Answer to Question #167982 in Differential Equations for Randal Rodriguez

Question

(rounded to 4 decimal places)

1. Air flow steadily at the rate of 0.75 kg/s through air compressor, entering at 7 m/s speed, 100 Kpa

pressure and 0.95 m 3 /kg specific volume and leaving 15 m/s, 700 kpa and 0.19 m 3 /kg. the internal

energy of the air leaving is 100 KJ/kg greater than that of the air entering. Cooling water in the

compressor jacket absorbs heat from the air at the rate of 60 kJ/s. compute the steady flow work in kW.

2. The work done by substance in a reversible non-flow manner in accordance with V=700/P where V is

volume in m 3 , P is absolute pressure in Kpaa. Evaluate the work done on or by the substance as the

pressure increases from 80 to 800 KPa

3. During the execution of a reversible non-flow process the work is -156.2 KJ if V 1 =0.845 and the

pressure varies as P=-730 V +690 KPa, where volume V is in m 3 . Find the final volume.

4. A fluid system undergoes a non-flow frictionless process from 4.5 ft 3 to a final volume 1.5 ft 3 in

accordance with the defining relation P= 60/V + 30 psi, where volume in ft 3 . During the process the

system reject 30 BTU of heat. Determine the change in enthalpy and internal energy.

5. A steady state, steady flow compressor draws in 240 liters per second of air whose density is 1.26

kg m /m 3 and discharges it with a density of 4.9 kg m /m 3 . At suction P 1 = 104 kPaa; at discharge, P 2 = 551

kPaa. The increase of specific internal energy is 78.5 kJ/kg m and the heat from the air by cooling is 30

kJ/kg m . Neglecting the change in potential and kinetic energies, determine the work in kJ/min.

6. Two cubic-meters of CO 2 gas (MW = 44 kg/mol) measured at 100 kPaa, 30 o C are withdrawn from a 1.7

m diameter spherical tank containing CO 2 gas initially at 700 kPaa and 60 o C. What is the pressure

(kPaa) of the gas remaining inside the tank if its temperature is 55 o C?

7. A drum 0.5 ft in diameter and 45 inches long contained acetylene at 350 psia, 90 o F. After some of the

acetylene was used, the pressure was 200 psia and 85 o F. a) determine the amount of acetylene that

was used (lb m ). b) What volume would the used acetylene occupy at 14.7 psia, 80 o F (ft 3 ) ? R for this

gas is 59.35 ft-lb f /lb m- R.

8. A spherical balloon is 50 ft in diameter and surrounded by air at 60 F and 29.92 in Hg abs. (a) if the

balloon is filled with hydrogen at temp of 70 F and atmospheric pressure, what total load can it lift. (b) if

it contains helium instead of hydrogen, what load can it lift (lb f ) if gravitational acceleration is 31.85 ft/s 2 ?

9. A 6m 3 tank contains helium at 127 C and in evacuated from atmospheric pressure to a pressure of 740mm Hg vacuum. Determine (a) mass of helium remaining in the tank (b) mass of helium pumped out (c)

the temp of the remaining helium falls to 10 C, what is the pressure in Kpa? R of helium is 2077.67

J/kg-K.

10. A rigid insulated tank is separated into two rooms by a stiff plate. Room A of 0.5 m 3 contains air at 250

Kpa, 300 K and room B is 1 m 3 has air at 200 Kpa, 1000 K. the plate is removed and the air comes to a uniform state without any heat transferred. find the final temp and pressure of the room

Answer 1

1.

$m=0.75\ kg/s,\ p_1=100\ kPa,\ v_1=0.95\ m^3/kg$

$Q=-60\ kW,\ p_2=700\ kPa,\ v_2=0.19\ m^3/kg$

$\Delta u=100\ kJ/kg,\ V_1=7\ m/s,\ V_2=15\ m/s$

$p_1v_1=100\cdot0.95=95\ kJ/kg$

$p_2v_2=700\cdot0.19=133\ kJ/kg$

$\Delta pv=133-95=38\ kJ/kg$

$Q-W=m(\Delta h+\Delta KE+\Delta PE)$

$\Delta h=\Delta(u+pv)=100+38=138\ kJ/kg$

$\Delta KE=(V_2^2-V_1^2)/2=(15^2-7^2)/2=88\ kJ/kg$

$\Delta PE=0$

$-60-W=0.75(138+88)$

$W=-229.5\ kW$

2.The work done by substance:

$W=\intop PdV$

$V=700/P$ , $dV=(-700/P^2)dP$

$W=-700\displaystyle\intop_{80}^{800}\frac{dP}{P}=-700(ln800-ln80)=-1612\ kJ$

3.

