Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.


Answer 1


The power input, in kW is -86.396 kW



initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

v_1 = (R'T_1)/(p_1) = ((0.28699.(kJ)/(kg.K) )(300 k))/((1.05 bar *\ (10^5 N/m^2)/(1 bar) *(1kJ)/(1000N.m) )) \n\nv_1 = 0.81997 \ m^3/kg

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh


Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

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The pressure in an automobile tire depends on thetemperature of the air in the tire. When the air temperature is25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire whenthe air temperature in the tire rises to 50°C. Also, determinethe amount of air that must be bled off to restore pressure toits original value at this temperature. Assume the atmosphericpressure to be 100 kPa.



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The amount of air that must be bled off to restore pressure 0.007 Kg



T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025 m^(3)

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

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After the temperature rise

P2 = (m2 * R * T2)/(V2) = (0.091 *287*323 )/(0.025) = 337.43 KPa

after bleeding the pressure and the volume returns  to its first value

P1 = P2 and V1 = V2


(m2 * R * T2)/(V) = (m1 * R * T1)/(V)

m2 = (m1*T1)/(T2)

m2 = (0.091*298)/(332) = 0.084 Kg\n\n

mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg

P2 = 337.43 KPa

mbleed = 0.007 Kg

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See attachment for the detailed step by step solution of the given problem.

A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following: a. The amount by which this specimen will elongate in the direction of the applied stress.
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.



1)ΔL = 0.616 mm

2)Δd = 0.00194 mm


We are given;

Force; F = 52900 N

Initial length; L_o = 207 mm = 0.207 m

Diameter; d_o = 19.2 mm = 0.0192 m

Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²

Now, from Hooke's law;

E = σ/ε

Where; σ is stress = force/area = F/A

A = πd²/4 = π × 0.0192²/4

A = 0.00009216π

σ = 52900/0.00009216π

ε = ΔL/L_o

ε = ΔL/0.207

Thus,from E = σ/ε, we have;

61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)

Making ΔL the subject, we have;

ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)

ΔL = 0.616 × 10^(-3) m

ΔL = 0.616 mm

B) Poisson's ratio is given as;

υ = ε_x/ε_z

ε_x = Δd/d_o

ε_z = ΔL/L_o


υ = (Δd/d_o) ÷ (ΔL/L_o)

Making Δd the subject gives;

Δd = (υ × d_o × ΔL)/L_o

We are given Poisson's ratio to be 0.34.


Δd = (0.34 × 19.2 × 0.616)/207

Δd = 0.00194 mm

A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm



tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa


Given data in question

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

to find out

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point


in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield =  yield load / area

tensile stress at yield =  80 ×10³ / \pi /4 × D²

tensile stress at yield =  80 ×10³ / \pi /4 × 0.020²

tensile stress at yield = 254 MPa

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³   / \pi /4 × 0.020²

ultimate stress = 477 MPa

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³  / \pi /4 × d²

average stress = 70 ×10³  / \pi /4 × 0.010²

average stress = 892 MPa