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77julia77 [94]
3 years ago
15

Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

20 kg; the mass of the liquid is 1.85 kg. Balance D reads 3.10 kg and balance E reads 7.50 kg. The volume of block A is 4.15 × 10−3 m3.
a) What is the density of the liquid?
b) What will the balance D read if block A is pulled up out of the liquid?
c) What will the balance E read if block A is pulled up out of the liquid?
Engineering
1 answer:
nikitadnepr [17]3 years ago
6 0

Answer:

a)  m_e= 3.05 Kg

b)  \rho=1072.3kg/m^3

c)  m_e= 3.05 Kg

Explanation:

From the question we are told that:

Beaker Mass m_b=1.20

Liquid Mass m_l=1.85

Balance D:

Mass m_d=3.10

Balance E:

Mass m_e=7.50

Volume v=4.15*10^{-3}m^3

a)

Generally the equation for Liquid's density is mathematically given by

m_e=m_b+m_l+(\rho*v)

\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}

\rho=1072.3kg/m^3

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

m_d=m- (\rho *v )

m=3.10+ (1072.30 *4.15*10^{-3}m^3 )

m=18.10kg

c)

Generally the equation for E's Reading at A pulled is mathematically given by

m_e=m_b+m_l

m_e = 1.20 + 1.85

m_e= 3.05 Kg

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Given

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For better understanding, it's better to convert both measurements to decimal.

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Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

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AT INLET:

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AT OUTLET:

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P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

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x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

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