$W=P_2V_2-P_1V_1$

$V_2=\frac{W+P_1V_1}{P_2}=\frac{-156.2-730\cdot0.845}{690}=-1.120\ m^3$

4.

$W=\displaystyle\intop^{1.5}_{4.5}(60/V + 30)dV=60(ln1.5-ln4.5)-90=-156\ BTU$

Change in internal energy:

$\Delta U=Q-W=-30+156=126\ BTU$

Change in enthalpy:

$\Delta H=\Delta U+\Delta P\Delta V$

$\Delta P=60/1.5-60/4.5=26.67\ psi$

$\Delta H=126-3\cdot26.67=46\ BTU$

5.

$W=Q-\Delta U-\Delta (pV)$

$W=Q-\Delta U- (p_2/ \rho_2-p_1/\rho_1)$

$W=-78.5-30-(551/4.9-104/1.26)=-78.6\ kJ/kg$

$m=\rho V=1.26\cdot240\cdot10^{-3}=0.3024\ kg/min$

$W$ in $kJ/min$ :

$W=-78.6\cdot0.3024=23.77\ kJ/min$

6.Gas pressure:

$p=mRT/V$

$R=\overline{R}/MW=8314/44=189\ J/kg\cdot K$

Volue of tank:

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \cdot0.85^3=2.57\ m^3$

Initial mass of gas:

$m_0=\frac{p_0V}{RT_0}=\frac{700\cdot2.57}{189\cdot(60+273)}=28.58\ kg$

Mass of withdrawn gas:

$m_1=\frac{p_1V_1}{RT_1}=\frac{100\cdot2}{189\cdot(30+273)}=3.49\ kg$

Remaining mass:

$m=28.58-3.49=24.09\ kg$

Pressure of remaining gas:

$p=24.09\cdot189\cdot(55+273)/2.57=581\ kPa$

7.Volume of drum:

$V_0=\pi r^2l=3.14\cdot(0.25\cdot12)^2\cdot45=1272\ in^3$

$90\degree F=5/9(90-32)+273=305\ K$

$85\degree F=5/9(85-32)+273=302\ K$

$80\degree F=5/9(80-32)+273=300\ K$

$m_0=\frac{350\cdot1272}{59.35\cdot12\cdot305}=2.05\ lb$

$m_1=\frac{200\cdot1272}{59.35\cdot12\cdot302}=1.12\ lb$

a)

$m=m_0-m_1=8.2-4.7=0.875\ lb$

b)

$V=mRT/p=0.875\cdot59.35\cdot12\cdot300/14.7=12718\ in^3$

$V=50871/12^3=9.86\ ft$

8.Volume of balloon:

$V=\frac{4}{3}\pi\cdot25^3=65449\ ft^3$

$29.92\ Hg=14.7\ psi$

$R_{air}=10.73/29=0.37, R_{hydr}=10.73/2=5.36$

$60\degree F=5/9(60-32)+273=288.5\ K$

$70\degree F=5/9(70-32)+273=294\ K$

a) Baloon can lift:

$W=V\rho_{air}g-V\rho_{hydr}g$

$\rho_{air}=\frac{p}{RT}=\frac{14.7}{0.37\cdot288.5}=0.137$

$\rho_{hydr}=\frac{14.7}{5.36\cdot294}=0.009$

$W=31.85\cdot65449(0.137-0.009)=267\cdot10^3\ lb\cdot ft$

b)

$R_{hel}=10.73/4=2.68$

$\rho_{hel}=14.7/(2.68\cdot294)=0.019$

$W=31.85\cdot65449(0.137-0.019)=246\cdot10^3\ lb\cdot ft$

9. Initial mass:

$m_0=\frac{100\cdot10^3\cdot6}{2077.67\cdot400}=0.722\ kg$

Evacuated mass:

$m_1=\frac{98.66\cdot10^3\cdot6}{2077.67\cdot400}=0.712\ kg$

a) Remaining mass:

$m=m_0-m_1=0.722-0.712=0.01\ kg$

b)

$m_1=0.712 \ kg$

c)

$p=mRT/V=0.01\cdot2077.67\cdot283/6=980\ kPa$

10. A rigid insulated tank is separated into two rooms by a stiff plate. Room A of 0.5 m 3 contains air at 250

Kpa, 300 K and room B is 1 m 3 has air at 200 Kpa, 1000 K. the plate is removed and the air comes to a uniform state without any heat transferred. find the final temp and pressure of the room

$\frac{p_1V_1}{T_1}+\frac{p_2V_2}{T_2}=\frac{p(V_1+V_2)}{T}$

$T=\frac{T_1+T_2}{2}=(300+1000)/2=650\ K$

$p=(250\cdot0.5/300+200/1000)\cdot650/1.5=267\ kPa$