a)
The final temperature of the helium
a)

Answer to Problem 175RP
The final temperature of the helium is
Explanation of Solution
Write the relation between the pressure and temperature in the isentropic process:
Here, temperature at state 2 is
Conclusion:
From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the specific heat ratio
Substitute
b)
The final volume of nitrogen
b)

Answer to Problem 175RP
The final volume of nitrogen is
Explanation of Solution
Write the expression for calculating the initial volume of helium
Here, mass of the helium is
Write the expression for calculating the initial volume of helium
Here, mass of the helium is
Write the expression for the final volume of the nitrogen
Here, volume of the nitrogen at initial state 1 is
Conclusion:
From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant for the helium gas as
Substitute
Substitute
Substitute
Thus, the final volume of nitrogen is
c)
The heat transferred to the nitrogen.
c)

Answer to Problem 175RP
The heat transferred to the nitrogen is
Explanation of Solution
Write the expression for calculating the mass of the nitrogen
Write the expression for calculating the temperature of the nitrogen
Write the formula for the change in internal energy of the nitrogen.
Here, specific heat capacity of nitrogen at constant volume is
Write the formula for the change in internal energy of the Helium.
Here, specific heat capacity of Helium at constant volume is
Write the formula for the energy balance equation for the system:
Conclusion:
From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant for the helium gas and specific heat constant at constant volume for the helium gas as
From the Table A-2, “Ideal-gas specific heats of various common gases table”, select the gas constant for the Nitrogen gas and specific heat constant at constant volume for the helium gas as
Substitute
Substitute
Substitute
Substitute
Substitute
Thus, the heat transferred to the nitrogen is
d)
The entropy generation during the process
d)

Answer to Problem 175RP
The entropy generation during the process is
Explanation of Solution
Write the expression for the entropy generation for the isentropic process:
Here, specific heat capacity of nitrogen at constant pressure is
Conclusion:
Substitute 0.2185 kg for
Thus, the entropy generation during the process is
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Chapter 7 Solutions
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
- A rotating shaft is made of 42 mm by 4 mm thick cold-drawn round steel tubing and has a 6 mm diameter hole drilled transversely through it. The shaft is subjected to a pulsating torque fluctuating from 20 to 160 Nm and a completely reversed bending moment of 200 Nm. The steel tubing has a minimum strength of Sut = 410 MPa (60 ksi). The static stress-concentration factor for the hole is 2.4 for bending and 1.9 for torsion. The maximum operating temperature is 400˚C and a reliability of 99.9% is to be assumed. Find the factor of safety for infinite life using the modified Goodman failure criterion.arrow_forwardI need help with a MATLAB code. This code just keeps running and does not give me any plots. I even reduced the tolerance from 1e-9 to 1e-6. Can you help me fix this? Please make sure your solution runs. % Initial Conditions rev = 0:0.001:2; g1 = deg2rad(1); g2 = deg2rad(3); g3 = deg2rad(6); g4 = deg2rad(30); g0 = deg2rad(0); Z0 = 0; w0 = [0; Z0*cos(g0); -Z0*sin(g0)]; Z1 = 5; w1 = [0; Z1*cos(g1); -Z1*sin(g1)]; Z2 = 11; w2 = [0; Z2*cos(g2); -Z2*sin(g2)]; [v3, psi3, eta3] = Nut_angle(Z2, g2, w2); plot(v3, psi3) function dwedt = K_DDE(~, w_en) % Extracting the initial condtions to a variable % Extracting the initial condtions to a variable w = w_en(1:3); e = w_en(4:7); Z = w_en(8); I = 0.060214; J = 0.015707; x = (J/I) - 1; y = Z - 1; s = Z; % Kinematic Differential Equations dedt = zeros(4,1); dedt(1) = pi*(e(3)*(s-w(2)-1) + e(2)*w(3) + e(4)*w(1)); dedt(2) = pi*(e(4)*(w(2)-1-s) + e(3)*w(1) - e(1)*w(3)); dedt(3) = pi*(-e(1)*(s-w(2)-1) - e(2)*w(1) + e(4)*w(3));…arrow_forwardalpha 1 is not zero alpha 1 can equal alpha 2 use velocity triangle to solve for alpha 1 USE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, (leaving a stator in axial flow) R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',...…arrow_forward
- 3. Find a basis of eigenvectors and diagonalize. 4 0 -19 7 a. b. 1-42 16 12-20 [21-61arrow_forward2. Find the eigenvalues. Find the corresponding eigenvectors. 6 2 -21 [0 -3 1 3 31 a. 2 5 0 b. 3 0 -6 C. 1 1 0 -2 0 7 L6 6 0 1 1 2. (Hint: λ = = 3)arrow_forwardUSE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, (leaving a stator in axial flow) R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),...…arrow_forward
- USE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ...…arrow_forwardThe answer should equal to 1157. Please sent me the solution. Thank you!arrow_forwardBONUS: If the volume of the 8cm x 6.5cm x 6cm Block of Aluminum was 312cm3 before machining, find how much material was removed when the fixture below was machined. +2 2.00 cm 6.00 cm 2.50 cm 6.50 cm 1.00 cm 2.50 cm 11.00 cm 8.00 cm 30 CP 9411 FL.4) (m² 1157 Area of triangle = 1/2*B*H Area of circle = лR² Circumference of a circle = 2πR 6.00 cm 6.50 cm 1.50 cm Radius 1.50 cm 1.00 cmarrow_forward
- Consider a 5m by 5m wet concret patio with an average water film thickness of .2mm. Now wind at 50 km/h is blowing over the surface. If the air is at 1 atm, 15oC and 35 percent relative humidity, determine how long it will take for the patio to completely dry.arrow_forward70. Compute the number of cubic centimeters of iron required for the cast-iron plate shown. The plate is 3.50 centimeters thick. Round the answer to the nearest cubic centimeter. 50.0 cm 40.0 cm Radius 150° 115.0 cm- 81.0 cmarrow_forwardLaw of Sines Solve the following problems using the Law of Sin 7. Find side x. All dimensions are in inches. -°-67°-37° 81° x Sin A 8.820 X 67°00' 32°00' a sin A b C sin B sin Carrow_forward
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