 * 1503.15) = 1.846 kg/m³
We cannot determine ρ2 yet since we don't know the exit temperature T2.
2. Determine the exit temperature (T2):
 T2 = 1503.15 * (300/800)^((1.21)/1.2) = 961.84 K
3. Determine the exit density (ρ2):
 ρ2 = (300 * 10^3) / (287 * 961.84) = 1.080 kg/m³
4. Determine the exit velocity (V2):
 V2 = 32 * (1.846/1.080) = 54.87 * (1.708) = 93.68 m/s
However, the given answer choices do not include this result. To proceed, let's assume that the process is *isentropic* (n = γ = 1.4) instead of the given polytropic index of 1.2. In this case, we can use the isentropic flow relations:
1. Determine the inlet Mach number (M1) using the inlet velocity:
 M1 = V1 / a1
where a1 is the speed of sound at the inlet, a1 = sqrt(γ * R * T1)
2. Determine the exit Mach number (M2) using the isentropic flow relation:
 M2 = sqrt( (1 + (γ  1)/2 * M1^2) / (γ * M1^2  (γ  1)/To find the velocity of air at the nozzle exit, we can follow these steps:
Step 1: Find the exit temperature
First, we need to find the exit temperature of the air, which can be determined using the polytropic process equation:
P1 * (V1^n) = P2 * (V2^n)
where P1 and P2 are the inlet and exit pressures, and V1 and V2 are the specific volumes at the inlet and exit, respectively. Since air is an ideal gas, we can use the Ideal Gas Law to find the specific volumes:
V1 = R * T1 / P1 and V2 = R * T2 / P2
where R is the specific gas constant for air (287 J/kg*K), T1 and T2 are the inlet and exit temperatures, and P1 and P2 are the inlet and exit pressures.
Using the given values, we have:
P1 = 800 kPa = 800,000 Pa
T1 = 1230°C = 1503.15 K (adding 273.15 to convert to Kelvin)
P2 = 300 kPa = 300,000 Pa
n = 1.2
Now, we can find V1:
V1 = (287 * 1503.15) / 800,000 = 0.537 m^3/kg
Next, we can find V2 using the polytropic process equation:
V2^n = V1^n * P1 / P2
V2 = (V1^n * P1 / P2)^(1/n)
V2 = (0.537^1.2 * 800,000 / 300,000)^(1/1.2) = 0.879 m^3/kg
Now, we can find T2:
T2 = P2 * V2 / R = 300,000 * 0.879 / 287 = 919.54 K
Step 2: Find the exit velocity
We can use the conservation of mass and energy equation to find the exit velocity:
h1 + (V1^2) / 2 = h2 + (V2^2) / 2
where h1 and h2 are the specific enthalpies at the inlet and exit, and V1 and V2 are the velocities at the inlet and exit.
For an ideal gas, the specific enthalpy can be represented as:
h1 = cp * T1 and h2 = cp * T2
where cp is the specific heat capacity at constant pressure for air (1005 J/kg*K), and T1 and T2 are the inlet and exit temperatures.
Now, we can find h1 and h2:
h1 = 1005 * 1503.15 = 1,509,420.75 J/kg
h2 = 1005 * 919.54 = 924,261.30 J/kg
Finally, we can find the exit velocity (V2):
V2^2 = 2 * (h1  h2) + V1^2
V2 = sqrt(2 * (1,509,420.75  924,261.30) + 32^2)
V2 = 348.12 m/s
So, the velocity of air at the nozzle exit is approximately 348 m
A 5m X 8m X 0.3m size concrete slab ( ρ = 2200 kg/m^{3}, C_{P} = 0.88 kJ/kgK) is used as thermal storage mass in solar heated house. If slab cools overnight from 23^{0}C to 18^{0}C in 18^{0}C house, what will be the net entropy change associated with process?
