Question #272767

1. The hot-air balloon contains air having a temperature of 80 degrees Celsius, while the surrounding air has a temperature of 20 degrees Celsius. Determine the volume of hot air if the total mass of the balloon and the load is 770 kg. 


2.Water in the container is originally at a height of h = 1m. If a block having a density of 800 kg/m3 is placed in the water, determine the new level h of the water. The base of the block is 600 mm square, and the base of the container is 1.2 m square.


1
Expert's answer
2021-11-29T11:43:14-0500

1.m=770kg1.\newline m= 770kg\newline

T1=20°CT2=80°Cg=9.8ms2fromT_1= 20\degree C \newline T_2= 80\degree C \newline g =9.8\frac{m}{s^2}\newline \text{from}\newline

https://www.me.psu.edu/cimbala/me433/Links/Table_A_9_CC_Properties_of_Air.pdf

ρ1=1.204kgm3\rho_1= 1.204\frac{kg}{m^3}

ρ2=0.9994kgm3\rho_2= 0.9994\frac{kg}{m^3}

Fa=ρ1gVF_a = \rho_1gV

let m1 air mass in a balloon\text{let }m_1 \text{ air mass in a balloon}

Famgm1g0F_a-mg-m_1g\ge0

m1=ρ2Vm_1 = \rho_2V

ρ1gVmgρ2gV0\rho_1gV-mg-\rho_2gV\ge0

Vmρ1ρ2=7701.2040.99943763m3V\ge \frac{m}{\rho_1-\rho_2}=\frac{770}{1.204-0.9994}\approx3763m^3

Answer: V3763m3\text{Answer: }V\ge 3763m^3

2.2.

ρw=1000kgm3\rho_w= 1000\frac{kg}{m^3}

ρb=800kgm3\rho_b= 800\frac{kg}{m^3}

ab=600mm=0.6ma_b= 600mm= 0.6m

ac=1.2ma_c=1.2m

hc=1mh_c= 1m

Vc=ac2h=1.44m3V_c = a_c^2h= 1.44m^3

Vb=ab3=0.216m3V_b =a_b^3= 0.216m^3

mb=ρbVb=172.8kgm_b= \rho_bV_b= 172.8kg

let V1 volume of the block immersed in water\text{let }V_1\text{ volume of the block immersed in water}

Fa=ρwgV1F_a = \rho_wgV_1

Fambg=0F_a - m_bg = 0

ρwgV1mbg=0\rho_wgV_1-m_bg=0

V1=mbρw=0.1728m3V_1=\frac{m_b}{\rho_w}=0.1728m^3

h1=Vc+V1ac2=1.12mh_1 = \frac{V_c+V_1}{a_c^2}=1.12m


Answer: 1.12m\text{Answer: }1.12m



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