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Question : (35 pts.) A 0.5 m internal diameter spherical tank 10 mm thi...

  1. (35 pts.) A 0.5 m internal diameter spherical tank 10 mm thick stainless steel (k=15W/m.K) is used to store liquid nitrogen at 77 K. Tank is insulated with silica powder (k=0.0017 W/m.K) and the insulation is 25 mm thick. Temperature of the outer surface of the tank was measured as 299 K and it is exposed to ambient air of 300 K. The outer surface of the tank is painted white and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection coefficient is known to be 20W/m2.K20 W/m^2.K. The latent heat of vaporization and the density of liquid nitrogen are 2×103J/Kg 2×10^3 J/Kg and 804Kg/m3804 Kg/m^3, respectively. (Neglect the resistance to heat transfer from the container wall to the liquid nitrogen.)
    a. What is the rate of heat transfer to the liquid nitrogen?
    b. What is the rate of the liquid boil-off (L/day)
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Qcond=299.77kT4Q_{cond} =\frac{299.77}{kT_4}

 

RT4=r2−r14πkr1r2+r3−r24πkr3r2⇒R_{T_4}=\frac{r_2-r_1}{4πkr_1r_2}+\frac{r_3-r_2}{4πkr_3r_2}⇒

 

RT4=14π(10×10315×250×260+25×10717×260×285)R_{T_4}=\frac{1}{4π}(\frac{10×10^3}{15×250×260}+\frac{25×10^7}{17×260×285})

 

⇒RT4=15.75K/W⇒R_{T_4}=15.75 K/W

 

⇒Qcond=299.7715.75=14.059W⇒Q_{cond}=\frac{299.77}{15.75}=14.059W

Actual heat transfer=26.63 W

b) Remining heat is used in liquid boiling :

26.63-14.059=12.5W

ρ=804kg/m3         qh=2×105J/kgρ=804 kg/m^3                  q_h=2×10^5 J/kg

 

ρ.qh=1608×105J/m3=1608×102J/Lρ.q_h=1608×10^5 J/m^3 =1608×10^2 J/L

from 1.  we have 12.57304 is per second

⇒per  day=12.57×24×60×60⇒per   day =12.57×24×60×60

 

=1086.048KJ=1086.048 KJ

⇒ Rate of liquid boil_off =6.754L/day=6.754 L/day

Q.1: Internal diameter =0.5 m

a)r1=0.25m=250mma) r_1=0.25m=250mm

 

t=10mmt=10mm

 

r2=250+10=260mmr_2=250+10=260mm

 

ksteel=15W/m.kk_{steel}=15 W/m.k

 

ts=25mm⇒r3=260+25=285mmt_s=25 mm⇒r_3=260+25=285mm

 

Ksilica=0.0017W/m.kK_{silica}=0.0017 W/m.k

 

h=0.20w/m2Kh=0.20 w/m^2K

rate of heat transfer to liquid nitrogen

Q=Qconducted=Qconveted+QradiatedQ=Q_{conducted}=Q_{conveted}+Q_{radiated}

 

Q=hAΔT+TA(T4−T∝4)Q=h_A \Delta T+T_A(T^4-T_∝^4)

 

=A(20(300−299))+5.67×10−8(3004−2994)=A(20(300-299))+5.67×10^{-8}(300^4-299^4)

 

=4πr32(20+5.67×10−8)(3004−2994)=4πr_3^2(20+5.67×10^{-8})(300^4-299^4)

 

⇒Q=26063W⇒Q=26063W

 

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