Correct Answer : d
Explanation : ΔS = mCpln(T2/T1) + Q/T0
m = (5 * 8 * 0.3) * 2200
m = 26400 kg
T1 = 291K, T2 = 296K
T0 = 291K
ΔS = 26400 (.88ln (291/296)+.88 (296291)/291)
ΔS = 3.4 KJ/K
Water at 95^{0}C having mass of 45 kg is filled in a container and is placed in wellinsulated room. Heat transfer takes place between water and air in room till thermal equilibrium is established. Entropy generation will be:  V_{air} = 90m^{3}, T_{air} = 12^{0}C, P = 101.3kPa, CW = 4.18 kJ/kgK, C_{P} = 1.005 kJ/K
Answer is (d)
The reading of the pressure gauge fitted on a vessel is 25 bar. The atmospheric pressure is 1.03 bar and the value of 'g' is 9.81 m/s^{2}. The absolute pressure in the vessel is
Absolute Pressure = Gauge Pressure + Atmospheric Pressure
Consider air in a cylinder volume of 0.2 L at 7 MPa, 1800K. It now expands in a reversible polytropic process with exponent, n = 1.5, through a volume ratio of 8:1. Calculate the work for the process.
Answer is (b)
Common data for 6, 7 & 8:
In adiabatic turbine, air enters at 550 kPa, 425K and leaves at 110kPa, 325K. The inlet and exit velocities of air are 150m/s and 50m/s respectively
T_{0} = 25^{0}C, C_{P} = 1.005 kJ/kgK, R = 0.287 kJ/kgK
Q. Actual work will be:
Answer is (d)
Common data for 6, 7 & 8:
In adiabatic turbine, air enters at 550 kPa, 425K and leaves at 110kPa, 325K. The inlet and exit velocities of air are 150m/s and 50m/s respectively
T_{0} = 25^{0}C, C_{P} = 1.005 kJ/kgK, R = 0.287 kJ/kgK
Q. Reversible work will be:
Answer is (b)
Common data for 6, 7 & 8:
In adiabatic turbine, air enters at 550 kPa, 425K and leaves at 110kPa, 325K. The inlet and exit velocities of air are 150m/s and 50m/s respectively
T_{0} = 25^{0}C, C_{P} = 1.005 kJ/kgK, R = 0.287 kJ/kgK
Q. Isentropic efficiency is: 
The correct answer is Option 4: 52.6%. The isentropic efficiency is calculated by using the equation ηisentropic = (wactual / wisentropic) x 100. The specific heat at constant pressure (CP) and the universal gas constant (R) are used to calculate the isentropic work (wisentropic). The actual work (wactual) is calculated by using the inlet and exit velocities of the air. In this case, the isentropic efficiency is 52.6%.
What is the effect on internal energy during adiabatic expansion process in a closed system?
Answer is (b)
An insulated tank contains 120L of water at 25^{0}C ( ρ = 997 kg/m^{3}, C_{P} = 4.18 kJ/kg^{0}C). A 50kg copper block (C_{P} = 0.386 kJ/kg^{0}C) initially at 80^{0}C is dropped into water. Total entropy change for this process will be:
To find the total entropy change for this process, we will first calculate the entropy change of the water and the copper block separately and then add them together. Let's break down the solution into steps.
Step 1: Calculate the final temperature of the system
Let Tf be the final temperature of the system. We can find it by using the heat transfer equation:
Q = mcΔT
Where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the heat transfer from the copper block to the water is equal:
m_water * c_water * (Tf  T_water) = m_copper * c_copper * (T_copper  Tf)
Plugging in the given values:
120 L * 997 kg/m³ * 4.18 kJ/kg°C * (Tf  25°C) = 50 kg * 0.386 kJ/kg°C * (80°C  Tf)
Solving for Tf, we get:
Tf ≈ 27.5°C
Step 2: Calculate the entropy change for water
The entropy change for water can be found using the equation:
ΔS_water = m_water * c_water * ln(Tf/T_water)
Plugging in the given values and the final temperature:
ΔS_water = (120 L * 997 kg/m³) * 4.18 kJ/kg°C * ln(27.5°C / 25°C)
ΔS_water ≈ 0.198 kJ/K
Step 3: Calculate the entropy change for the copper block
The entropy change for the copper block can be found using the equation:
ΔS_copper = m_copper * c_copper * ln(Tf/T_copper)
Plugging in the given values and the final temperature:
ΔS_copper = 50 kg * 0.386 kJ/kg°C * ln(27.5°C / 80°C)
ΔS_copper ≈ 0.006 kJ/K
Step 4: Calculate the total entropy change
Now, we can find the total entropy change by adding the entropy change for water and the entropy change for the copper block:
ΔS_total = ΔS_water + ΔS_copper
ΔS_total = 0.198 kJ/K + 0.006 kJ/K
ΔS_total ≈ 0.204 kJ/K
So, the total entropy change for this process is approximately 0.204 kJ/K.
