A house is on a hill 10 meters above the base of a 50 meter tall water tower. Calculate the gauge pressure at a tap on the ground floor of the house.

Density of water (rho) = 1000 kg/m^3

Acceleration of gravity (g) = 9.8 m/sec^2

Pressure of a fluid column ΔP=ρgΔh

SI units for pressure Pascal = 1 N/m^2 = 1 kg /ms^2


A) 98,000 N/M^2

B) 490,000 N/m^2

C) 588,000 N.m^2

D) 392,000 N/m^2

Answers

Answer 1

Answer:

Explanation:

Given: Pressure of a fluid column ΔP=ρgΔh

Density of water (ρ) = 1000 kg/m^3

Acceleration of gravity (g) = 9.8 m/sec^2

Δh = difference in height / elevation of water tower and tap

= 50 - 10

= 40

Substitute back into equation:

Pressure = 1000*9.8*40

= 392000 N/m^2

The answer is D.

Answer 2

Answer:

Explanation:

Pressure of a fluid column ΔP=ρgΔh

delta h = 50-10 = 40

P = 1000x9.8x40

= 392000

ans is D


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A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler and 125 kPa in the condenser. Saturated steam enters the turbine. Irreversibilities in the turbine cause the steam quality at the outlet of the turbine to be 70 percent. Determine the isentropic efficiency of the turbine and the thermal efficiency of the cycle. Use steam tables. The isentropic efficiency of the turbine is %. The thermal efficiency of the cycle is

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Answer:

The right solution is "28.45%".

Explanation:

The given values are:

[tex]P_4=50\ kPa[/tex]

[tex]h_4=0.7(2304.7)+340.5[/tex]

    [tex]=1953.83 \ KJ/Kg[/tex]

and,

[tex]P_3=15 \ mPa[/tex]

[tex]h_3=hg[/tex]

    [tex]=2610.8 \ KJ/Kg[/tex]

[tex]s_3=sg[/tex]

    [tex]=5.3108 \ KJ/Kgh[/tex]

At 45,

⇒ [tex]x_{45} = \frac{5.3108-1.0912}{6.5019}[/tex]

          [tex]=0.66[/tex]

At [tex]P_4=50 \ Kpa[/tex],

[tex]h_f=340.54[/tex]

or,

[tex]V_f=0.001030 \ m^3/Kg[/tex]

then,

⇒ [tex]h_2=340.54+0.001030(15\times 10^{3}-50)[/tex]

        [tex]=355.94 \ kJ/kg[/tex]

hence,

The isentropic efficiency of turbine will be:

⇒ [tex]n_T=\frac{h_3-h_4}{h_3-h_{45}}[/tex]

         [tex]=\frac{2610.8-1953.83}{2610.8-1836.26}[/tex]

         [tex]=84.818[/tex] (%)

The thermal efficiency of cycle will be:

⇒ [tex]n_C=\frac{W_T-W_P}{2_{in}}[/tex]

         [tex]=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}[/tex]

         [tex]=28.45[/tex] (%)  

The isentropic efficiency of the turbine is; η_t = 76.35%

The thermal efficiency of the cycle is; η_th = 27.08%

We are given;

P₃ = 15 mPa

P₄ = 125 kPa

At P₃ = 15 mPa = 15000 kPa, from the first table attached, we have;

Enthalpy of saturation vapour; h_g = h₃ = 2610.8 kJ/kg

Entropy of saturation vapour; s_g = s₃ = 5.3108 kJ/kg.k

Similarly, At P₄ = 125 kPa, from the second table attached, we have;

Enthalpy of evaporation; h_fg = 2240.6 kJ/kg

Enthalpy of saturation; h_f = 444.36 kJ/kg

Entropy of saturation; s_f = s₄ = 1.3741

Specific volume; v_f = 0.001048 m³/kg

Since the the steam quality at the outlet of the turbine to be 70 percent, then;

h₄ = 0.7h_fg + h_f

h₄ = 0.7(2240.6) + 444.36

h₄ = 2012.78 kJ/kg

Formula for the quality of the steam is;

x,₄₋₅  = (s₃ - s₄)/(s₃ + s₄)

x,₄₋₅  = (5.3108 - 1.3741)/(5.3108 + 1.3741)

x,₄₋₅ = 0.5889

Formula for h₂ is;

h₂ = h_f + V_f(P₃ - P₄)

h₂ = 444.36 + 0.001048(15000 - 125)

h₂ = 459.959 kJ/kg

Formula for Isentropic efficiency is;

η_t = (2610.8 - 2012.78)/(2610.8 - (0.7*2610.8))

η_t = 0.7635

η_t = 76.35%

Formula for thermal efficiency here is;

η_th = [(h₃ - h₄) - (h₂ - h_f)]/(h₃ - h₂)

η_th = [(2610.8 - 2012.78) - (459.959 - 444.36)]/(2610.8 - 459.959)

η_th = 0.2708

η_th = 27.08%

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