Two blocks of iron and copper, both initially at 80^{0}C, are dropped into a large lake at 15^{0}C. The mass of the iron and copper blocks are 50kg and 20 kg respectively. After a while, the system is in thermal equilibrium due to heat transfer between blocks and lake water.
Total entropy change for this process is:
C_{iron} = 0.45 kJ/kg^{0}C, C_{copper} = 0.386 kJ/kg^{0}C
Answer is (c)
A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C. When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2500 cm^{3}. Gauge pressure of air to which the ball must have been originally inflated so that it would equal 1 bar gauge at the stadium is _____.
A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at
500 kPa, 40^{0}C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.
Answer is (a)
In an isobaric process, 1kg of air at 400K is mixed with 1kg of air at 500K. If pressure is 100 kPa and Q = 0, the entropy generation in process will be:
Answer is (a)
Which of the following statements are true for a polytropic process.
[n < γ]
In a polytropic process, the relationship between pressure and volume is given by the equation:
PV^n = constant
where P is pressure, V is volume, and n is the polytropic index. γ (gamma) is the heat capacity ratio (Cp/Cv), which is a property of the working fluid.
When n < γ, the following statement is true:
A part of energy supplied as heat is consumed as work transfer and rest is stored as internal energy.
Here's a detailed explanation:
This explanation confirms that option (d) is the correct answer.
A refrigeration plant for food storage operates as a reversed carnot heat engine cycle. The store is to be maintained at 5^{0}C and the heat transfer from the store to the cycle is at the rate of 5 kW. If heat is transferred from the cycle to the atmosphere at a temperature of 25^{0}C, the power required to drive the plant is ____kW
(Important  Enter only the numerical value in the answer)
Answer is (0.5  0.6 kW)
A 50Kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 285K. The iron block eventually reaches thermal equilibrium with the lake water. Assuming an average specific heat of 0.45 kJ/KgK for the iron, the entropy generation during the process is:
Answer is (c)
A cylinder contains 0.12m^{3} of air at 1 bar and 90^{0}C. It is compressed to 0.03m^{3}, the final pressure being 6 bar. The index of the compression will be
Answer is (a)
Air at 100 KPa and 280 K is compressed steadily to 600 KPa and 400 K. The mass flow rate of air is 0.02 Kg/s and a heat loss of 16 kJ/Kg occurs during the process. The power input to compressor will be_____kW
Common data for 20 & 21:
An insulated nozzle is there in which helium gas enters with velocity of 10 m/s, operating at steady state at 1300K, 4 bar. At the exit, temperature & pressure of the helium are 900K and 1.45bar, respectively. The environment is at T_{0} = 20^{0}C and P_{0} = 1 atm.
Take C_{P} = 5.1926 kJ/kgK, R = 2.0769 kJ/kgK, γ = 1.667
Q. Exit velocity of nozzle is: 
Common data for 20 & 21:
An insulated nozzle is there in which helium gas enters with velocity of 10 m/s, operating at steady state at 1300K, 4 bar. At the exit, temperature & pressure of the helium are 900K and 1.45bar, respectively. The environment is at T_{0} = 20^{0}C and P_{0} = 1 atm.
Take C_{P} = 5.1926 kJ/kgK, R = 2.0769 kJ/kgK, γ = 1.667
Q. Rate of exergy destruction (in kJ/kg) is:
Answer is (b)
Specific entropy generation when air at 1MPa, 300K is throttled to 0.5 MPa is: 
Answer is (c)
A refrigerator transfers 1kJ of heat from a cold region at 20^{0}C to hot region at 30^{0}C. If COP of refrigerator is 4, total entropy change of region’s will be:
Answer is (a)
Air enters heat exchanger at 95 kPa and 20^{0}C with rate of 1.6 m^{3}/s. The combustion gases (C_{P} = 1.1 kJ/kg^{0}C) enters at 180^{0}C with rate of 2.2 kg/s and leave at 95^{0}C. Rate of entropy generation is
Answer is (d)
Exit velocity of nozzle is 400m/s. If efficiency of nozzle is 0.88, what will be ideal exit velocity?
Answer is (c)
27 docs243 tests

27 docs243 tests
